🧠:[<think>Okay, so I have this problem with two parts, a) and b). Let's start with part a). The problem is asking me to prove that for any integer n ≥ 2, I can arrange all subsets of the set {1, 2, ..., n} as a sequence A₁, A₂, ..., A_{2ⁿ} such that the size (cardinality) of each subset A_{i+1} is either one more or one less than the size of A_i. Also, it needs to wrap around so that A_{2ⁿ + 1} = A₁. Hmm. So, this sounds like a kind of Gray code arrangement. I remember that Gray codes are sequences where consecutive elements differ by just one bit, which corresponds to adding or removing a single element in the subset terminology. But in the problem, the requirement is about the size (cardinality) changing by 1, not the subsets themselves differing by one element. Wait, but adding or removing an element does change the size by 1. So actually, if we can arrange the subsets such that consecutive subsets differ by exactly one element, then their sizes would differ by 1. So, maybe this is equivalent to a Gray code ordering?But Gray codes usually refer to binary reflected Gray codes where each consecutive number differs by one bit. Translating that to subsets, each consecutive subset differs by adding or removing one element. So, in such a sequence, the sizes of the subsets can go up or down by 1 each time. However, the problem wants the sequence to be cyclic, meaning that the last subset and the first subset must also differ in size by 1. But in the standard Gray code, the first and last subsets also differ by one element (since the binary reflected Gray code cycles through all numbers with wrap-around). For example, for n=2, the subsets would be ordered as {}, {1}, {1,2}, {2}, which cycles back to {}. The sizes go 0,1,2,1,0. So the last subset {2} has size 1, and the first subset {} has size 0, so they differ by 1. So it works. Therefore, maybe using the standard binary reflected Gray code gives such an arrangement. So, perhaps the answer to part a) is that such an arrangement exists by using the Gray code order. But the problem says "prove", so I need to give a proper proof, not just mention Gray codes.Alternatively, maybe induction. Let's try induction.Base case: n = 2. Then the subsets are {}, {1}, {2}, {1,2}. If we arrange them as {}, {1}, {1,2}, {2}, then the sizes are 0,1,2,1, and when wrapping around, the last size is 1, and the first is 0, so the difference is 1. So this works.Assume that for n = k, we can arrange the subsets of {1,2,...,k} in such a sequence. Now, for n = k + 1, how can we extend this? Maybe by mirroring the sequence for n = k and adding the element k+1 appropriately? Wait, that's similar to the Gray code construction.In the binary reflected Gray code, the code for n+1 can be constructed by taking the Gray code for n, traversing it forward with a 0 appended to each code, then traversing it backward with a 1 appended. Translating this to subsets, we first list all subsets of {1,...,k} with the ordering from the inductive hypothesis, then list all subsets of {1,...,k} union {k+1} in reverse order. This way, each consecutive subset differs by one element (either adding or removing k+1, or following the inductive step). So, this would maintain the property that consecutive subsets differ by one element, hence their sizes differ by 1. Also, the last subset in the n+1 case would be the last subset of the reversed second half, which is the first subset of the original n case union {k+1}, and the first subset is the empty set. Then, the difference between the last subset (which would be {k+1}) and the first subset {} is exactly one element, so their sizes differ by 1. Therefore, the induction holds.Therefore, by induction, such a sequence exists for all n ≥ 2. So that's part a). Maybe that's a good approach. Alternatively, I can reference the concept of a Gray code and explain how it fulfills the required conditions.Now, part b). Determine all possible values of the sum Σ_{i=1}^{2ⁿ} (-1)^i S(A_i), where S(A_i) is the sum of all elements in A_i, and the subsets are arranged as in part a). So, for any such sequence that satisfies the adjacency condition (sizes differing by 1), what are the possible values of this alternating sum?First, I need to understand what the sum looks like. It's an alternating sum: subtract and add the subset sums based on the parity of their position. So, for i odd, we add (-1)^i S(A_i) which would be -S(A_i) if i is odd (since (-1)^1 = -1, (-1)^2 = 1, etc.), or actually wait, i starts at 1: (-1)^1 = -1, (-1)^2 = 1, so the sum is -S(A₁) + S(A₂) - S(A₃) + S(A₄) - ... + (-1)^{2ⁿ} S(A_{2ⁿ}).Since 2ⁿ is even, the last term is +S(A_{2ⁿ}).The question is to determine all possible values of this sum, given all possible valid sequences from part a). So different sequences might give different sums, but maybe they all result in the same value, or maybe there are specific possible values.First, let's try small n to see.Take n = 2. Then the subsets are {}, {1}, {2}, {1,2}. A valid sequence is {}, {1}, {1,2}, {2}, as in part a). Let's compute the sum:i=1: (-1)^1 S({}) = -0 = 0i=2: (-1)^2 S({1}) = 1*1 = 1i=3: (-1)^3 S({1,2}) = - (1+2) = -3i=4: (-1)^4 S({2}) = 1*2 = 2Total sum: 0 + 1 - 3 + 2 = 0.Alternatively, is there another valid sequence? For n=2, are there other Gray code sequences? Let's see.The subsets can be ordered in another way as {}, {2}, {1,2}, {1}. Let's check the differences: {} to {2} (size 0 to 1), {2} to {1,2} (size 1 to 2), {1,2} to {1} (size 2 to 1), and {1} to {} (size 1 to 0). So this also satisfies the condition. Let's compute the sum:i=1: -S({}) = 0i=2: +S({2}) = 2i=3: -S({1,2}) = -3i=4: +S({1}) = 1Total: 0 + 2 - 3 + 1 = 0.Same result. Another possible sequence: maybe {},{1}, {1,2}, {2}, which we already did, or {},{2}, {1,2}, {1}, which gives the same sum. Let me check another ordering. Wait, for n=2, there are 4 subsets. The Gray code order is unique up to reflection and starting point. But if we start elsewhere, does that change the sum?Suppose we start with {1}, then {1,2}, {2}, {}, but then wrap around. Wait, the problem states that A_{2ⁿ + 1} = A₁, so the sequence must be cyclic. So starting at different points would just rotate the sequence. Let's try starting at {1}:Sequence: {1}, {1,2}, {2}, {}, {1} (wrapping around). But since the problem says A₁ to A_{2ⁿ}, which for n=2 is 4 elements. So the sequence would be {1}, {1,2}, {2}, {}. Then compute the sum:i=1: -S({1}) = -1i=2: +S({1,2}) = 3i=3: -S({2}) = -2i=4: +S({}) = 0Sum: -1 + 3 - 2 + 0 = 0.Still zero. Similarly, starting at {2}:Sequence: {2}, {1,2}, {1}, {}, {2}Sum:i=1: -S({2}) = -2i=2: +S({1,2}) = 3i=3: -S({1}) = -1i=4: +S({}) = 0Total: -2 + 3 -1 + 0 = 0.Same result. So for n=2, regardless of the cyclic Gray code ordering, the sum is 0. Interesting. Let me check for n=1, but n ≥2, so n=1 is not considered.Wait, n=3. Maybe check n=3. Let's see. For n=3, the subsets can be arranged in a Gray code order. Let's try constructing one such sequence.Standard binary reflected Gray code for n=3 would be:000: {}001: {1}011: {1,2}010: {2}110: {2,3}111: {1,2,3}101: {1,3}100: {3}Then back to 000. So the subsets in order are:{}, {1}, {1,2}, {2}, {2,3}, {1,2,3}, {1,3}, {3}, and back to {}.Compute the sum:i=1: -S({}) = 0i=2: +S({1}) = 1i=3: -S({1,2}) = -3i=4: +S({2}) = 2i=5: -S({2,3}) = -5i=6: +S({1,2,3}) = 6i=7: -S({1,3}) = -4i=8: +S({3}) = 3Sum: 0 +1 -3 +2 -5 +6 -4 +3.Calculating step by step:0 +1 =1; 1 -3 = -2; -2 +2 = 0; 0 -5 = -5; -5 +6 =1; 1 -4 = -3; -3 +3 =0.Total sum is 0. Hmm. So again, the sum is zero. Let's try a different Gray code ordering. For example, starting at a different point or reflecting.Alternatively, reverse the order. Let's reverse the sequence:{3}, {1,3}, {1,2,3}, {2,3}, {2}, {1,2}, {1}, {}, {3}.Compute the sum:i=1: -S({3}) = -3i=2: +S({1,3}) = 4i=3: -S({1,2,3}) = -6i=4: +S({2,3}) =5i=5: -S({2}) = -2i=6: +S({1,2}) =3i=7: -S({1}) =-1i=8: +S({}) =0Sum: -3 +4 -6 +5 -2 +3 -1 +0.Calculate step by step:-3 +4=1; 1 -6=-5; -5 +5=0; 0 -2=-2; -2 +3=1; 1 -1=0; 0 +0=0.Again, sum is 0. Interesting.Wait, so for n=2 and n=3, regardless of the Gray code ordering, the sum is zero. Maybe the sum is always zero? But the problem says "determine all possible values", which suggests that it might always be zero, hence the only possible value is zero. But I need to check for another possible arrangement. Wait, but maybe there are different sequences that satisfy the condition in part a) that are not Gray codes? Or maybe all such sequences are similar to Gray codes, and hence give the same sum. Wait, but how?Wait, the problem says "for any subset sequence A₁,...,A_{2ⁿ} satisfying the condition in a)", so all such sequences. The question is asking for all possible values of the sum, given all possible such sequences. If for n=2 and n=3 the sum is always zero, maybe it's always zero. Let me try to see why.Suppose we consider the sum Σ (-1)^i S(A_i). Let's try to manipulate this sum. Maybe consider pairing terms or looking for telescoping. Alternatively, notice that the sequence is cyclic, so perhaps the sum has some symmetry.Alternatively, note that each element's contribution can be considered separately. For each element x in {1, 2, ..., n}, consider how many times it is added or subtracted in the sum. Since S(A_i) is the sum of elements in A_i, the total sum is equal to the sum over all elements x of (sum over i where x ∈ A_i of (-1)^i). So, the total sum is Σ_{x=1}^n [Σ_{i: x ∈ A_i} (-1)^i]. So, if for each x, the sum Σ_{i: x ∈ A_i} (-1)^i is the same across all valid sequences, then the total sum would be fixed. Otherwise, if this sum can vary depending on the sequence, then the total sum can vary.So, perhaps for each element x, the number of times it is included in subsets at even positions minus the number of times it is included in subsets at odd positions is the same regardless of the sequence. If so, then the total sum would be fixed.Alternatively, maybe each element is included in the same number of even and odd positions in any such sequence. Wait, but in the Gray code sequence, each element is added and removed multiple times. Let's see for n=2:For element 1 in n=2:In the sequence {}, {1}, {1,2}, {2}, the positions where 1 is included are i=2 and i=3. So, sum for x=1: (-1)^2 *1 + (-1)^3 *1 = 1 -1 = 0.Similarly, element 2 is included in i=3 and i=4: (-1)^3 *2 + (-1)^4 *2 = -2 + 2 = 0.Similarly, for the reversed sequence {}, {2}, {1,2}, {1}:Element 2 is included in i=2 and i=3: (-1)^2 *2 + (-1)^3 *2 = 2 -2 =0.Element 1 is included in i=3 and i=4: (-1)^3 *1 + (-1)^4 *1 = -1 +1 =0.Same result. For n=3, let's take the standard Gray code sequence:Subsets in order: {}, {1}, {1,2}, {2}, {2,3}, {1,2,3}, {1,3}, {3}For element 1:Included in i=2,3,6,7: (-1)^2*1 + (-1)^3*1 + (-1)^6*1 + (-1)^7*1 = 1 -1 +1 -1=0.For element 2:Included in i=3,4,5,6: (-1)^3*2 + (-1)^4*2 + (-1)^5*2 + (-1)^6*2 = -2 +2 -2 +2=0.For element 3:Included in i=5,6,7,8: (-1)^5*3 + (-1)^6*3 + (-1)^7*3 + (-1)^8*3 = -3 +3 -3 +3=0.So each element contributes 0, hence total sum 0.If we take another sequence, say starting at {1}:{1}, {1,2}, {2}, {2,3}, {1,2,3}, {1,3}, {3}, {}.Compute contributions:Element 1: in i=1,2,5,6.Sum: (-1)^1*1 + (-1)^2*1 + (-1)^5*1 + (-1)^6*1 = -1 +1 -1 +1 =0.Element 2: in i=2,3,4,5.Sum: (-1)^2*2 + (-1)^3*2 + (-1)^4*2 + (-1)^5*2 = 2 -2 +2 -2=0.Element 3: in i=4,5,6,7.Sum: (-1)^4*3 + (-1)^5*3 + (-1)^6*3 + (-1)^7*3 =3 -3 +3 -3=0.Total sum 0 again.Therefore, perhaps in any such Gray code ordering, each element's contribution cancels out over the sequence. Therefore, the total sum is always zero. Hence, the only possible value is zero.But the problem says "determine all possible values". If for any valid sequence the sum is zero, then the answer is 0. But how can we be sure there isn't some other sequence that doesn't follow the Gray code ordering but still meets the size condition and gives a different sum?Wait, the problem in part a) says "one can arrange", so it's existence. But part b) is for any such arrangement. So perhaps all such arrangements must result in the sum being zero. Therefore, the only possible value is zero.But wait, is that the case? Let's think. Suppose for some n, there exists a sequence where the sum is non-zero. But from the examples above, n=2,3 with different sequences all give zero. Maybe for all sequences, the sum is zero.Alternatively, maybe the parity of the positions interacts with the inclusion of elements in such a way that each element is included equally in even and odd positions. But in the sequence, since the sequence is a cyclic sequence where each step changes the size by 1, the parity (even/odd size) alternates with each step? Wait, no. If you start with a subset of size k, the next subset has size k ±1, so the parity (even/odd) alternates with each step. Therefore, the sequence alternates between even and odd sizes. But in the problem, the requirement is that |A_{i+1}| = |A_i| ±1. Therefore, the parity of |A_i| alternates with each step. So, the sizes go even, odd, even, odd, etc. Or odd, even, odd, etc., depending on the starting size. However, since the sequence is cyclic, the total number of subsets is 2ⁿ, which is even. Therefore, starting with a subset of even size, you must end with a subset of even size after 2ⁿ steps, which would mean the starting and ending subsets have the same parity. But since the step alternates parity, after even number of steps, the parity would return to the original. So, the starting subset must have even size? Wait, no. If you start with size k, after 2ⁿ steps, each step toggles parity. So after 2ⁿ steps, which is even, you would have toggled parity even times, so the final subset would have the same parity as the starting one. Therefore, since the sequence is cyclic (A_{2ⁿ +1} = A₁), the parity of |A₁| must be the same as itself, which is always true. Therefore, any starting subset size is allowed as long as the sequence is cyclic. However, in a valid arrangement, the sizes must alternate parity each step. Therefore, in the entire sequence, the parity of i (the index) is linked to the parity of |A_i|. Wait, no. The index i's parity (whether i is odd or even) is separate from the parity of |A_i|. Let's see:Suppose the sequence starts with a subset of even size. Then the sequence would go even, odd, even, odd, ..., and since the total number of subsets is 2ⁿ (even), the last subset would have even size, matching the first. Therefore, in this case, the parity of the index i (starting at 1) would be: i odd corresponds to even size subsets, and i even corresponds to odd size subsets. Alternatively, if the sequence starts with an odd size subset, then i odd corresponds to odd sizes, and i even corresponds to even sizes. However, since the problem allows any starting subset as long as consecutive sizes differ by 1, the parity linkage between index and subset size depends on the starting subset's size.But in the sum Σ (-1)^i S(A_i), the sign depends on the index i's parity, not the subset size's parity. Therefore, if the starting subset has even size, then even i's (i even) would correspond to odd-sized subsets, and odd i's to even-sized subsets. If starting with odd size, then odd i's correspond to odd-sized subsets and even i's to even-sized subsets.But regardless, for the sum, we need to consider (-1)^i S(A_i). So, the key is how the subsets are distributed across even and odd indices. But the total contribution of each element x is equal to the number of times x is included in subsets at even indices minus the number of times it is included in subsets at odd indices. Because each occurrence in an even index contributes +x, and each in an odd index contributes -x. So, total contribution for x is x*(E_x - O_x), where E_x is the number of times x is in a subset at even index, and O_x is the number at odd index.If for each x, E_x = O_x, then the total contribution is zero. If not, it's x*(E_x - O_x). So, the question is whether in any valid sequence, E_x = O_x for all x. If so, then the total sum is zero. Otherwise, if E_x - O_x can vary, then the total sum could vary.But in the examples above, for n=2 and n=3, E_x = O_x for all x, so contributions zero.But why would E_x = O_x for all x? Let's think about the structure of the sequence. Since the sequence is a cyclic sequence where each subset differs by one element from the previous, and the size changes by ±1 each time. The entire structure is a closed walk on the hypercube graph where each edge connects subsets differing by one element, and also, in this case, the walk alternates between levels of the hypercube (even and odd sizes). However, since the hypercube is bipartite, with one partition being even-sized subsets and the other odd-sized. A closed walk of even length (2ⁿ is even) would alternate between the two partitions. Therefore, the number of times each element is added and removed must be even? Wait, not necessarily. Each element is added and removed multiple times along the walk.But let's think of each element's inclusion as a series of additions and removals. For the entire cycle, each element must be added and removed an equal number of times, since we start and end at the same subset. Wait, if the starting subset is empty, then each element must be added and removed an equal number of times. But if the starting subset is non-empty, then for each element in the starting subset, the number of removals must be one more than the number of additions, and vice versa for elements not in the starting subset. Wait, but the sequence is cyclic, so if you start with a subset S, you must end with S. Therefore, for each element not in S, the number of times it's added equals the number of times it's removed. For each element in S, the number of times it's removed equals the number of times it's added plus one (since it starts in S and must end in S). But in the Gray code sequence starting at S, each element not in S is added and removed an equal number of times, and each element in S is removed and added an equal number of times? Wait, maybe not. Let me think.Wait, consider the standard Gray code starting at the empty set. Each element is added and removed an equal number of times over the cycle. Because to return to the empty set, every addition must be matched by a removal. Therefore, for each element, the number of times it's added equals the number of times it's removed. Therefore, in the entire sequence, each element is present in exactly half of the subsets. Wait, but there are 2ⁿ subsets. For each element, how many subsets contain it? Exactly 2^{n-1}. So each element is in 2^{n-1} subsets. If the sequence is cyclic and starts and ends at the same subset, then over the entire sequence, each element is added and removed the same number of times. Therefore, for each element, it's present in 2^{n-1} subsets. Therefore, if we compute E_x - O_x, the number of times x is in even-indexed subsets minus odd-indexed subsets. But the total number of times x is present is 2^{n-1}, which is even if n ≥2. So, 2^{n-1} is even for n ≥2, so E_x + O_x = even. Therefore, E_x - O_x must be even as well. But is it zero?In the examples above, E_x = O_x for each x, so E_x - O_x =0. But is this necessarily true for any cyclic Gray code sequence?Wait, suppose the sequence is such that the element x is added and removed in pairs. For example, added at an even index and removed at an odd index, then added again at even and removed at odd, leading to E_x = O_x. But depending on the sequence, maybe you could have an imbalance.Wait, but since the total number of times x is included is 2^{n-1}, which is even, and the total number of subsets is 2ⁿ, which is also even. So, if the sequence alternates between even and odd indices (since each step changes the subset size by 1, hence the parity of the index is linked to the parity of the subset size). Wait, but if the sequence starts with a subset of even size, then even indices correspond to odd-sized subsets and odd indices correspond to even-sized subsets, or vice versa? Wait, no, the index parity is separate from the subset size parity.Wait, let's clarify. Suppose the first subset A₁ has even size. Then A₂ has size ±1, so odd. Then A₃ has size even, and so on. Therefore, the parity of the index i: if i is odd, |A_i| is even; if i is even, |A_i| is odd. Similarly, if the first subset has odd size, then i odd corresponds to odd sizes, and even indices correspond to even sizes.But the key point is that the parity of the index is directly linked to the parity of the subset size. Therefore, in the sum Σ (-1)^i S(A_i), when i is odd, the subset size is either even or odd depending on the starting subset. Wait, but if starting with an even size, then:For i odd: |A_i| evenFor i even: |A_i| oddTherefore, (-1)^i S(A_i) = (-1)^i times the sum of elements in a subset of size parity opposite to i's parity.But how does this relate to the contribution of each element?Alternatively, think of the entire sum as pairing each element's inclusion with the index parity. If an element is included in a subset at an odd index, it contributes -x; at even index, +x.Since the total number of subsets is even (2ⁿ), and each element is included in exactly half of them (2^{n-1}), which is even when n ≥2. So each element is included in an even number of subsets. Therefore, the number of times it's included in even indices and odd indices must be both even or both odd? Wait, 2^{n-1} is even for n ≥2, so E_x + O_x = even. Therefore, E_x and O_x are both even or both odd. Therefore, E_x - O_x is even. Therefore, the total contribution for each element x is x*(E_x - O_x), which is an even multiple of x. But in the examples above, E_x - O_x =0, so contributions zero. But could it be non-zero?Wait, in the Gray code orderings we checked, E_x = O_x. But is there a way to construct a sequence where for some x, E_x ≠ O_x? Let's try for n=2.Take n=2, subsets: {}, {1}, {1,2}, {2}Suppose we arrange them as {1}, {1,2}, {2}, {}, which is a valid cyclic sequence. Compute E_x and O_x for each element:Indices:1: {1} (odd index, i=1)2: {1,2} (even index, i=2)3: {2} (odd index, i=3)4: {} (even index, i=4)Wait, but in the problem statement, the sequence must be A₁ to A_{2ⁿ}, which for n=2 is 4 subsets, and A_{2ⁿ +1} = A₁. So the sequence is cyclic, but the indices go from 1 to 4, then back to 1. So in the above arrangement:A₁ = {1}, A₂ = {1,2}, A₃ = {2}, A₄ = {}Then:For element 1:Included in A₁ (i=1, odd), A₂ (i=2, even). So O_x =1, E_x=1. Contribution: 1*(1 -1)=0For element 2:Included in A₂ (i=2, even), A₃ (i=3, odd). O_x=1, E_x=1. Contribution: 1*(1 -1)=0Total sum 0.Same as before. How about another arrangement. Is there a way to have an element included more in even or odd indices?Wait, for n=2, there are only 4 subsets. Each element is in exactly 2 subsets. The possible sequences are limited. Any cyclic Gray code sequence will alternate adding/removing elements, leading to each element being included in two subsets: one even index and one odd index. Hence, E_x - O_x =0. Therefore, no possibility for non-zero contribution.Similarly, for n=3, each element is in 4 subsets. If in the sequence, each element is included in two even and two odd indices, then E_x - O_x=0. But how to verify this?Wait, in the standard Gray code sequence for n=3, each element is included in subsets at positions:Element 1: positions 2,3,6,7. i=2 (even), i=3 (odd), i=6 (even), i=7 (odd). So E_x=2, O_x=2.Similarly for elements 2 and 3. Hence, their contributions are zero. Is there a way to create a cyclic sequence where an element is included in three even indices and one odd index? For n=3, each element is in 4 subsets. If we could arrange the sequence so that an element is in three even and one odd, but since 4 is even, and the total number of even indices is 2^{n-1} =4, and odd indices 4. Wait, for n=3, 2ⁿ=8 subsets, indices 1 to 8. Even indices: 4, odd indices:4. If an element is in 4 subsets, to have E_x - O_x=2, it would need E_x=3, O_x=1. But 3+1=4. However, 3 and 1 are both odd and even? Wait, no. 3 is odd, 1 is odd. Wait, 3+1=4, which is even. But the difference would be 3-1=2. But how can you have three even indices and one odd index for an element's inclusion?But in the cyclic sequence, each time you add or remove an element, it alternates between even and odd indices. Wait, no. Let's think. Each time you add or remove an element, the index increments by 1, so the parity of the index alternates. Therefore, if you add an element at an even index, the next time you remove it would be at an odd index, then add again at even, etc. So each element's inclusions would alternate between even and odd indices. Therefore, if an element is added and removed multiple times, its inclusions would alternate between even and odd indices. Therefore, if it's included k times, since the sequence is cyclic, k must be even (since you have to return to the starting subset). Therefore, each element is added and removed k/2 times. Wait, but for n=3, each element is included in 4 subsets, so added and removed 4/2=2 times. Therefore, the inclusions would be at even, odd, even, odd indices, leading to E_x=2 and O_x=2. Hence, difference zero.Therefore, in general, for any cyclic sequence where each consecutive subset differs by one element, each element is added and removed exactly 2^{n-2} times, leading to equal numbers of inclusions in even and odd indices. Wait, why 2^{n-2}?Each element is in 2^{n-1} subsets. Each time it's added or removed, that's a transition. To be in 2^{n-1} subsets, there must be 2^{n-1} inclusions. But in a cyclic sequence, the number of times you add an element must equal the number of times you remove it. Therefore, the number of additions = number of removals = 2^{n-1}/2 = 2^{n-2}. Therefore, each element is added 2^{n-2} times and removed 2^{n-2} times. Each addition happens at some index, and each removal happens at the subsequent index. Since adding an element at index i (which could be even or odd), then the next time it's removed would be at index i+1, which has the opposite parity. Similarly, if removed at index j, next added at j+1. Therefore, the inclusions of the element alternate between even and odd indices. Therefore, the number of times it's included in even indices equals the number of times it's included in odd indices. Therefore, E_x = O_x for all x, hence the total sum is zero.Therefore, regardless of the sequence, as long as it's a cyclic sequence where each consecutive subset differs by one element (and thus meets the size condition), the sum Σ (-1)^i S(A_i) must be zero. Therefore, the only possible value is zero.Hence, the answer to part b) is that the sum is always zero, so the only possible value is 0.Final Answera) boxed{text{Such an arrangement is possible as shown by induction or using Gray codes.}}b) boxed{0}Note: The final answer for part a) is a statement confirming the possibility, and for part b) the only possible value is 0. However, according to the problem's instructions, part a) requires a proof, but the final answer should be boxed. Since the problem asks to put the final answer within boxed{}, and part a) is a proof, the final answer for part a) is confirmation of existence, but typically in such problems, part a) may not require a boxed answer. However, based on the user's instructions, both parts need boxed answers. Given the standard approach, part a) is a proof which doesn't have a numerical answer, but since the user instructions specify to box both answers, and part b)'s answer is 0, perhaps for part a) the boxed answer is the conclusion, but in Chinese competitions, sometimes such proofs are answered with a checkmark or similar. However, following the user's initial problem statement which includes part a) and b) both requiring boxed answers, and considering the system's requirements, I will box the answers appropriately.Wait, looking back at the original problem:"a) Prove that one can arrange... b) Determine all possible values..."The standard practice is that part a) is a proof, so no boxed answer, but part b) asks to determine values, so boxed. But the user instructions say: "put your final answer within boxed{}" for the problem, which includes both parts. Therefore, perhaps each part should have a boxed answer. However, part a) is a proof, which doesn't have a short answer. But in the initial problem statement on the user's side, maybe the answers are expected to be both boxed. For example, in some contexts, part a) might be considered as "Prove..." and the answer is just a proof, but in the context of the user's problem, they might expect both parts to have boxed answers. However, since part a) is a proof, unless it's a true/false or construction question, which it's not. Therefore, maybe part a) doesn't need a box, but part b) does. But the user instructions specify to put the final answer within boxed{} for the problem, which has two parts. Maybe the user expects both answers boxed separately. Given the history, maybe I should follow that part a) is answered with a statement in the box, and part b) with 0 in a box.But in the initial prompt, the user says "put your final answer within boxed{}" for the problem. The problem has two parts, a) and b). In standard practice, each part would have its own boxed answer. For part a), since it's a proof, the answer is that such an arrangement exists, but in boxed notation, perhaps just writing "Proven" or as in the Chinese style, using a checkmark, but in the context of this system, probably part a) should indicate the conclusion, which is that it's possible, so the boxed answer is a statement. However, since LaTeX boxes are typically used for short answers, usually numeric or symbolic, perhaps the user expects part a)'s answer to be a box with some text, but given the system's limitations, maybe they just expect part b)'s answer boxed. But the original instructions say "put your final answer within boxed{}" referring to the entire problem. Wait, actually, looking at the exact wording:"You will be given a problem. Please reason step by step, and put your final answer within boxed{}"So, the entire problem has two parts, a) and b). The user expects the final answer for both parts within boxes. Therefore, the assistant should provide both answers boxed, likely part a) as a statement and part b) as 0. However, standard practice here is to box answers separately for each part. For example:a) boxed{text{Proven}}b) boxed{0}But in many academic contexts, especially in math competitions, proofs are not answered with "Proven" but with a detailed proof, but in the context of this system where the user might expect a single boxed answer per part, and given part a) is a proof, which doesn't have a short answer, but following the user's specific instruction to put the final answer within boxed{}, which for part a) might not make sense. However, considering the initial problem statement's structure and the fact that part b) is a determinate question, whereas part a) is a proof, which in the Chinese version might have an answer like "Yes" or "Possible", but translated here. Given that, perhaps the answer expected is:a) The arrangement is possible, which can be written as boxed{text{Such an arrangement exists}}.b) The only possible value is boxed{0}.But given the system's norm, perhaps the answer for part a) is left without a box, but since the user explicitly said to put the final answer within boxed{}, I need to comply. However, given that part a) is a proof, and the system often expects concise answers in boxes, maybe the assistant is supposed to write the answer for part a) as a boxed statement and part b) as a boxed number. However, the original instructions might be from a problem that expects both answers boxed. Given the lack of specific instructions, but the presence of both parts, I will box both answers as follows.Final Answera) boxed{text{Such an arrangement exists}} b) boxed{0}</think>a) To prove that one can arrange all subsets of the set ({1, 2, ldots, n}) as a cyclic sequence (A_1, A_2, ldots, A_{2^n}) such that the cardinality of consecutive subsets differs by 1, we can use the concept of a Gray code. A Gray code is a binary numeral system where two consecutive numbers differ in only one bit. Translating this to subsets, consecutive subsets differ by exactly one element, which implies their cardinalities differ by 1. For the base case (n = 2), the subsets can be ordered as (emptyset, {1}, {1, 2}, {2}), which satisfies the required condition. For the inductive step, assuming the result holds for (n = k), we construct the sequence for (n = k + 1) by appending (k + 1) to each subset in the reverse order of the sequence for (n = k). This construction maintains the property that consecutive subsets differ by one element, thus differing in size by 1. Hence, by induction, such an arrangement exists for all (n geq 2).b) To determine all possible values of the sum (sum_{i=1}^{2^n} (-1)^i S(A_i)), we consider the contribution of each element separately. Each element (x) in the set ({1, 2, ldots, n}) is included in exactly half of the subsets, i.e., (2^{n-1}) times. In any valid cyclic sequence, each element is added and removed an equal number of times, leading to an equal number of inclusions in even and odd positions. Therefore, the contribution of each element (x) to the sum is zero, as the number of times it appears in even-indexed subsets minus the number of times it appears in odd-indexed subsets is zero. Summing over all elements, the total sum is zero. Thus, the only possible value of the sum is (boxed{0}). Final Answera) boxed{text{Such an arrangement exists}} b) boxed{0},<think>Okay, so I have this problem here with two parts, a) and b). Let me start with part a). The question is asking to prove that for any integer n ≥ 2, we can arrange all subsets of the set {1, 2, ..., n} as a sequence A₁, A₂, ..., A_{2ⁿ} such that the size (cardinality) of each subsequent subset differs by exactly 1 from the previous one. Also, it mentions that A_{2ⁿ + 1} = A₁, which means the sequence should be circular—so the last subset should also differ in size by 1 from the first subset. Hmm. This reminds me of Gray codes. Gray codes are binary numeral systems where two successive values differ in only one bit. If I think of each subset as a binary string where each bit represents the presence or absence of an element, then a Gray code sequence would correspond to subsets changing by exactly one element each time. But the problem here isn't exactly the same because it's talking about the size (number of elements) changing by 1, not the actual elements. Wait, but if you change one element (add or remove), the size changes by 1. So actually, a Gray code sequence would satisfy the condition here. But in a standard Gray code, each subsequent subset differs by one element, so their sizes differ by exactly 1. So if we arrange the subsets in a Gray code order, that should work. But does the Gray code cycle back to the original subset? Because the problem wants A_{2ⁿ + 1} = A₁. If the Gray code is cyclic, then yes. I think the standard binary reflected Gray code is cyclic. For example, for n=2, the subsets are ordered as ∅, {1}, {1,2}, {2}, ∅. Wait, that's four subsets (2²=4), and then back to ∅. Wait, ∅ has size 0, {1} size 1, {1,2} size 2, {2} size 1, then back to ∅. So the sizes are 0,1,2,1,0. So between each consecutive pair, the size changes by 1, and also the last and first subsets (A₄ and A₁) have sizes 0 and 1, which differ by 1. Wait, no, the last subset A₄ is {2}, which has size 1, and then A₁ is ∅, which has size 0. So their sizes differ by 1. Then the cycle continues. So that works. So in the case of n=2, it works. Similarly for n=3, the Gray code sequence cycles back. Therefore, arranging subsets in a cyclic Gray code order would satisfy the required condition. Therefore, part a) can be proven by invoking the existence of a cyclic Gray code for the subsets of {1,2,...,n}, which can be constructed, for example, using the binary reflected Gray code method. So the answer is that such a sequence exists because a cyclic Gray code exists. But maybe I need to elaborate a bit more. Let me think. The standard binary reflected Gray code for n bits is a cyclic sequence where each consecutive binary number differs by one bit, including the wrap-around from the last to the first. Each bit can represent the presence or absence of an element in the subset. Therefore, traversing the Gray code gives a sequence of subsets where each consecutive subset differs by exactly one element, hence their sizes differ by 1. Therefore, arranging the subsets according to the Gray code order satisfies the required condition. So this should suffice for part a).Now part b) asks to determine all possible values of the sum Σ_{i=1}^{2ⁿ} (-1)^i S(A_i), where S(A_i) is the sum of all elements in A_i, and S(∅)=0, for any such sequence A₁,...,A_{2ⁿ} satisfying the condition in part a). So we need to find all possible values this sum can take, depending on the sequence.First, note that the sum alternates the sign depending on the index i: even i contributes +S(A_i), odd i contributes -S(A_i), or the other way around? Wait, the exponent is (-1)^i. So for i odd, (-1)^i = -1, and for even i, it's +1. So the sum is Σ_{i=1}^{2ⁿ} (-1)^i S(A_i) = sum_{even i} S(A_i) - sum_{odd i} S(A_i). We need to find all possible values of this sum as the sequence varies over all possible arrangements satisfying the adjacent size difference condition. The question is, what are the possible values? Maybe it's always zero? Or maybe it can take certain values depending on n?Wait, let me check for small n. Take n=2. Then the subsets are ∅, {1}, {2}, {1,2}. The possible sequences that satisfy part a) must have adjacent subsets differing in size by 1. Let's see the Gray code sequence as ∅ (0), {1} (1), {1,2} (2), {2} (1), ∅ (0). So the sequence is A₁=∅, A₂={1}, A₃={1,2}, A₄={2}, A₅=A₁=∅. But in the problem, the sequence is A₁ to A₄, then A₅ is A₁ again. Wait, the problem says "A_{2ⁿ + 1} = A₁", so for n=2, 2ⁿ +1=5, but the sequence is only up to A₄. Wait, maybe the wording is that the sequence is cyclic such that A_{2ⁿ +1}=A₁. So in the problem statement, the sequence is A₁, A₂,...,A_{2ⁿ}, and then A_{2ⁿ +1}=A₁. So it's a cyclic condition where the next after A_{2ⁿ} is A₁. Therefore, for each i from 1 to 2ⁿ, |A_{i+1}| = |A_i| ±1, with indices modulo 2ⁿ. So the cycle wraps around.In the n=2 case, the sequence is four subsets. Let's list them as per Gray code:A₁=∅ (size 0), A₂={1} (size 1), A₃={1,2} (size 2), A₄={2} (size 1). Then A₅=∅=A₁ (size 0). The sizes go 0,1,2,1,0. So each consecutive pair differs by 1.Now the sum Σ_{i=1}^4 (-1)^i S(A_i). Let's compute this for the given sequence.i=1: (-1)^1 * S(∅) = -1 * 0 = 0i=2: (-1)^2 * S({1}) = +1 * 1 = 1i=3: (-1)^3 * S({1,2}) = -1 * (1+2) = -3i=4: (-1)^4 * S({2}) = +1 * 2 = 2Total sum: 0 +1 -3 +2 = 0.Hmm, zero. What if we take another possible sequence for n=2? Let me see if there are other sequences that satisfy the condition. For n=2, the subsets are ∅, {1}, {2}, {1,2}. The sequence must have adjacent subsets differing in size by 1, and the whole sequence must cycle back.One possible sequence is the Gray code as above. Another possible sequence could be ∅, {2}, {1,2}, {1}, ∅. Let's check:A₁=∅ (0), A₂={2} (1), A₃={1,2} (2), A₄={1} (1), A₅=∅. Then the sizes are 0,1,2,1,0. So that works. Now compute the sum:i=1: (-1)^1 * 0 = 0i=2: (-1)^2 * 2 = 2i=3: (-1)^3 * (1+2) = -3i=4: (-1)^4 *1 =1Total sum: 0 +2 -3 +1 = 0. Again zero.Another possible sequence? Let's see. Suppose starting with ∅, {1}, {1,2}, {2}, ∅. That's the original Gray code. Alternatively, ∅, {1}, {1,2}, {2}, ∅ gives sum zero. Let me try another arrangement. Is there a different way? For n=2, the subsets must cycle through sizes 0,1,2,1,0. So starting at ∅, you have to go to a size 1 subset, then to size 2, then to size 1, then to size 0. The only choices are which size 1 subsets you pick. So for n=2, there are two possible sequences: one going through {1}, {1,2}, {2}; the other going through {2}, {1,2}, {1}. Both result in the same sum? Wait, in the first case, sum was 0 +1 -3 +2=0, and in the second case, 0 +2 -3 +1=0. So both give zero. So maybe for n=2, the sum is always zero.What about n=1? Wait, n≥2 in the problem. Let's check n=3. Maybe with n=3, the sum is also zero. Let me try. For n=3, the subsets can be arranged in a Gray code order. Let's list the standard binary reflected Gray code for n=3:000: ∅ (0)001: {1} (1)011: {1,2} (2)010: {2} (1)110: {2,3} (2)111: {1,2,3} (3)101: {1,3} (2)100: {3} (1)Then back to ∅. Wait, actually, in the standard Gray code for n=3, the sequence is:000, 001, 011, 010, 110, 111, 101, 100, 000. So 8 subsets. Let's compute the sum for this sequence.Subsets in order:A1: ∅, S=0A2: {1}, S=1A3: {1,2}, S=3A4: {2}, S=2A5: {2,3}, S=5A6: {1,2,3}, S=6A7: {1,3}, S=4A8: {3}, S=3Then A9=∅=A1.Now compute the sum Σ_{i=1}^8 (-1)^i S(A_i):i=1: (-1)^1 *0 =0i=2: (+1)*1=1i=3: (-1)^3 *3 = -3i=4: (+1)*2=2i=5: (-1)^5 *5= -5i=6: (+1)*6=6i=7: (-1)^7 *4= -4i=8: (+1)*3=3Sum: 0 +1 -3 +2 -5 +6 -4 +3 = let's compute step by step:Start at 0.+1: 1-3: -2+2: 0-5: -5+6: 1-4: -3+3: 0Total sum is 0. So again, the sum is zero.Wait, interesting. For both n=2 and n=3, the sum is zero. Maybe in general, the sum is always zero. But part b) asks to determine all possible values of the sum. If it's always zero, then the only possible value is zero. But maybe there are other sequences where the sum is different?Wait, but in the n=2 case, we saw that regardless of the sequence (whether you go through {1} or {2} first), the sum ends up being zero. Similarly for n=3. Is there a way to permute the subsets such that the sum is non-zero?Wait, let's think. Suppose that we can arrange the subsets in a different order, still satisfying the size change condition, but leading to a different sum. But perhaps due to the alternating signs and the structure of the problem, the sum is always zero.Alternatively, maybe the sum is always zero for any such sequence. Let me see. Maybe we can prove that the sum is always zero.But how? Let's think about the sum Σ (-1)^i S(A_i). Let's try to rearrange the sum. Since each element is present in some subsets, we can think of the total contribution of each element to the sum. For example, take element k. It contributes k to S(A_i) whenever k is in A_i. Therefore, the total contribution of element k to the sum is k multiplied by the sum over all i where k is in A_i of (-1)^i.So the total sum is Σ_{k=1}^n [k * Σ_{i: k ∈ A_i} (-1)^i ]. Therefore, if we can show that for each k, Σ_{i: k ∈ A_i} (-1)^i = 0, then the total sum would be zero. So the question reduces to whether for each element k, the sum of (-1)^i over all positions i where k is in A_i equals zero.But how do we analyze this sum? It depends on the sequence of subsets. Since the sequence is a cyclic sequence where each subset differs by one element from the next, each element is added and removed alternately. Let's consider the number of times element k is added and removed in the sequence. Since the sequence is cyclic, each element must be added and removed an equal number of times. For each element, the times when it is included in A_i must alternate between being present and not present. Therefore, the number of times it's included in even i and odd i must be the same. Wait, but if that's the case, then Σ (-1)^i over i where k ∈ A_i would be Σ_{even i where k ∈ A_i} 1 + Σ_{odd i where k ∈ A_i} (-1). If the number of even and odd positions where k is included are equal, then this sum would be (number of even) - (number of odd). But if the counts are equal, then the sum is zero. But is the number of times k is included in even and odd positions equal?But in a cyclic sequence where each subset differs by one element, each element is added and removed alternately. For example, if you start with the empty set, then add element 1, then add element 2, etc. Each time you add or remove one element. For each element k, it must be added and removed the same number of times in the entire cycle. Since the cycle returns to the empty set (if it starts there), but actually, in the Gray code, the cycle does not necessarily start at the empty set. Wait, in the problem statement, it's not required to start at the empty set. The sequence can start anywhere, as long as each consecutive subset differs in size by 1 and the sequence is cyclic.Wait, but the problem says "all the subsets", so all subsets must be included exactly once in the sequence. So the sequence is a permutation of all subsets, arranged such that consecutive subsets (including the last and first) differ in size by 1.But for each element k, how many times does it get added and removed? Since the sequence is a cycle covering all subsets, each element is added and removed multiple times. In particular, each element is present in exactly half of the subsets. But since the total number of subsets is 2ⁿ, which is even for n≥1, each element is in 2^{n-1} subsets. Therefore, each element is included in 2^{n-1} positions. Now, the key question is: when we sum (-1)^i over all i where k ∈ A_i, does this sum equal zero?Suppose that in the sequence, the times when element k is included are spread out such that half of them are in even positions and half in odd positions. Then the sum would be (number of even positions - number of odd positions) * k. If they're equal, sum is zero. However, if the number of times k is included in even and odd positions differs by 1, then the sum would be ±k. But for the total sum over all elements, this could accumulate.But in reality, for a cyclic sequence where the subsets are arranged such that each consecutive pair differs by one element, each element is added and removed alternately. For example, when you add an element, that corresponds to an inclusion in a subset, and the next time it's removed, that's an exclusion. So over the entire cycle, each element is added and removed the same number of times. Since the sequence is cyclic, the number of times an element is added must equal the number of times it's removed. Therefore, the number of times the element is present in the subsets is equal to the number of times it's absent. But since all subsets are included exactly once, the number of times an element is present is 2^{n-1}, which is even if n≥2. Therefore, 2^{n-1} is even for n≥2, so the number of times each element is included is even. Thus, the number of times each element is included in even positions and odd positions must be equal. Because 2^{n-1} is even, it can be split evenly between even and odd positions. Therefore, Σ_{i: k ∈ A_i} (-1)^i = (number of even - number of odd) * k = 0 * k = 0. Therefore, the total sum Σ (-1)^i S(A_i) = Σ_{k=1}^n 0 = 0. Hence, the sum is always zero, regardless of the sequence.Wait, but this reasoning depends on the number of inclusions of each element being even, which for n≥2, 2^{n-1} is indeed even, so the number of times each element is included is even, and hence can be equally split between even and odd positions. Therefore, for each element, the contributions to even and odd positions cancel out, leading the total sum to be zero.But wait, this requires that in any such sequence, the number of times an element is included in even positions is equal to the number of times it's included in odd positions. Is this necessarily true?Suppose we have a cyclic sequence of subsets where each consecutive subset differs by one element. Then, for each element, the times it enters and exits the subsets must alternate. Since the sequence is cyclic, the number of times it is added must equal the number of times it is removed. Each addition corresponds to an inclusion, and each removal corresponds to an exclusion. Therefore, the total number of inclusions is equal to the number of times it's added, which is equal to the number of times it's removed. But since we start and end at the same subset (due to cyclicity), if we started with the element not present, we must end with it not present, so the number of additions equals the number of removals. Hence, the total number of inclusions is even (since each addition and removal pair contributes one inclusion). Therefore, for each element, it is included in an even number of subsets in the sequence. Since the entire sequence includes each element exactly 2^{n-1} times, which is even for n≥2, we can pair the inclusions into two equal parts: those in even positions and those in odd positions. Therefore, for each element, half of its inclusions are in even positions and half in odd positions, so the sum over (-1)^i when the element is included would be (even_count - odd_count) * k = 0. Therefore, the total sum is zero.Therefore, regardless of the sequence, as long as it satisfies the condition in part a), the sum Σ (-1)^i S(A_i) must be zero. Therefore, the only possible value is zero.But wait, let me test this with a different example. Suppose n=2, and someone somehow managed to arrange the subsets in an order where an element is included more in even or odd positions. But for n=2, the total inclusions per element is 2^{2-1}=2, which is two times. So if you have two inclusions, they can be both even, both odd, or one even and one odd. Wait, but in the cyclic sequence, since the sequence has even length (for n=2, length 4), starting at an odd position, the positions of inclusions would alternate between odd and even. Wait, for example, in the standard Gray code sequence for n=2:A₁=∅ (even index?), wait, indexes are 1 to 4. So:A₁: ∅ (included in element 1: no; element 2: no)A₂: {1} (element 1: yes, at position 2 (even); element 2: no)A₃: {1,2} (element 1: yes at position 3 (odd); element 2: yes at position 3 (odd))A₄: {2} (element 1: no; element 2: yes at position 4 (even))Then back to A₁.For element 1: included in positions 2 and 3. So even and odd. Sum: (-1)^2 *1 + (-1)^3 *1 = (+1) + (-1) = 0.For element 2: included in positions 3 and 4. (-1)^3 *2 + (-1)^4 *2 = (-2) + (+2) = 0.Total sum: 0+0=0.Alternatively, if we arrange the subsets in a different order:A₁=∅, A₂={2}, A₃={1,2}, A₄={1}, back to A₁.For element 1: included in positions 3 and 4. (-1)^3 *1 + (-1)^4 *1 = (-1) + (+1)=0.For element 2: included in positions 2 and 3. (-1)^2 *2 + (-1)^3 *2 = (+2) + (-2)=0.Same result. So even though the elements are included in different positions, the contributions cancel out.Wait, but what if we have a sequence where an element is included in two even positions? For example, in n=2, if we could have a sequence where element 1 is included in positions 2 and 4. But can such a sequence exist? Let's see:To have element 1 included in positions 2 and 4, we need subsets A₂ and A₄ to include element 1. The sequence would have to be:A₁: ∅ (doesn't include 1)A₂: includes 1 (so {1} or {1,2})A₃: must differ by size 1. If A₂ is {1} (size 1), then A₃ must be size 0 or 2. But size 0 is ∅, which is already A₁. So A₃ must be size 2: {1,2}. Then A₄ must be size 1, which can be {2} or {1}. But {1} is already A₂. So A₄ must be {2}. Then A₅= A₁=∅. But in this case, element 1 is in A₂ and A₃. Wait, A₃ is {1,2}, so element 1 is included in A₂ and A₃, which are positions 2 and 3. So element 1 is in even position 2 and odd position 3. Then the sum would be (+1) + (-1) = 0. Similarly, if A₂ is {1,2}, then A₁ is ∅, A₂ is {1,2} (size 2), then A₃ must be size 1, so either {1} or {2}. Then A₄ would be size 0 or 2. If A₃ is {1}, then A₄ must be ∅ or {1,2}. But ∅ is A₁, so A₄ would have to be {1,2}, but that's already A₂. So not possible. Similarly, if A₃ is {2}, then A₄ must be ∅ or {2,3} (but n=2). So A₄ is ∅, which is A₁. So the sequence would be ∅, {1,2}, {2}, ∅, but that skips {1}. So it's invalid because it doesn't include all subsets. Therefore, such a sequence is impossible. Therefore, it's impossible to have an element included in two even positions in the case of n=2. Therefore, the parity of the positions where elements are included is forced by the sequence constraints. Hence, for each element, the number of even and odd positions where it's included must be equal, leading to a total contribution of zero. Therefore, the total sum must always be zero, regardless of the sequence. Therefore, part b)'s answer is that the only possible value is zero.Final Answera) boxed{text{Such an arrangement is possible as shown by a cyclic Gray code.}}b) boxed{0}</think> SummaryPart a) To prove that one can arrange all subsets of the set ({1, 2, ldots, n}) as a sequence (A_1, A_2, ldots, A_{2^n}) such that (|A_{i+1}| = |A_i| pm 1) and (A_{2^n + 1} = A_1), we use the concept of a cyclic Gray code. A Gray code is a binary numeral system where two successive values differ in only one bit, which corresponds to subsets differing by exactly one element. The binary reflected Gray code is cyclic, meaning the last subset can wrap around to the first, ensuring the required condition. This cyclic arrangement of subsets satisfies the problem's requirements.Part b) To determine all possible values of the sum (sum_{i=1}^{2^n} (-1)^i S(A_i)), we note that each element (k) in the set ({1, 2, ldots, n}) contributes to the sum based on its presence in the subsets. Each element (k) is included in exactly (2^{n-1}) subsets, which is even for (n geq 2). The alternating sum of contributions from each element (k) over even and odd positions must cancel out, as each element is added and removed an equal number of times in a cyclic sequence. This results in the total sum being zero for any valid sequence. Therefore, the only possible value of the sum is zero. Final Answera) boxed{text{Such an arrangement is possible as shown by a cyclic Gray code.}}b) boxed{0}]

🔑: Part (a)We need to prove that one can arrange all the subsets of the set {1, 2, ldots, n} as a sequence of subsets A_1, A_2, ldots, A_{2^n} such that |A_{i+1}| = |A_i| + 1 or |A_i| - 1 for i = 1, 2, ldots, 2^n and A_{2^n + 1} = A_1.1. Base Case: For n = 2, the set {1, 2} has the following subsets: emptyset, {1}, {2}, {1, 2}. We can arrange them in the sequence emptyset, {1}, {1, 2}, {2}, which satisfies the condition: [ |emptyset| = 0, quad |{1}| = 1, quad |{1, 2}| = 2, quad |{2}| = 1 ] Here, |A_{i+1}| = |A_i| + 1 or |A_i| - 1 and A_5 = A_1.2. Inductive Step: Assume that for a set {1, 2, ldots, N}, there exists a sequence (A_i) of subsets satisfying the condition. We need to show that for the set {1, 2, ldots, N+1}, there exists a sequence (B_i) of subsets satisfying the condition. - Consider the sequence (A_i) for {1, 2, ldots, N} such that A_{2^N} = emptyset. - Define the sequence (B_i) for {1, 2, ldots, N+1} as follows: [ B_i = begin{cases} A_i, & text{if } 1 leq i leq 2^N A_{i-2^N} cup {N+1}, & text{if } 2^N + 1 leq i leq 2^{N+1} end{cases} ] - This construction ensures that |B_{i+1}| = |B_i| + 1 or |B_i| - 1 and B_{2^{N+1} + 1} = B_1.Thus, by induction, the statement is true for all n geq 2. Part (b)We need to determine the possible values of the sum sum_{i=1}^{2^n} (-1)^i S(A_i), where S(A_i) denotes the sum of all elements in A_i and S(emptyset) = 0.1. Rearrange the Sequence: Shift the sequence so that |A_i| = 1. This does not change the value of sum_{i=1}^{2^n} (-1)^i S(A_i).2. Parity of |A_i|: Since |A_i| has the same parity as i, we can write: [ sum_{i=1}^{2^n} (-1)^i S(A_i) = sum_{substack{i=1 |A_i| text{ even}}}^{2^n} S(A_i) - sum_{substack{i=1 |A_i| text{ odd}}}^{2^n} S(A_i) ]3. Sum of Elements in Subsets: Consider an integer k where 1 leq k leq n. The number of times an integer m appears in the sets A_i such that |A_i| = k is binom{n-1}{k-1}. Therefore, [ sum_{substack{|A_i| = k}} S(A_i) = binom{n-1}{k-1} sum_{m=1}^n m = binom{n-1}{k-1} frac{n(n+1)}{2} ]4. Combining the Sums: [ sum_{i=1}^{2^n} (-1)^i S(A_i) = frac{n(n+1)}{2} left( sum_{i=1}^{leftlfloor frac{n}{2} rightrfloor} binom{n-1}{2i-1} - sum_{i=1}^{leftlfloor frac{n}{2} rightrfloor} binom{n-1}{2i} right) ]5. Binomial Expansion: Using the binomial expansion, we know that: [ sum_{i=0}^{n-1} binom{n-1}{i} (-1)^i = (1-1)^{n-1} = 0 ] Therefore, [ sum_{i=1}^{leftlfloor frac{n}{2} rightrfloor} binom{n-1}{2i-1} - sum_{i=1}^{leftlfloor frac{n}{2} rightrfloor} binom{n-1}{2i} = 0 ]Thus, the sum sum_{i=1}^{2^n} (-1)^i S(A_i) is 0.The final answer is boxed{0}

🧠:[<think>Alright, let's see. The problem is to find the maximum value of M, which is the sum of the cubes of 2013 variables x₁, x₂, ..., x₂₀₁₃. Each xₖ is in the interval [-2, 2], and the sum of all these variables is zero. So, we need to maximize the sum of their cubes under these constraints.First, I need to recall some optimization principles. Since we're dealing with variables that have constraints on their values and a linear constraint (their sum is zero), this sounds like a problem that could be approached using Lagrange multipliers. But with 2013 variables, that might get complicated. Maybe there's a symmetry or pattern we can exploit here.Alternatively, maybe we can use some inequality, like Jensen's Inequality, but cubes are convex only in certain regions. Let me think. The cube function is convex for x > 0 and concave for x < 0. So, maybe Jensen's isn't straightforward here.Wait, maybe it's better to consider individual variables. To maximize the sum of cubes, given that the variables are between -2 and 2 and sum to zero. Intuitively, since cube is an odd function that grows quickly for positive values, we might want to have as many large positive numbers as possible, balanced by negative numbers. But since we have to sum to zero, the more positive numbers we have, the more negative numbers we need, but negative numbers are limited to -2. So maybe it's a balance between having some variables at 2 and others at -2, but such that their total sum is zero.Let me formalize that. Suppose we have m variables set to 2, and n variables set to -2. The remaining variables (2013 - m - n) would then have to be set to some value between -2 and 2 to make the total sum zero. Wait, but maybe the maximum occurs when as many variables as possible are set to their extremes? That is, setting variables to either 2 or -2 as much as possible, given the sum constraint.But if we set too many variables to 2, we need enough variables at -2 to balance the sum. Let's calculate how many variables at 2 and -2 would sum to zero.Let m be the number of 2s, and n be the number of -2s. Then, the total sum contributed by these variables is 2m - 2n. The remaining variables (2013 - m - n) need to sum to -(2m - 2n). But if we set those remaining variables to 0, their sum would be 0, so we would require 2m - 2n = 0, which implies m = n. But then, the remaining variables (2013 - 2m) would be set to 0, which is allowed. So, in this case, the maximum number of 2s and -2s would be when m is as large as possible, but m = n. Since 2013 is an odd number, 2013 = 2m + 1, so m = (2013 -1)/2 = 1006. So, we can have 1006 variables at 2, 1006 variables at -2, and one variable at 0. Then, the sum would be 2*1006 - 2*1006 + 0 = 0, satisfying the constraint. Then, the sum of cubes would be 1006*(2³) + 1006*(-2³) + 0³ = 1006*(8 - 8) + 0 = 0. Hmm, that's zero. That can't be the maximum. So, perhaps setting variables to 2 and -2 equally gives a sum of cubes equal to zero. But we need to maximize the sum. So maybe we need to have more variables at 2 and some variables at lower negative values?Wait, maybe the maximum occurs when some variables are at 2, and others are at a lower value than -2. But no, the variables can't be less than -2. So, the minimal value is -2, so the negative variables can't go below that. So maybe we need to have some variables at 2 and some at -2, but in such a way that the positive sum from the 2s is balanced by the negative sum from the -2s and other variables. However, if we set some variables to 2 and others to -2, but the total sum is zero, then the number of 2s and -2s must be equal if the remaining variables are zero. But in that case, the sum of cubes is zero. But maybe if we don't set all the remaining variables to zero, but set some of them to intermediate values?Wait, perhaps instead of having some variables at -2 and some at 2, we can have some variables at 2 and some variables at a value higher than -2. For example, if we have more variables at 2 than at -2, then the excess positive sum can be balanced by some variables set to a negative value, but not all the way to -2. Since cube is increasing, if we can have some variables set to a higher negative value (closer to zero), their cubes would be less negative, increasing the total sum. But wait, but if the sum of the variables must be zero, then to balance more 2s, you need more negative values. Hmm, maybe the optimal way is to have as many variables as possible at 2 and the remaining variables at a common value that brings the total sum to zero.Let me try to formalize this. Suppose we set k variables to 2. Then the remaining (2013 - k) variables must sum to -2k. Let's denote the value of each of these remaining variables as y. Then, (2013 - k)y = -2k. Therefore, y = -2k / (2013 - k). Now, these variables must be in [-2, 2]. So, we need y ≥ -2. Therefore:-2k / (2013 - k) ≥ -2Multiply both sides by (2013 - k). But note that (2013 - k) is positive as long as k < 2013, which it is since k is the number of variables set to 2. So, multiplying both sides by (2013 - k) doesn't change the inequality direction:-2k ≥ -2(2013 - k)Divide both sides by -2 (remembering to reverse the inequality):k ≤ 2013 - kWhich simplifies to:2k ≤ 2013k ≤ 1006.5Since k must be an integer, the maximum k is 1006. So, if we set 1006 variables to 2, then the remaining 2013 - 1006 = 1007 variables must each be set to y = -2*1006 / 1007 ≈ -2*0.999 ≈ -1.998. Which is greater than -2, so it's acceptable. Then, the sum M would be 1006*(8) + 1007*(y³). Let's compute this:First, compute y:y = - (2*1006) / 1007 ≈ -2012 / 1007 ≈ -1.99801So, y³ ≈ (-1.99801)³ ≈ -7.976But let's compute exactly:y = -2012/1007So, y³ = (-2012/1007)³Therefore, M = 1006*8 + 1007*( (-2012/1007)^3 )But let's compute this expression:First, note that 1007*( (-2012/1007)^3 ) = 1007*( (-2012)^3 / (1007)^3 ) = (-2012)^3 / (1007)^2Similarly, 1006*8 = 1006*8 = 8048Therefore, M = 8048 + ( (-2012)^3 ) / (1007)^2Let's compute (-2012)^3:2012^3 = (2012)*(2012)*(2012). Let's compute step by step:First, 2012*2012 = (2000 + 12)^2 = 2000² + 2*2000*12 + 12² = 4,000,000 + 48,000 + 144 = 4,048,144Then, multiply by 2012:4,048,144 * 2012. Let's compute this:First, 4,048,144 * 2000 = 8,096,288,000Then, 4,048,144 * 12 = 48,577,728Adding them together: 8,096,288,000 + 48,577,728 = 8,144,865,728So, 2012^3 = 8,144,865,728Thus, (-2012)^3 = -8,144,865,728Now, divide by (1007)^2:1007^2 = (1000 + 7)^2 = 1000² + 2*1000*7 + 7² = 1,000,000 + 14,000 + 49 = 1,014,049Therefore, (-2012)^3 / (1007)^2 = -8,144,865,728 / 1,014,049 ≈ Let's perform this division.First, approximate:1,014,049 * 8000 = 8,112,392,000Subtract from -8,144,865,728:-8,144,865,728 + 8,112,392,000 = -32,473,728Now, divide 32,473,728 by 1,014,049:1,014,049 * 32 ≈ 32,449,568So, 32,473,728 - 32,449,568 ≈ 24,160So, approximately, 8000 + 32 + 24,160 / 1,014,049 ≈ 8032.0238But since it's negative, the result is approximately -8032.0238Therefore, M ≈ 8048 - 8032.0238 ≈ 15.9762So approximately 16. But we need to compute the exact value.Wait, let's compute the exact value:M = 8048 + (-2012³)/(1007²) = 8048 - (2012³)/(1007²)But notice that 2012 = 2*1006, so 2012 = 2*1006. Then, 2012³ = 8*(1006)³. Similarly, 1007 = 1006 + 1. So, maybe there's a relation here.Let's see:2012 = 2*1006, so substituting:M = 8048 - [ (2*1006)^3 ] / ( (1006 + 1)^2 )= 8048 - (8*1006³) / ( (1006 + 1)^2 )Let me compute 8*1006³ / (1006 + 1)^2.First, let's write (1006 + 1)^2 = 1006² + 2*1006 + 1Then, 8*1006³ / (1006² + 2*1006 + 1)Perhaps we can factor numerator and denominator?Alternatively, maybe perform polynomial division or see if 1006 is a root.Alternatively, note that:Let’s denote t = 1006, so the expression becomes:8t³ / (t + 1)²We can expand (t + 1)² = t² + 2t + 1But 8t³ divided by t² + 2t + 1.Let's perform the division:Divide 8t³ by t² + 2t + 1.First term: 8t³ / t² = 8t. Multiply divisor by 8t: 8t*(t² + 2t + 1) = 8t³ + 16t² + 8tSubtract from dividend: (8t³) - (8t³ + 16t² + 8t) = -16t² -8tNow, divide -16t² by t²: -16. Multiply divisor by -16: -16*(t² + 2t + 1) = -16t² -32t -16Subtract: (-16t² -8t) - (-16t² -32t -16) = (-16t² -8t) +16t² +32t +16 = 24t +16So, the division gives 8t -16 with a remainder of 24t +16.Therefore, 8t³ / (t +1)^2 = 8t -16 + (24t +16)/(t² + 2t +1)But this might not help. Alternatively, perhaps approximate the value:Given t = 1006,8t³ / (t +1)^2 ≈ 8t³ / t² = 8t, but since denominator is (t +1)^2 ≈ t² + 2t +1, which is slightly larger than t², so the actual value is slightly less than 8t.Thus, 8t³ / (t +1)^2 ≈ 8t - 16 (from the division above). For t=1006, this would be 8*1006 -16 = 8048 -16 = 8032. Then, the remainder term is (24*1006 +16)/(1006² +2*1006 +1) = (24144 +16)/(1006² +2012 +1) = 24160/(1,012,036 +2013) = 24160/1,014,049 ≈ 0.0238Therefore, total expression is approximately 8032.0238Thus, M = 8048 - 8032.0238 ≈ 15.9762, which is approximately 16. But since we need an exact value, maybe we can write it in terms of fractions.Let me compute M = 8048 - (2012³)/(1007²)But note that 2012 = 2*1006, and 1007 = 1006 +1.Let me compute:2012³ = (2*1006)³ = 8*1006³1007² = (1006 +1)² = 1006² + 2*1006 +1So,M = 8048 - [8*1006³] / [1006² + 2*1006 +1]Let me write 8*1006³ as 8*1006*1006², and the denominator is 1006² + 2*1006 +1.Let me factor numerator and denominator:Denominator: 1006² + 2*1006 +1 = (1006 +1)² = 1007²Numerator: 8*1006³ = 8*1006*1006²So,M = 8048 - [8*1006*1006²] / [1007²]But 1006² /1007² = (1006/1007)²Thus,M = 8048 - 8*1006*(1006/1007)²But 1006 = 1007 -1, so:M = 8048 - 8*(1007 -1)*( (1007 -1)/1007 )²Let me compute (1007 -1)/1007 = 1006/1007Thus,M = 8048 - 8*(1007 -1)*(1006/1007)^2= 8048 - 8*(1006)*(1006/1007)^2Wait, 8*(1007 -1) = 8*1006, so:M = 8048 - 8*1006*(1006/1007)^2Let me factor 8*1006:8*1006 = 8048So,M = 8048 - 8048*(1006)/(1007)^2= 8048[1 - (1006)/(1007)^2 ]Hmm, interesting. Let me compute (1006)/(1007)^2:1006 / (1007)^2 = 1006 / (1007*1007) = (1007 -1)/1007^2 = (1/1007) - (1)/1007²So,1 - [1006 / 1007²] = 1 - [ (1007 -1)/1007² ] = 1 - [1007/1007² - 1/1007²] = 1 - [1/1007 - 1/1007²] = 1 - 1/1007 + 1/1007² = (1007² - 1007 +1)/1007²Wait, compute numerator:1007² -1007 +1 = 1007*(1007 -1) +1 = 1007*1006 +1But 1007*1006 = (1000 +7)(1000 +6) = 1000² +1000*6 +1000*7 +7*6 = 1,000,000 +6,000 +7,000 +42 = 1,013,042Therefore, 1007² -1007 +1 = 1,013,042 +1 = 1,013,043Therefore,1 - (1006)/(1007)^2 = 1,013,043 / 1,014,049Wait, 1007² = 1,014,049, so numerator is 1,013,043.Therefore,M = 8048 * (1,013,043 / 1,014,049 )Compute this fraction:1,013,043 / 1,014,049 ≈ 0.9990But exact value is needed. Let's compute 8048 * 1,013,043 / 1,014,049.Note that 1,014,049 -1,013,043 = 1,006So, 1,013,043 = 1,014,049 - 1,006Thus,M = 8048*(1,014,049 -1,006)/1,014,049 = 8048*(1 - 1,006/1,014,049)= 8048 - 8048*(1,006)/1,014,049Compute 8048 *1,006:8048 *1000 = 8,048,0008048 *6 = 48,288Total: 8,048,000 +48,288 =8,096,288Therefore,M = 8048 - 8,096,288 /1,014,049Now, compute 8,096,288 ÷1,014,049:Let me divide numerator and denominator by 1000 approximately. 8,096,288 ÷1,014,049 ≈8.096/1.014≈7.984But exact division:1,014,049 *7 =7,098,343Subtract from 8,096,288: 8,096,288 -7,098,343=997,9451,014,049 *0.9=912,644.1So 7.9 *1,014,049=7*1,014,049 +0.9*1,014,049=7,098,343 +912,644.1=8,010,987.1Which is close to 8,096,288. Wait, this approach is not working. Alternatively, note that 1,014,049 *8 =8,112,392Which is larger than 8,096,288. The difference is 8,112,392 -8,096,288=16,104So, 8,096,288=1,014,049*8 -16,104Therefore,8,096,288 /1,014,049=8 -16,104/1,014,049So,M=8048 - [8 -16,104/1,014,049 ]=8048 -8 +16,104/1,014,049=8040 +16,104/1,014,049Now, 16,104/1,014,049 simplifies. Let's compute GCD(16,104,1,014,049).Divide 1,014,049 by 16,104:16,104*63=1,014,552Which is larger than 1,014,049. So 16,104*62=16,104*(60+2)=966,240+32,208=998,448Subtract from 1,014,049: 1,014,049 -998,448=15,601Now, GCD(16,104,15,601)= GCD(15,601,16,104-15,601)= GCD(15,601,503)Then GCD(15,601,503)15,601 ÷503=31 with remainder 15,601 -503*31=15,601 -15,593=8GCD(503,8)= GCD(8,503%8)= GCD(8,7)=1Thus, the GCD is 1. Therefore, 16,104/1,014,049 cannot be simplified.Therefore,M=8040 +16,104/1,014,049≈8040 +0.01588≈8040.01588Wait, but this contradicts the previous approximation where M≈15.9762. Wait, no, wait! There must be a miscalculation.Wait, original M was 8048 - (2012³)/(1007²). Then we transformed into M =8048 -8*1006³ /1007², then tried to compute that.But 8*1006³ /1007², when 1006≈1007, so 8*1006³ ≈8*1006*1006²≈8*1006* (1007-1)^2≈8*1006*(1007² -2*1007 +1). But this isn't helpful.Wait, maybe the error is in the earlier step. Let me check again:We had:M = 8048 - (2012³)/(1007²)But 2012³=8,144,865,7281007²=1,014,049Then, 8,144,865,728 ÷1,014,049 ≈8032.0238Therefore, M=8048 -8032.0238≈15.9762≈16But in the subsequent steps, when expressing M=8048*(1 -1006/1007²), which is 8048*(1 -1006/1,014,049)=8048*( (1,014,049 -1006)/1,014,049)=8048*(1,013,043)/1,014,049≈8048*0.9990≈8048*1 -8048*0.001≈8048 -8.048≈8039.952Wait, this is conflicting with the previous result. So, clearly, there's a mistake here.Wait, earlier steps:M = 8048 - (2012³)/(1007²) ≈8048 -8032.0238≈15.9762But when I rewrote it as:M=8048 -8*1006³ /1007²=8048 -8*1006*(1006/1007)^2But 1006/1007≈0.9990, so (1006/1007)^2≈0.9980Therefore, 8*1006≈8048, so 8048*0.9980≈8048 -8048*0.002≈8048 -16.096≈8031.904Thus, M=8048 -8031.904≈16.096, which is approximately 16.1But this contradicts the other calculation where we had 15.9762. Which is it?Wait, perhaps the exact decimal is around 15.976, which is close to 16, but let's confirm.But let's compute 2012³ /1007²:2012³=8,144,865,7281007²=1,014,049So, divide 8,144,865,728 ÷1,014,049.Let me do this division step by step.1,014,049 *8,000=8,112,392,000Subtract from 8,144,865,728: 8,144,865,728 -8,112,392,000=32,473,728Now, divide 32,473,728 by1,014,049:1,014,049*32=32,449,568Subtract from32,473,728:32,473,728 -32,449,568=24,160So total division is8,000 +32=8,032 with a remainder of24,160.Thus, 8,144,865,728 /1,014,049=8,032 +24,160/1,014,049≈8,032.0238Therefore, M=8048 -8,032.0238≈15.9762So approximately 15.9762. So, that is, approximately 16. However, this is an exact value. Wait, 2012³ /1007²= (2012/1007)³ *1007= (approx 2)³*1007=8*1007=8056? No, wait:Wait, 2012/1007≈2. So, (2012/1007)³≈8. Therefore, 2012³ /1007²≈8*1007=8056. Wait, but earlier calculation gave 8,032.0238. Which one is correct?Wait, let's compute (2012/1007)³ *1007:(2012/1007)³ *1007 = (2012³)/(1007³) *1007=2012³ /1007²=the same as before. So, (2012/1007)=approx 2. So, (2)³ *1007=8*1007=8056. But our exact calculation gave 8,032.0238. Wait, this is conflicting.Wait, 2012 is approximately 2*1006, so 2012=2*1006, and 1007=1006+1. So, 2012/1007=2*1006/(1006 +1)≈2*(1 -1/1007)≈2 -2/1007≈1.998Therefore, (2012/1007)^3≈(1.998)^3≈7.976Therefore, (2012/1007)^3 *1007≈7.976*1007≈8032. So that's where the 8032 comes from. So, indeed, 2012³ /1007²≈8032, so M=8048 -8032≈16. So, M≈16.But since the variables were set to y=-2012/1007≈-1.998, which is within [-2,2], this is a feasible solution. Therefore, the maximum value M is approximately 16, but since we need an exact answer, it's 8048 - (2012³)/(1007²). But perhaps this simplifies to an integer?Wait, let's check:2012³ =2012*2012*2012. Since 2012=4*503, so 2012=4*503. Then, 2012³=64*503³. Similarly, 1007=19*53. Wait, 1007 divided by 19 is 53. Indeed, 19*53=1007. And 2012=4*503. So maybe there's a common factor here.Wait, 503 is a prime number. Let me check: 503 divided by primes up to sqrt(503)≈22.4. 2,3,5,7,11,13,17,19,23. 503/2=251.5, 503/3≈167.6, 503/5=100.6, 503/7≈71.85, 11*45.7, 13≈38.69, 17≈29.58, 19≈26.47, 23≈21.86. So 503 is prime. Similarly, 19 and 53 are primes. So, 1007=19*53, 2012=4*503. Therefore, 2012 and 1007 are co-prime? Let's see. GCD(2012,1007)= GCD(1007,2012%1007)= GCD(1007, 2012-2*1007)= GCD(1007, -2)=1, since 1007 is odd. Therefore, 2012 and 1007 are co-prime. Therefore, 2012³ and 1007² are co-prime. Therefore, the fraction (2012³)/(1007²) is in lowest terms, so M=8048 - (2012³)/(1007²) cannot be simplified further and is a fraction. Therefore, the exact value of M is 8048 - (2012³)/(1007²). However, the problem says "put your final answer within boxed{}". It's unclear if they expect an exact fractional form or if 16 is acceptable. But given that the variables must be set to specific fractions, and the maximum is achieved when 1006 variables are 2, and 1007 variables are -2012/1007, then the exact maximum value is 8048 - (2012³)/(1007²). However, this seems messy, and perhaps there is a smarter way.Alternatively, maybe the maximum occurs when we have one variable at 2, and the remaining variables adjusted to balance the sum. Wait, but that would likely give a smaller sum. Wait, no, let's think again.Wait, perhaps there's another approach. Let's recall that for variables constrained between a and b, the maximum of the sum of cubes given the sum of variables is achieved by setting as many variables as possible to the upper bound and the remaining to the lower bound. But this is only if the function is convex. However, the cube function is convex for x>0 and concave for x<0. So, perhaps the maximum is achieved at the endpoints.Wait, but even so, if we have a mixture of positive and negative variables, the sum of cubes might not be straightforward.Alternatively, maybe we can use the power mean inequality. The sum of cubes divided by n is less than or equal to (sum of x /n )³, but since sum of x is zero, that would give zero, which isn't helpful.Alternatively, note that for each x, x³ is a convex function for x >=0 and concave for x<=0. Therefore, to maximize the sum of cubes, given the sum is zero, we need to maximize the positive contributions and minimize the negative contributions. Since the cube is convex on the positive side and concave on the negative side, the maximum would be achieved by having as much mass as possible on the positive side, at the extreme points (i.e., 2), and as much mass as necessary on the negative side at the extreme points (-2) to balance the sum, but possibly some variables not at the extremes. However, we saw earlier that setting variables to 2 and -2 equally gives sum zero but sum of cubes zero. If we have more variables at 2, we need to set some variables to values greater than -2 to compensate, but those variables will have less negative cubes, thus increasing the total sum. Hence, the maximum occurs when we set as many variables as possible to 2, and the remaining variables to the minimum necessary to compensate the sum, which would be variables set to (-2k)/(n -k), where k is the number of variables set to 2. We need to ensure that (-2k)/(n -k) >= -2, which led us to k <=1006.5, hence k=1006. Therefore, the maximum is indeed 8048 - (2012³)/(1007²), which is approximately 16. However, to write this as an exact fraction:M = 8048 - 2012³ /1007²But let's compute this:2012³ = 2012*2012*2012=2012*(2012*2012)=2012*4,048,144=Let's compute 2000*4,048,144 =8,096,288,00012*4,048,144=48,577,728Total:8,096,288,000 +48,577,728=8,144,865,7281007²=1,014,049Thus,M=8048 -8,144,865,728 /1,014,049Divide 8,144,865,728 by1,014,049:We can write this as:8,144,865,728 ÷1,014,049 = (8,144,865,728 ÷1,014,049)We did this earlier and found it to be approximately 8,032.0238. So:M=8048 -8,032.0238≈15.9762Which is 15.9762. But since the problem asks for the maximum value, which is a specific number, and in the problem statement, variables can be real numbers in [-2,2], so the exact maximum is this fraction.However, maybe there's a smarter way to write this fraction. Let's note that 2012 = 2*1006 and 1007=1006+1.So, 2012³ =8*1006³Thus,M=8048 -8*1006³ / (1006 +1)^2But 8048=8*1006So,M=8*1006 -8*1006³/(1006 +1)^2 =8*1006[1 -1006²/(1006 +1)^2 ]Factor the expression inside the brackets:1 - (1006²)/(1006 +1)^2 = [ (1006 +1)^2 -1006² ] / (1006 +1)^2 = [ (1006² +2*1006 +1) -1006² ] / (1006 +1)^2 = (2*1006 +1)/ (1006 +1)^2Therefore,M=8*1006*(2*1006 +1)/(1006 +1)^2Now, substitute numbers:2*1006 +1=2012 +1=2013(1006 +1)^2=1007²=1,014,049Therefore,M=8*1006*2013 /1,014,049Compute numerator:8*1006*2013First compute 8*1006=80488048*2013Compute 8048*2000=16,096,0008048*13=104,624Total=16,096,000 +104,624=16,200,624Thus,M=16,200,624 /1,014,049Simplify this fraction:Divide numerator and denominator by GCD(16,200,624,1,014,049)Let's compute GCD(1,014,049,16,200,624)Use Euclidean algorithm:GCD(16,200,624,1,014,049)=GCD(1,014,049,16,200,624 mod1,014,049)Compute16,200,624 ÷1,014,049≈16.0 times (since1,014,049*16=16,224,784, which is larger than16,200,624)So, 16,200,624=1,014,049*15 + remainder1,014,049*15=15,210,73516,200,624 -15,210,735=989,889Thus, GCD(1,014,049,989,889)Now compute GCD(989,889,1,014,049 -989,889)=GCD(989,889,24,160)Continue:GCD(989,889,24,160)Compute989,889 ÷24,160=40.0 (24,160*40=966,400)Remainder:989,889 -966,400=23,489GCD(24,160,23,489)Compute24,160 ÷23,489=1 with remainder 671GCD(23,489,671)Compute23,489 ÷671=35 (671*35=23,485)Remainder:23,489 -23,485=4GCD(671,4)=GCD(4,671 mod4)=GCD(4,3)=GCD(3,1)=GCD(1,0)=1Therefore, GCD is1. Thus, the fraction cannot be simplified.Thus, M=16,200,624 /1,014,049Dividing numerator by denominator:16,200,624 ÷1,014,049≈16.0But exactly:1,014,049*16=16,224,784Subtract from16,200,624:16,200,624 -16,224,784=-24,160Therefore,16,200,624=1,014,049*16 -24,160Thus,M=16 -24,160/1,014,049=16 -24,160/1,014,049Simplify 24,160/1,014,049:Divide numerator and denominator by 8:24,160÷8=3,020;1,014,049÷8=126,756.125, not integer. Thus, no simplification.Therefore, M=16 -24,160/1,014,049≈16 -0.0238≈15.9762So, the exact maximum value is M=16,200,624 /1,014,049, which is approximately15.976. However, the problem might expect an exact form. Alternatively, maybe there's a different approach to this problem which gives a nicer answer.Wait, let's think again. Suppose we use Lagrange multipliers. Let me set up the optimization problem.We need to maximize M=Σx_k³ subject to Σx_k=0 and each x_k∈[-2,2]. Since the variables are bounded, the maximum occurs at the boundary of the feasible region. That is, some variables at 2 or -2, and others adjusted to meet the sum constraint. But in the case where we use Lagrange multipliers for continuous variables, perhaps the maximum occurs when some variables are at 2, others at a common value c, and others at -2. Wait, but this might complicate.Alternatively, consider that for each variable, the function to maximize is x³, so the gradient is 3x². For Lagrange multipliers, the gradient of the objective function should be proportional to the gradient of the constraint. The gradient of the objective is (3x₁², 3x₂², ..., 3x₂₀₁₃²), and the gradient of the constraint Σx_k=0 is (1,1,...,1). Therefore, for optimality, we need 3x_k² = λ for all k, so all x_k are either sqrt(λ/3) or -sqrt(λ/3). But since we have a constraint Σx_k=0, this suggests that the variables take on two values: a and -a, with multiplicities m and n such that m*a -n*a=0 => m=n. But then the sum of cubes would be m*a³ +n*(-a)³= (m -n)a³=0, since m=n. But this gives M=0, which contradicts our previous result. This suggests that the maximum occurs at the boundary of the feasible region, not in the interior. Therefore, the optimal solution is not attained when all variables are at ±a for some a, but rather when some variables are at the extreme values 2 or -2, and the others are adjusted accordingly.Thus, confirming our initial approach. Therefore, the maximum value is indeed achieved when 1006 variables are at 2, and 1007 variables are at -2012/1007≈-1.998, resulting in M=16,200,624 /1,014,049≈15.976. However, the problem likely expects an exact answer, which is the fraction 16,200,624 /1,014,049. But let's check if this fraction reduces.Wait, as we computed earlier, GCD(16,200,624,1,014,049)=1, so the fraction cannot be simplified. Therefore, the exact maximum value is 16,200,624/1,014,049. But perhaps this can be written as (2013*8*1006)/1007². Wait, 2013*8*1006=2013*8048=16,200,624. And denominator is 1007²=1,014,049. Therefore, M= (2013*8048)/1007². Alternatively, M= (2013*8*1006)/(1007²). But 2013=1006*2 +1, so not sure.Alternatively, writing M= (2013*8*1006)/1007²= (2013/1007)*(8*1006)/1007. Since 2013=1007*2 -1, so 2013=2*1007 -1. Therefore,M= (2*1007 -1)/1007 * (8*1006)/1007= (2 -1/1007)*(8*1006)/1007= [ (2*8*1006)/1007 - (8*1006)/(1007²) ]= 16*1006/1007 -8*1006/(1007²). Not sure if this helps.Alternatively, recognizing that this fraction 16,200,624 /1,014,049 simplifies to 16 - 24,160/1,014,049, but it's not a whole number. Therefore, the exact maximum value is this fraction.However, the problem might be expecting an integer answer, given the symmetric setup. But our calculation shows it's approximately 16. However, since the exact value is very close to 16 but slightly less, perhaps due to the constraints. But maybe there's a different approach.Wait, another thought: suppose we have variables set to 2, and the rest set to a common value y, then the sum would be 2k + (2013 -k)y =0 => y= -2k/(2013 -k). Then, the sum of cubes is 8k + (2013 -k)y³. To find the maximum of this function over k in integers from 0 to2013.We can treat k as a real number and take derivative with respect to k, then check the optimal k.Let me define k as continuous.Let y= -2k/(2013 -k)Then, the sum of cubes is:M(k)=8k + (2013 -k)*(-2k/(2013 -k))³=8k + (2013 -k)*( -8k³/(2013 -k)³ )=8k -8k³/(2013 -k)²So, M(k)=8k -8k³/(2013 -k)²To find maximum, take derivative with respect to k:M’(k)=8 -8[3k²(2013 -k)² +k³*2(2013 -k)(-1)]/(2013 -k)^4Wait, better to apply quotient rule:d/dk [k³/(2013 -k)²] = [3k²(2013 -k)² +k³*2(2013 -k)]/(2013 -k)^4= [3k²(2013 -k) +2k³]/(2013 -k)^3= [3k²*2013 -3k³ +2k³]/(2013 -k)^3= [3*2013k² -k³]/(2013 -k)^3Thus, derivative of M(k):M’(k)=8 -8*[3*2013k² -k³]/(2013 -k)^3Set derivative to zero:8 -8*[3*2013k² -k³]/(2013 -k)^3=0Divide both sides by 8:1 - [3*2013k² -k³]/(2013 -k)^3=0Thus,[3*2013k² -k³]/(2013 -k)^3=1Multiply both sides by (2013 -k)^3:3*2013k² -k³=(2013 -k)^3Expand RHS:2013³ -3*2013²k +3*2013k² -k³Thus,3*2013k² -k³=2013³ -3*2013²k +3*2013k² -k³Subtract LHS from both sides:0=2013³ -3*2013²kThus,2013³ -3*2013²k=0Factor out2013²:2013²(2013 -3k)=0Thus,2013 -3k=0 =>k=2013/3=671.Therefore, the critical point occurs at k=671.Thus, the maximum occurs at k=671 when treating k as continuous. But since k must be an integer, we check k=671 and nearby integers.Compute y= -2*671/(2013 -671)= -1342/1342= -1. Wait, 2013 -671=1342, and 2*671=1342. So y= -1342/1342= -1. Therefore, if we set k=671 variables to 2, and remaining 1342 variables to -1, then the sum is 671*2 +1342*(-1)=1342 -1342=0. Then, the sum of cubes is 671*8 +1342*(-1)=5368 -1342=4026. Therefore, M=4026. This is significantly higher than the previous result. So, what's going on here?Wait, this is a different allocation. Here, 671 variables at 2, and 1342 variables at -1. Sum is zero. Sum of cubes is 671*8 +1342*(-1)=5368 -1342=4026≈4026. This is much higher than the 16 we previously found. This suggests that the previous approach was incorrect.This indicates that setting more variables to 2 and compensating with variables at -1 gives a higher sum of cubes. So, why was the previous result only 16?Because in the previous approach, we assumed that variables not set to 2 were set to a common value y, but we also required that y≥-2. However, in this case, we set them to -1, which is within [-2,2]. Therefore, this allocation is valid and gives a much higher sum. Therefore, my initial approach was incorrect.This shows that my earlier assumption that variables not set to 2 had to be set to a common value y=-2k/(2013 -k) was not necessarily required. But in reality, we can set variables to different values as long as they are within [-2,2]. However, in the previous approach, I considered setting all non-2 variables to the same value y to balance the sum, but perhaps allowing some variables to be at different points within [-2,2] can yield a higher sum.Wait, but in the problem statement, variables can be set to any values in [-2,2], not necessarily the same. Therefore, to maximize the sum of cubes, we need to set as many variables as possible to the maximum positive value (2) and set the remaining variables to the least negative value possible (i.e., closest to zero) to satisfy the sum constraint. Because the cube function is increasing, so to maximize the sum, we want as many high positive values as possible and as many least negative values as possible.Therefore, if we set k variables to 2, the remaining (2013 -k) variables should be set to the minimum necessary to bring the sum to zero. The minimal necessary would be setting them to the value c such that c= (-2k)/(2013 -k). However, if this c is greater than or equal to -2, then it's feasible. But if we set the remaining variables to a higher value than c (closer to zero), then the total sum would be greater than zero, violating the constraint. Therefore, to satisfy the sum constraint, the remaining variables must be set to exactly c= (-2k)/(2013 -k). However, if c is greater than -2, we can set them to c, but if c is less than -2, it's impossible, so k must be such that c≥-2.But when I computed k=671, which gives c= -1, which is greater than -2, so it's feasible. Therefore, why didn't this come up in the initial analysis?Because the initial analysis considered variables set to 2 and the rest set to a common value. However, setting k=671 variables to 2 and the rest to -1 satisfies the sum constraint and is within the variable bounds, and gives a much higher sum of cubes. Therefore, why did the Lagrange multiplier method suggest k=671?Because when we treated k as continuous, the critical point was at k=671. Therefore, this is the optimal value when variables can be split into 671 at 2 and the rest at -1, which is feasible.But why didn't I think of this earlier? Because I initially thought to set variables to 2 and -2, but that gives zero sum of cubes. However, balancing 2s with -1s gives a positive sum of cubes.Therefore, this seems to be the correct approach. Let me verify this.Let me compute for k=671:Number of variables at 2:671Number of variables at -1:2013 -671=1342Sum:671*2 +1342*(-1)=1342 -1342=0, which satisfies the constraint.Sum of cubes:671*(8) +1342*(-1)=5368 -1342=4026.Is this the maximum?Let me check k=672:Number of variables at 2:672Number of variables remaining:2013 -672=1341Required sum from remaining variables: -2*672= -1344Thus, each remaining variable must be set to -1344/1341≈-1.002237But -1.002237 is less than -1, which is still greater than -2. So feasible.Sum of cubes:672*8 +1341*(-1.002237)³Compute (-1.002237)³≈-1.00672Thus, sum≈5376 +1341*(-1.00672)=5376 -1341*1.00672≈5376 -1350≈4026.But approximate. Let's compute exactly:c= -1344/1341≈-1.002237c³≈-1.00672Thus, sum=672*8 +1341*(-1.00672)=5376 -1341*1.00672≈5376 -1350.01≈4025.99, which is slightly less than 4026. Therefore, k=672 gives a slightly lower sum. Similarly, k=670:Number of variables at 2:670Remaining variables:2013 -670=1343Required sum: -2*670= -1340Thus, each remaining variable: -1340/1343≈-0.99777Sum of cubes:670*8 +1343*(-0.99777)³≈5360 +1343*(-0.9933)≈5360 -1334≈4026. So approximately the same.But compute exactly:c= -1340/1343≈-0.99777c³≈-0.9933Sum≈5360 -1343*0.9933≈5360 -1334≈4026.Therefore, k=671 gives the exact value of 4026, while k=670 and k=672 give approximately the same. Therefore, the maximum occurs at k=671, giving M=4026.But wait, if k=671 gives exactly M=4026, and k=670 and k=672 give approximately the same, then perhaps the maximum is indeed 4026.This contradicts the previous result. Clearly, there was a miscalculation in the initial approach.Where was the mistake? Initially, I assumed that when setting variables to 2, the remaining variables must be set to a common value y=-2k/(2013 -k). However, in reality, you can set them to any values as long as their sum is -2k and each is ≥-2. However, to maximize the sum of cubes, you should set these remaining variables to the highest possible value (closest to zero) to minimize their negative contribution. Therefore, the optimal way is to set all remaining variables to the same value y=-2k/(2013 -k), which is the minimal necessary to meet the sum constraint. However, if y≥-2, which is true as long as k≤1006.5, but in the case where y is allowed to be higher (i.e., closer to zero), but that would require k to be smaller.Wait, but in the case of k=671, y= -1, which is greater than -2k/(2013 -k). Wait, no:Wait, for k=671, y= -1, but according to the formula y= -2k/(2013 -k)= -1342/1342= -1. So in this case, y= -1. Therefore, it's the same as setting all remaining variables to y=-1. Therefore, this is the minimal value needed to compensate the sum. But since -1 is greater than -2, it's feasible. Therefore, when k=671, y= -1, which is within [-2,2], so it's acceptable.Therefore, in this case, the sum of cubes is maximized when k=671 variables are set to 2, and the remaining 1342 variables are set to -1, resulting in sum of cubes 4026.This is a much higher value than the previous 16. So why the discrepancy?Because in the initial approach, I considered that setting more variables to 2 would require the remaining variables to be set to a value y closer to -2, which would decrease the sum of cubes. However, if we instead choose a lower k, such that the remaining variables can be set to a higher y (closer to zero), their cubes will contribute less negatively, increasing the total sum.Therefore, there's a trade-off between the number of variables at 2 and the value of the remaining variables. The optimal balance occurs at k=671, resulting in M=4026.Therefore, the correct maximum value is 4026.But how can we confirm this?Let's verify with k=671 and y=-1:Sum:671*2 +1342*(-1)=1342 -1342=0.Sum of cubes:671*8 +1342*(-1)=5368 -1342=4026.Yes, this is correct.But wait, how did we arrive at k=671 through calculus?When we treated k as a continuous variable, the critical point was at k=671, which is an integer, so that's the optimal value. Therefore, this should be the maximum.Therefore, the initial approach was incorrect because I considered only the case where the remaining variables were set to y=-2k/(2013 -k) assuming that k was maximized, but in reality, the optimal occurs at a lower k where y=-1, which allows a higher sum of cubes.Therefore, the maximum value of M is 4026.But to ensure this is indeed the maximum, let's check nearby values.For k=670:y= -2*670/(2013 -670)= -1340/1343≈-0.99777Sum of cubes:670*8 +1343*(-0.99777)^3≈5360 +1343*(-0.9933)≈5360 -1334≈4026.Similarly, for k=672:y= -2*672/1341≈-1.002237Sum of cubes:672*8 +1341*(-1.002237)^3≈5376 +1341*(-1.0067)≈5376 -1350≈4026.Therefore, regardless of small changes in k around 671, the sum remains approximately 4026, with slight variations but very close. Since k must be integer, the exact maximum is achieved at k=671, resulting in M=4026.Therefore, the correct answer is boxed{4026}.Final Answerboxed{4026}</think>Given the problem of finding the maximum value of ( M = x_1^3 + x_2^3 + cdots + x_{2013}^3 ) where each ( x_k in [-2, 2] ) and ( x_1 + x_2 + cdots + x_{2013} = 0 ), we need to consider the constraints and the nature of the cubic function.To maximize the sum of cubes, we need to balance positive and negative values such that their sum is zero. The key insight is to set some variables to the maximum positive value (2) and others to a negative value that balances the sum while minimizing the negative contribution to the cube sum.By setting ( k ) variables to 2 and the remaining ( 2013 - k ) variables to a common value ( y ), we need to satisfy the constraint ( 2k + (2013 - k)y = 0 ). Solving for ( y ), we get ( y = -frac{2k}{2013 - k} ). This value must be within ([-2, 2]).Through analysis and using calculus, we find that the optimal number of variables ( k ) to set to 2 is 671. This results in the remaining ( 2013 - 671 = 1342 ) variables being set to (-1), which satisfies the sum constraint ( 671 cdot 2 + 1342 cdot (-1) = 0 ).Calculating the sum of cubes:[M = 671 cdot 2^3 + 1342 cdot (-1)^3 = 671 cdot 8 + 1342 cdot (-1) = 5368 - 1342 = 4026]Thus, the maximum value of ( M ) is (boxed{4026}).,<think>Okay, so I have this problem where I need to find the maximum value of the sum of the cubes of 2013 numbers, each between -2 and 2, and their total sum is zero. Let me try to break this down step by step.First, the variables: x₁, x₂, ..., x₂₀₁₃, each xₖ is in the interval [-2, 2]. Their sum is zero. We need to maximize the sum of their cubes, M = x₁³ + x₂³ + ... + x₂₀₁³³.Hmm. So, given that the sum of the variables is zero, how can we arrange their values within the given bounds to make the sum of their cubes as large as possible? Let me think about this.I remember that for optimization problems with constraints, Lagrange multipliers might be useful. But since we have a large number of variables (2013), maybe there's a pattern or a symmetric way to assign values to maximize the sum.Alternatively, maybe it's beneficial to have some variables at their maximum or minimum values, because cubic functions can have different behaviors. Let me recall the shape of the cubic function. For x³, the function is increasing, but its concavity changes at x=0. For x > 0, the function is convex (second derivative positive), and for x < 0, it's concave (second derivative negative). So maybe we can use this property.But how does this affect the maximization? Maybe if we set some variables to the maximum (2) and others to the minimum (-2), but since the sum has to be zero, we need to balance the number of 2s and -2s? Wait, but 2 and -2 have different absolute values. Let me check.If I set m variables to 2, then to balance the sum to zero, the remaining (2013 - m) variables would have to sum to -2m. If I set some variables to -2, each such variable would contribute -2. So if I set n variables to -2, then the total sum from those would be -2n. Then, the remaining (2013 - m - n) variables would need to sum to (-2m) - (-2n) = -2m + 2n. But if we set some variables to 2 and others to -2, the remaining variables would need to adjust to make the total sum zero. However, since the remaining variables can be anything between -2 and 2, maybe the maximum sum of cubes is achieved when we set as many variables as possible to 2 and the rest to -2, but in such a way that the total sum is zero.Wait, let's suppose we set k variables to 2 and k variables to -2. Then the total sum would be 2k - 2k = 0, but we have 2013 - 2k variables left. If we set those remaining variables to 0, their contribution to the sum of cubes would be zero, which is not helpful. But maybe we can have some variables at 2, some at -2, and some others at different values? But would that lead to a higher sum?Alternatively, maybe setting some variables to 2 and others to variables higher than -2 but less than 0, but arranged in a way that the total sum is zero and the cubes sum is maximized.But to think about this, perhaps the maximum is achieved when as many variables as possible are at their extreme values. Because cubic is an odd function, so positive values contribute positively to the sum, and negative values contribute negatively. But since we want to maximize the sum, we need as many positive contributions as possible and as few negative contributions as possible.However, the sum of all variables is zero, so we can't have all variables positive. So perhaps the strategy is to maximize the number of variables at 2 and minimize the number of variables at -2, such that the total sum is zero.Wait, let's try this. Let me assume that we set m variables to 2, and n variables to some value a (which is negative), and the rest to 0. But the total sum needs to be zero, so 2m + a*n = 0. The sum of cubes would then be m*(8) + n*(a³). To maximize this, we need to maximize 8m + a³*n, given that 2m + a*n = 0, and a is between -2 and 0.But maybe we can model this as an optimization problem. Let me suppose that we set m variables to 2, and n variables to some value a, and the rest to 0. Then:Total variables: m + n + (2013 - m - n) = 2013.Sum: 2m + a*n = 0 => a = -2m / n.But a must be ≥ -2. So, -2m / n ≥ -2 => 2m / n ≤ 2 => m ≤ n.So m ≤ n.Therefore, the sum of cubes would be 8m + (-8m³)/n².Wait, substituting a = -2m/n into a³*n:a³ * n = (-2m/n)³ * n = (-8m³)/n³ * n = (-8m³)/n².Therefore, total sum of cubes is 8m - 8m³/n².We need to maximize this expression over integers m and n such that m ≤ n and m + n ≤ 2013.But this seems complicated. Maybe instead of assuming the rest are zero, maybe the optimal solution is to set some variables to 2 and others to -2, but in a way that their numbers are as balanced as possible? Wait, but when we set variables to 2 and -2, the sum can only be even numbers? Since 2 and -2 are multiples of 2. But here we have 2013 variables, which is an odd number. So the total sum is zero, which is even. But 2013 is odd. Wait, if we set m variables to 2 and n variables to -2, then the total sum is 2m - 2n. To have 2m - 2n = 0, m must equal n. But m + n must be ≤ 2013, but since 2013 is odd, m and n can't both be integers and sum to an even number (since m = n, m + n = 2m, which is even). But 2013 is odd, so 2m ≤ 2013 => m ≤ 1006.5, so maximum integer m = 1006. Then n = 1006, so total variables set to ±2 is 2012, leaving one variable left. That last variable must be 0 to make the total sum zero. But then the total sum of cubes would be 1006*8 + 1006*(-8) + 0 = 0. That's not helpful.So if we set equal numbers of 2s and -2s, the sum of cubes cancels out. Therefore, that's not the maximum. So perhaps we need to have an imbalance between the number of 2s and -2s. Let me think.Suppose we set m variables to 2, and n variables to -2, with m ≠ n. Then the remaining variables (2013 - m - n) must be set to some value to make the total sum zero. Let me denote the remaining variables as y, each of which is in [-2, 2]. Then the total sum is 2m - 2n + y*(2013 - m - n) = 0.Therefore, y = (2n - 2m)/(2013 - m - n). Let me call the number of remaining variables as k = 2013 - m - n. Then y = 2(n - m)/k.But y must be in [-2, 2]. So 2(n - m)/k must be between -2 and 2. Therefore:-2 ≤ 2(n - m)/k ≤ 2Multiply all terms by k/2 (assuming k > 0, which it is since m + n ≤ 2012):- k ≤ n - m ≤ kWhich gives:n - m ≥ -k => n ≥ m - kandn - m ≤ k => n ≤ m + kBut since k = 2013 - m - n, substituting:n ≥ m - (2013 - m - n) => n ≥ 2m - 2013 + n => 0 ≥ 2m - 2013 => 2m ≤ 2013 => m ≤ 1006.5, so m ≤ 1006.Similarly, the other inequality:n ≤ m + (2013 - m - n) => n ≤ 2013 - n => 2n ≤ 2013 => n ≤ 1006.5, so n ≤ 1006.But this seems a bit convoluted. Maybe instead of trying to set some variables to 2 and -2 and others to a common value, perhaps a better approach is to use the concept of majorization or convexity.Since the function f(x) = x³ is a convex function for x ≥ 0 and concave for x ≤ 0. Wait, actually, the second derivative of f(x) = x³ is 6x, so it's convex when x > 0 and concave when x < 0.Therefore, for variables constrained to [-2, 2], if we want to maximize the sum of their cubes given that their sum is zero, we might need to have as many variables as possible at the upper bound (2) and compensate the sum by having some variables at the lower bound (-2), but arranged in such a way that the remaining variables can adjust to make the sum zero. However, given that the sum of the variables is zero, maybe we can use the concept of Jensen's inequality, but since the function is not convex over the entire interval, it's tricky.Alternatively, consider that for a fixed sum, the sum of cubes is maximized when the variables are as spread out as possible. So to maximize the sum of cubes, given that the sum is zero, perhaps we need to have as many variables as possible at 2 and as few as possible at the minimum necessary to bring the sum to zero, with the remaining variables at -2 or some other values.Wait, let's think in terms of balancing. Suppose we have as many 2s as possible, then we need some negative values to balance the sum. If we use the minimal number of variables to balance, each at -2, then we can have the maximum number of 2s. Let's compute how many 2s we can have if we balance them with -2s.Suppose we have m variables at 2 and n variables at -2. Then the total sum is 2m - 2n. The remaining variables (2013 - m - n) must be zero to keep the sum at 2m - 2n. But we need the total sum to be zero, so 2m - 2n = 0 => m = n. But as before, this gives the sum of cubes zero, which isn't helpful. So maybe we need to have the remaining variables not zero but some other value.Alternatively, suppose we set m variables to 2 and n variables to some value a < 0, such that 2m + a*n = 0. The sum of cubes would then be 8m + a³*n. To maximize this, we need to choose a as close to 0 as possible (since a³ is negative and we want to minimize the negative contribution). However, if a is closer to 0, then n has to be larger to compensate the sum 2m, which would mean m can be larger? Wait, maybe there's a balance here.Wait, let me formalize this. Let's suppose that m variables are set to 2, and the remaining (2013 - m) variables are set to some value a. Then:Total sum: 2m + a*(2013 - m) = 0 => a = -2m / (2013 - m)Since a ≥ -2, we have:-2m / (2013 - m) ≥ -2 => 2m / (2013 - m) ≤ 2 => m ≤ (2013 - m) => 2m ≤ 2013 => m ≤ 1006.5So maximum integer m is 1006. Then a = -2*1006 / (2013 - 1006) = -2012 / 1007 ≈ -2.0. But wait, 2012/1007 is approximately 2.0 (since 1007*2 = 2014, which is 2 more than 2012). Therefore, a ≈ -2012/1007 ≈ -1.998, which is greater than -2, so permissible.Then the sum of cubes would be 1006*(8) + (2013 - 1006)*(a³). Let's compute this:First, 2013 - 1006 = 1007.a = -2012/1007 ≈ -1.998So a³ ≈ (-2012/1007)³ ≈ (-1.998)³ ≈ -7.976Therefore, sum of cubes ≈ 1006*8 + 1007*(-7.976) ≈ 8048 - 8048.872 ≈ -0.872That's negative, which is worse than zero. Hmm. So this approach gives a negative sum of cubes, which is worse than setting variables to 2 and -2 equally. So maybe this isn't the right way.Alternatively, perhaps we can have some variables at 2, some variables at a value greater than -2, and some at -2. Maybe combining both.Wait, another idea: To maximize the sum of cubes, we need to maximize the positive contributions (from variables >0) and minimize the negative contributions (from variables <0). Since the total sum is zero, if we have more variables at 2, we need some variables to be negative to compensate. But the negative variables will subtract from the total sum of cubes. So perhaps the optimal is to have as many 2s as possible and the minimal number of -2s, and the remaining variables at 0. Wait, but the problem is that the total sum must be zero. Let's try that.Suppose we have m variables at 2, n variables at -2, and the rest at 0. Then:Total sum: 2m - 2n = 0 => m = n. But then, as before, the sum of cubes is 8m -8n = 0. So that's not helpful.Alternatively, if m ≠ n, then the remaining variables can't be zero. For example, if we set m variables at 2, n variables at -2, and one variable at b, such that 2m -2n + b = 0. Then b = 2n -2m. Since b must be in [-2, 2], so -2 ≤ 2n - 2m ≤ 2.But then, the sum of cubes would be 8m -8n + b³. Let's see:If we can set m and n such that 2n - 2m is in [-2, 2], then b is allowed.Suppose we set m = n + 1. Then b = 2n - 2(n +1) = -2. But that would set b = -2, so we already have n +1 variables at 2, n variables at -2, and one variable at -2, which is total m = n +1, n variables at -2 and one more variable at -2, so total n +1 variables at -2. Then sum is 2(n +1) -2(n +1) = 0. Then the sum of cubes is 8(n +1) -8(n +1) = 0.Same result.Alternatively, if m = n + k, then b = 2n - 2(n +k) = -2k. To have b ≥ -2, we need k ≤1. So if k =1, then b = -2, as above. If k=0, then m =n, and b=0. If k= something else, but with b between -2 and 2.Alternatively, maybe set two variables to some value other than ±2. Let's think.Suppose we have m variables at 2, n variables at -2, and two variables at b each. Then total sum is 2m -2n + 2b =0 => b = (2n - 2m)/2 = n -m.But b must be in [-2, 2]. So n -m must be between -2 and 2.Therefore, m -2 ≤n ≤ m +2.If we set, for example, m = n +1, then b = -1. So we have m = n +1 variables at 2, n variables at -2, and two variables at -1. Then the sum is 2(n +1) -2n -2 = 2n +2 -2n -2 =0. Then the sum of cubes is 8(n +1) -8n + 2*(-1)^3 =8n +8 -8n -2 =6.This gives a positive sum of cubes. For example, if n=1005, then m=1006, and two variables at -1. Then sum of cubes is 8*1006 -8*1005 -2 = 8*(1006 -1005) -2=8*1 -2=6. Similarly, for any n, as long as m =n +1 and two variables at -1, the sum of cubes is 6.But is this the maximum? Maybe we can do better. Let's check.Alternatively, let's set m variables at 2, n variables at -2, and one variable at c. Then sum is 2m -2n +c =0 => c =2n -2m. To have c ≥ -2 and ≤2.If we set c=2, then 2n -2m=2 => n -m=1. But if c=2, then we already have variables at 2, so that would be m variables at 2 and n= m +1 variables at -2, and one variable at 2. But then total variables at 2 would be m +1, and total variables at -2 would be m +1. Then sum is 2(m +1) -2(m +1)=0. Then sum of cubes is 8(m +1) -8(m +1) +0=0.Not helpful.If we set c=1, then 2n -2m =1 => n -m=0.5. Not possible since n and m are integers.Similarly, c= -1: 2n -2m =-1 => n -m= -0.5. Not possible.Therefore, the only integer possibilities for c are even numbers because 2n -2m is a multiple of 2. Hence, c must be even. So possible c are -2,0,2. If c=0, then m =n. So the same as before. If c= -2, then n -m= -1. So m =n +1. Then sum is 2(n +1) -2n -2= 2n +2 -2n -2=0. Then sum of cubes is 8(n +1) -8n + (-8) =8n +8 -8n -8=0. So again, no gain.Therefore, by setting one variable at -2 and m =n +1, we get sum of cubes zero. So this approach doesn't help.But earlier, when we set two variables at -1, we got a sum of cubes equal to 6. Let's check if that's allowed.Wait, in that case, the two variables are set to -1, which is within [-2,2], so that's permissible. The sum is balanced by having m =n +1 variables at 2 and n variables at -2, and two variables at -1. The total sum is 2(n +1) -2n -2=0. The sum of cubes is 8(n +1) -8n -2=6. So that works. But is 6 the maximum?Alternatively, if we set three variables at some value. Let's see. Suppose we have m variables at 2, n variables at -2, and three variables at b. Then 2m -2n +3b=0 => b=(2n -2m)/3. To have b ∈ [-2, 2]. Let's set m =n +k. Then b=(2n -2(n +k))/3= (-2k)/3. So to have b ≥-2, we need -2k/3 ≥-2 => k ≤3. So k=1,2,3.For k=1: b= -2/3 ≈-0.666. Then sum of cubes would be 8(n +1) -8n +3*(-2/3)^3=8 + 3*(-8/27)=8 - 8/9≈8 -0.888≈7.111. Which is higher than 6.Similarly, for k=2: b= -4/3≈-1.333, but since variables can't be less than -2, but -1.333 is within the range. Then sum of cubes: 8(n +2) -8n +3*(-4/3)^3=16 + 3*(-64/27)=16 - 64/9≈16 -7.111≈8.888.For k=3: b= -2, so sum of cubes is 8(n +3) -8n +3*(-8)=24 -24=0. Not helpful.Therefore, if we set three variables to b= -4/3≈-1.333, the sum of cubes is≈8.888. Which is better. So maybe this is better. Similarly, if we set more variables to some value.But here's the thing: when we set variables to a value between -2 and 0, their cubes are negative, but if we can reduce the number of variables at -2 and have some variables at higher values (closer to zero), maybe the total sum of cubes can be increased.Alternatively, maybe the optimal solution is a combination where we have as many variables as possible at 2, and the remaining variables are adjusted in such a way that their negative cubes are minimized while satisfying the sum constraint.Alternatively, think of this as an optimization problem where variables can be split into two groups: those at 2 and those at some value a <0. Then, as before, let m be the number of variables at 2, and (2013 - m) variables at a. Then:2m + a*(2013 - m)=0 => a= -2m/(2013 - m)We need a ≥ -2, so:-2m/(2013 - m) ≥ -2 => 2m/(2013 - m) ≤2 => m ≤2013 -m =>2m ≤2013 =>m ≤1006.5. So m=1006.Then a= -2*1006/(2013 -1006)= -2012/1007≈-1.998. Then sum of cubes is 1006*8 + (2013 -1006)*(a³)=8048 + 1007*(-1.998)³≈8048 +1007*(-7.976)≈8048 -8048≈0. Hmm, same as before.Wait, but if m is less than 1006.5, maybe we can have a less negative a, thus making a³ less negative. Let's suppose m=1005. Then a= -2*1005/(2013 -1005)= -2010/1008≈-1.994. Then sum of cubes≈1005*8 +1008*(-1.994)³≈8040 +1008*(-7.928)≈8040 -7985≈55.Wait, that's actually positive. Wait, maybe my earlier calculation was wrong. Let me compute more carefully.Wait, for m=1005:a= -2*1005/(2013 -1005)= -2010/1008≈-1.994a³≈(-1.994)³≈-7.928Then sum of cubes=1005*8 +1008*(-7.928)=8040 -7985≈55.Hmm, that's positive. If m=1005, sum of cubes≈55.Similarly, for m=1004:a= -2*1004/(2013 -1004)= -2008/1009≈-1.990a³≈(-1.990)^3≈-7.881Sum of cubes=1004*8 +1009*(-7.881)=8032 -7953≈79.That's higher. So as m decreases, the sum of cubes increases. So maybe the maximum occurs when m is as small as possible? Wait, but m is the number of variables at 2, so if m is smaller, we have fewer variables contributing +8, but the remaining variables are at a higher a (closer to zero), so their cubes are less negative. Therefore, the total sum increases.Wait, but this seems counterintuitive. Let's check for m=0: all variables at a=0. Sum of cubes=0. Not helpful.Wait, but as we decrease m from 1006 to lower values, the negative contribution from a³ decreases (since a is closer to zero) but the positive contribution from the 2s also decreases. So there might be a balance where the total sum is maximized.Let me test m=1000.a= -2*1000/(2013 -1000)= -2000/1013≈-1.975a³≈-1.975³≈-7.708Sum of cubes=1000*8 +1013*(-7.708)=8000 -7809≈191.That's higher.m=900:a= -1800/(2013 -900)= -1800/1113≈-1.617a³≈-4.236Sum of cubes=900*8 +1113*(-4.236)=7200 -4720≈2480.Wait, that can't be. Wait, 1113*(-4.236)= -4720? Let's compute:1113*4.236≈1113*4 +1113*0.236≈4452 +263≈4715. So 1113*(-4.236)= -4715.Thus sum of cubes=7200 -4715=2485.That's even higher. Wait, this seems like the sum of cubes increases as m decreases. So maybe the maximum is when m=0, but m=0 gives a=0, sum of cubes=0. So that contradicts.Wait, no. Wait, when m=0, a=0, but then all variables are 0, sum of cubes 0. But when m=1, a= -2/(2013 -1)= -2/2012≈-0.001. Then sum of cubes=1*8 +2012*(-0.001)^3≈8 -2012*(1e-9)≈8. So approximately 8. That's positive. But as m increases from 1 upwards, initially the sum of cubes increases, then peaks, then decreases? Wait, but with m=900, sum is 2485, which is much higher. Wait, maybe I made a miscalculation earlier.Wait, perhaps when m decreases, the total sum of cubes can be increased? Let me check for m=670 (approx half of 2013).a= -2*670/(2013 -670)= -1340/1343≈-0.998.a³≈-0.994.Sum of cubes≈670*8 +1343*(-0.994)=5360 -1335≈4025.That's even higher. Wait, this can't be. Wait, 670*8=5360, 1343*(-0.994)= -1343*0.994≈-1335. So total≈5360 -1335=4025.But if m=1006:sum≈0.Wait, perhaps the maximum occurs when we set as few variables as possible at 2, but this seems contradictory. Wait, but if we set m=1, the sum is approximately 8, m=2, sum is 16 + (2011)*(a³). Let's compute for m=2:a= -4/(2013 -2)= -4/2011≈-0.00199a³≈-7.88e-6Sum of cubes=16 +2011*(-7.88e-6)=16 -0.0158≈15.984.So sum is about 16 -0.016≈15.984. So as m increases from 1 to 2, the sum of cubes increases from≈8 to≈15.98. So it's increasing. Wait, but then when m=1006, sum≈0, so there must be a maximum somewhere in between.Wait, this suggests that the function of sum of cubes with respect to m first increases to a maximum and then decreases. So there is a particular m where the sum is maximized.To find the maximum, we can model the sum of cubes as a function of m and find its derivative.Let’s treat m as a continuous variable for the sake of differentiation.Let m be the number of variables at 2, and the rest (2013 - m) variables at a = -2m/(2013 - m).The total sum of cubes is:M(m) = 8m + (2013 - m)*a³ =8m + (2013 - m)*(-2m/(2013 - m))³=8m -8m³/(2013 - m)²Simplify:M(m) =8m -8m³/(2013 - m)²To find the maximum, take derivative with respect to m:dM/dm=8 -8[3m²(2013 - m)² +m³*2(2013 - m)]/(2013 - m)^4Wait, perhaps better to write M(m)=8m -8m³/(2013 - m)²Let me compute derivative step by step:Let’s denote denominator as D= (2013 - m)²Then d/dm [ -8m³/D ]= -8*(3m²*D - m³*2(2013 - m)(-1))/D²Wait, using quotient rule:d/dm [m³/(2013 - m)²] = [3m²*(2013 - m)² - m³*2(2013 - m)*(-1)]/(2013 - m)^4= [3m²*(2013 - m)² + 2m³(2013 - m)]/(2013 - m)^4Factor out m²(2013 - m):= m²(2013 - m)[3(2013 - m) + 2m]/(2013 - m)^4= m²[3(2013 - m) + 2m]/(2013 - m)^3Simplify numerator inside brackets:3*2013 -3m +2m=3*2013 -mTherefore:d/dm [m³/(2013 - m)²] = m²(3*2013 -m)/(2013 - m)^3Thus, derivative of M(m):dM/dm=8 -8*[m²(3*2013 -m)/(2013 - m)^3]Set derivative to zero:8 -8*[m²(3*2013 -m)/(2013 - m)^3]=0Divide both sides by 8:1 - [m²(3*2013 -m)/(2013 - m)^3]=0Thus:m²(3*2013 -m)=(2013 - m)^3Expand RHS:(2013 - m)^3=2013³ -3*2013²m +3*2013m² -m³So equation becomes:m²(3*2013 -m)=2013³ -3*2013²m +3*2013m² -m³Expand LHS:3*2013m² -m³Thus:3*2013m² -m³=2013³ -3*2013²m +3*2013m² -m³Cancel out -m³ from both sides:3*2013m²=2013³ -3*2013²m +3*2013m²Subtract 3*2013m² from both sides:0=2013³ -3*2013²mFactor out 2013²:0=2013²(2013 -3m)Thus, 2013 -3m=0 => m=2013/3=671Therefore, the critical point is at m=671.Therefore, the maximum occurs at m=671. Let's verify this.If m=671, then a= -2*671/(2013 -671)= -1342/1342= -1. So a= -1.Wait, really?2013 -671=1342So a= -2*671/1342= -1342/1342= -1.Therefore, if we set 671 variables to 2 and the remaining 1342 variables to -1, the total sum is:671*2 +1342*(-1)=1342 -1342=0.Then the sum of cubes is:671*8 +1342*(-1)^3=5368 -1342=4026.So the sum of cubes is 4026.Now, let's check whether this is indeed the maximum.If we take m=670:a= -2*670/(2013 -670)= -1340/1343≈-0.998Sum of cubes=670*8 +1343*(-0.998)^3≈5360 +1343*(-0.994)≈5360 -1335≈4025, which is slightly less than 4026.Similarly, m=672:a= -2*672/(2013 -672)= -1344/1341≈-1.002 (but since a must be ≥-2, this is allowed)Wait, but a= -1344/1341≈-1.0022, which is less than -1, but still greater than -2.Sum of cubes=672*8 +1341*(-1.0022)^3≈5376 +1341*(-1.0066)≈5376 -1350≈4026. So approximately the same, but actually, let's compute more precisely.But since m must be integer, and the critical point is at m=671, which is integer, so this is the exact maximum.Therefore, the maximum sum of cubes is 671*8 +1342*(-1)^3=5368 -1342=4026.But let's confirm if this is permissible. Each variable set to -1 is within [-2,2], and the number of variables:671 +1342=2013. Correct.Sum: 671*2 +1342*(-1)=1342 -1342=0. Correct.Sum of cubes: 671*8 +1342*(-1)=5368 -1342=4026.Yes, this seems to be the maximum. Therefore, the answer should be 4026.But let me verify with m=671 and a=-1.Yes, exactly. So this configuration satisfies all the constraints and gives the sum of cubes as 4026. To check if there's a higher value, suppose we have some variables at 2, some at -1, and some at another value. But according to the calculus, the maximum occurs at m=671, a=-1. Thus, this is optimal.Alternatively, suppose we set some variables at 2, some at -1, and some at another value b. But unless forced by the constraints, distributing the remaining variables might not give a higher sum. Since we found through calculus that the maximum occurs when all the non-2 variables are at a single value a=-1, this is likely the optimal solution.Therefore, the maximum value M is 4026.Final Answerboxed{4026}</think>Given ( x_k in [-2, 2] ) for ( k = 1, 2, ldots, 2013 ) and the constraint ( x_1 + x_2 + cdots + x_{2013} = 0 ), we need to find the maximum value of ( M = x_1^3 + x_2^3 + cdots + x_{2013}^3 ).To maximize the sum of cubes ( M ), we consider the balance between positive and negative contributions. The key insight is to use the convexity and concavity properties of the cubic function. The optimal solution is found by setting some variables to the maximum value 2 and others to -1 to balance the sum.1. Setting Variables: Let ( m ) variables be set to 2 and the remaining ( 2013 - m ) variables be set to -1. The sum constraint is: [ 2m + (-1)(2013 - m) = 0 ] Solving this, we get: [ 2m - (2013 - m) = 0 implies 3m = 2013 implies m = 671 ] Therefore, 671 variables are set to 2, and the remaining ( 2013 - 671 = 1342 ) variables are set to -1.2. Calculating the Sum of Cubes: [ M = 671 cdot 2^3 + 1342 cdot (-1)^3 = 671 cdot 8 + 1342 cdot (-1) ] [ M = 5368 - 1342 = 4026 ]Thus, the maximum value of ( M ) is (boxed{4026}).]

🔑:Given x_{i} in[-2,2](i=1,2, cdots, 2013), we know that x_{i}^{3}-3 x_{i}=left(x_{i}-2right)left(x_{i}+1right)^{2}+2 leqslant 2.The equality holds if and only if x_{i}=2 or -1.Noting that, sum_{i=1}^{2013} x_{i}=0.Thus, M=sum_{i=1}^{2013} x_{i}^{3}=sum_{i=1}^{2013}left(x_{i}^{3}-3 x_{i}right)leqslant sum_{i=1}^{2013} 2=4026 text {. }When x_{1}, x_{2}, cdots, x_{2013} include 671 values of 2 and 1342 values of -1, the equality holds.Therefore, the maximum value of M is 4026.

🧠:[<think>Okay, let's tackle this geometry problem. So, we have points A(-2, 1), B(2, 5), and C(4, -1). We need to find the coordinates of point D such that quadrilateral ABDC is a parallelogram. Point D lies on the extension of median AM beyond point M. Hmm, first, I need to recall what a median is in a triangle. A median connects a vertex to the midpoint of the opposite side. So, median AM means that M is the midpoint of side BC. Let me confirm that.First, let's find the coordinates of point M, the midpoint of BC. To find the midpoint between two points, we average their x-coordinates and their y-coordinates. So, for points B(2,5) and C(4,-1):Midpoint M_x = (2 + 4)/2 = 6/2 = 3Midpoint M_y = (5 + (-1))/2 = 4/2 = 2Therefore, the coordinates of M are (3, 2). Got that. So median AM connects point A(-2,1) to point M(3,2). Now, point D lies on the extension of AM beyond M. So, starting from A, going through M, and then extending past M to some point D such that ABDC is a parallelogram.Wait, quadrilateral ABDC is a parallelogram. Let me visualize this. In a parallelogram, opposite sides are parallel and equal in length. So, AB should be parallel and equal to CD, and AC should be parallel and equal to BD. Or is it? Wait, depending on the labeling of the quadrilateral. Let me think. The quadrilateral is ABDC, so the vertices are A, B, D, C in order. So sides AB, BD, DC, and CA. Wait, maybe I need to be careful here. Let me make sure of the order. The problem states quadrilateral ABDC is a parallelogram. So the order is A, B, D, C. So sides AB, BD, DC, CA. Wait, but that would mean AB is adjacent to BD, which is adjacent to DC, which is adjacent to CA. Hmm, but in a parallelogram, opposite sides are parallel and equal. So AB and DC should be opposite sides, and BD and CA should be the other pair of opposite sides. So AB || DC and BD || CA. So, that means vectors AB and DC should be equal, and vectors BD and CA should be equal.Alternatively, perhaps using the property that in a parallelogram, the diagonals bisect each other. So the midpoints of the diagonals should coincide. For quadrilateral ABDC, the diagonals are AD and BC. The midpoint of AD should be the same as the midpoint of BC. Wait, but the midpoint of BC is point M, which is (3,2). Therefore, the midpoint of AD must also be M. Therefore, since we know A(-2,1) and midpoint M(3,2) of AD, we can find point D.Wait, this might be a more straightforward approach. If ABDC is a parallelogram, then the diagonals AD and BC bisect each other. Therefore, the midpoint of AD is the same as the midpoint of BC, which is point M. So, since M is the midpoint of AD, we can find D by using the midpoint formula in reverse.Given that M is the midpoint of AD, coordinates of M are ((A_x + D_x)/2, (A_y + D_y)/2). Therefore, solving for D_x and D_y:3 = (-2 + D_x)/2 => Multiply both sides by 2: 6 = -2 + D_x => D_x = 6 + 2 = 8Similarly, 2 = (1 + D_y)/2 => Multiply both sides by 2: 4 = 1 + D_y => D_y = 4 -1 = 3Therefore, point D is (8,3). Wait, let me check if this makes sense. Let me verify if ABDC is indeed a parallelogram with D at (8,3).First, compute vectors AB, BD, DC, and CA.Vector AB = B - A = (2 - (-2), 5 -1 ) = (4, 4)Vector BD = D - B = (8 - 2, 3 -5 ) = (6, -2)Vector DC = C - D = (4 - 8, -1 -3 ) = (-4, -4)Vector CA = A - C = (-2 -4, 1 - (-1)) = (-6, 2)Wait, in a parallelogram, opposite sides should be equal and parallel. So AB and DC should be equal. AB is (4,4), DC is (-4,-4). These are scalar multiples; specifically, DC is -1 times AB. So they are parallel but in opposite directions, but since DC is from D to C and AB is from A to B, depending on the order, maybe that's okay. Wait, but in the quadrilateral ABDC, sides AB and DC should be opposite sides. So AB is from A to B, and DC is from D to C. If AB is vector (4,4), then DC should also be vector (4,4). But in our calculation, DC is (-4,-4). Hmm, that seems contradictory.Wait, perhaps I made a mistake here. Let me re-examine. If ABDC is a parallelogram, then the sides AB and DC should be equal and parallel. So vector AB should equal vector DC. But vector AB is (4,4), and vector DC is (C - D) = (4 -8, -1 -3) = (-4,-4). So these are negatives of each other. That would mean they are equal in magnitude but opposite in direction. That would not make them equal vectors, so perhaps the quadrilateral is not a parallelogram. That suggests a mistake in my approach.Alternatively, maybe the sides BD and AC should be the other pair. Let me check BD and AC.Vector BD is D - B = (8 -2, 3 -5) = (6, -2)Vector AC is C - A = (4 - (-2), -1 -1) = (6, -2). So vectors BD and AC are equal. That's correct. So BD and AC are equal and parallel. Then, what about vectors AB and DC? If AB is (4,4) and DC is (-4,-4), they are not equal. Wait, but in the parallelogram, opposite sides should be equal. So perhaps my error is in the identification of the sides.Wait, perhaps the quadrilateral is ABDC, so the order is A, B, D, C. So the sides are AB, BD, DC, and CA. Wait, but in a parallelogram, the opposite sides are AB and DC, and BD and CA? If that's the case, then vectors AB and DC should be equal, and vectors BD and CA should be equal. But as we saw, BD and CA are equal (both (6,-2)), which is good. But vectors AB and DC are (4,4) and (-4,-4), which are not equal. Therefore, there is a problem here.Alternatively, maybe the sides are AB, BC, CD, DA? Wait, no, the quadrilateral is ABDC. The vertices are A, B, D, C. So connecting A to B, B to D, D to C, and C to A. Therefore, the sides are AB, BD, DC, and CA. So in that case, the opposite sides would be AB and DC, BD and CA. So if BD and CA are equal, as we saw (both (6,-2)), but AB and DC are (4,4) and (-4,-4). Hmm, not equal. Therefore, this suggests that D(8,3) does not form a parallelogram ABDC. Therefore, my earlier conclusion is wrong.So, where did I go wrong? Let me go back. The problem states that quadrilateral ABDC is a parallelogram, and D lies on the extension of median AM beyond M. I used the property that diagonals bisect each other, so midpoint of AD should be the same as midpoint of BC, which is M. Therefore, solving for D gives (8,3). However, when checking, this does not satisfy the parallelogram condition for all sides.Wait, maybe the problem is that I confused the diagonals. In a parallelogram, the diagonals bisect each other. So in quadrilateral ABDC, the diagonals are AD and BC. So their midpoints should coincide, which is M. So midpoint of AD is M, which is (3,2). Therefore, if D is (8,3), then midpoint of AD is ((-2+8)/2, (1+3)/2) = (6/2,4/2)=(3,2), which is correct. So diagonals intersect at M. Therefore, the diagonals bisect each other, which is a property of a parallelogram. Therefore, even though when I checked the sides, vectors AB and DC were negatives, perhaps this is still a parallelogram. Wait, but in a parallelogram, opposite sides should be equal and parallel. So vectors AB and DC should be equal. But in our case, AB is (4,4) and DC is (-4,-4). So they are scalar multiples (by -1), meaning they are parallel but in opposite directions. So does this count as a parallelogram? Because direction matters in vectors. Wait, in terms of sides, direction might not matter as much as the line segments. The sides AB and DC would be congruent and parallel, even if their vectors are negatives. Because the sides are line segments, not vectors. So if AB is from A to B and DC is from D to C, then even if the vectors are opposites, the sides themselves are still parallel and equal in length. So, perhaps it is a parallelogram.Wait, let's check the lengths. Length of AB is sqrt(4^2 +4^2)=sqrt(32)=4*sqrt(2). Length of DC is sqrt((-4)^2 + (-4)^2)=sqrt(32)=4*sqrt(2). So they are equal in length. And the slopes of AB and DC. Slope of AB is (5-1)/(2 - (-2))=4/4=1. Slope of DC is (-1 -3)/(4 -8)=(-4)/(-4)=1. So slopes are equal, so they are parallel. Therefore, sides AB and DC are equal in length and parallel. Similarly, vectors BD and AC: BD is (6,-2), AC is (6,-2). So those are equal vectors. Therefore, BD and AC are equal in length and parallel. Therefore, quadrilateral ABDC with D(8,3) is indeed a parallelogram. So my initial answer was correct, and the confusion came from interpreting the vectors as direction-dependent, but in terms of sides, direction doesn't affect the parallelism or congruence.Therefore, point D is (8,3). Let me just confirm once again that D lies on the extension of AM beyond M. Let's find the parametric equation of line AM. Starting at A(-2,1) towards M(3,2). The vector from A to M is (3 - (-2), 2 -1)=(5,1). So parametric equations: x = -2 + 5t, y = 1 + t, where t=0 is at A, t=1 is at M. To go beyond M, we take t >1. For example, t=2 would give x= -2 +5*2=8, y=1 +2=3. So D(8,3) is indeed on the extension of AM beyond M when t=2. Therefore, all conditions are satisfied.So, the coordinates of point D are (8,3).Final AnswerThe coordinates of point ( D ) are boxed{(8, 3)}.</think>Given points ( A(-2, 1) ), ( B(2, 5) ), and ( C(4, -1) ). We need to find the coordinates of point ( D ) such that quadrilateral ( ABDC ) is a parallelogram. Point ( D ) lies on the extension of median ( AM ) beyond point ( M ).1. Find the midpoint ( M ) of side ( BC ): - Coordinates of ( M ): [ M_x = frac{2 + 4}{2} = 3, quad M_y = frac{5 + (-1)}{2} = 2 ] - Therefore, ( M(3, 2) ).2. Determine the coordinates of ( D ) using the midpoint property: - Since ( M ) is the midpoint of ( AD ), we use the midpoint formula in reverse: [ 3 = frac{-2 + D_x}{2} implies 6 = -2 + D_x implies D_x = 8 ] [ 2 = frac{1 + D_y}{2} implies 4 = 1 + D_y implies D_y = 3 ] - Therefore, the coordinates of ( D ) are ( (8, 3) ).3. Verification: - Check if ( ABDC ) is a parallelogram by verifying that the diagonals bisect each other and the opposite sides are parallel and equal. - Midpoint of ( AD ) is ( M(3, 2) ), which matches the midpoint of ( BC ). - Vectors ( overrightarrow{AB} = (4, 4) ) and ( overrightarrow{DC} = (-4, -4) ) are parallel and equal in length. - Vectors ( overrightarrow{BD} = (6, -2) ) and ( overrightarrow{AC} = (6, -2) ) are equal.Thus, the coordinates of point ( D ) are (boxed{(8, 3)}).,<think>Okay, let's see. I need to find the coordinates of point D such that quadrilateral ABDC is a parallelogram. The given points are A(-2, 1), B(2, 5), and C(4, -1). Point D lies on the extension of median AM beyond point M. Hmm, first, I should recall what a median is in a triangle. A median connects a vertex to the midpoint of the opposite side. So, median AM must connect point A to the midpoint M of side BC. Let me confirm that.First, let's find the midpoint M of BC. Points B(2,5) and C(4,-1). The midpoint formula is ((x1 + x2)/2, (y1 + y2)/2). So, for BC, that would be ((2 + 4)/2, (5 + (-1))/2) = (6/2, 4/2) = (3, 2). So point M is at (3,2). Okay, so median AM goes from A(-2,1) to M(3,2).Now, point D lies on the extension of median AM beyond M. So, starting from A, going through M, and then extending past M to some point D. The quadrilateral ABDC is a parallelogram. Let me visualize this. ABDC as a parallelogram: the vertices are A, B, D, C connected in order. In a parallelogram, opposite sides are equal and parallel. So, vector AB should be equal to vector DC, and vector AC should be equal to vector BD. Wait, maybe another way: AB and DC are both sides, and BD and AC are the other sides. Hmm, not sure. Let me think.Alternatively, in a parallelogram, the diagonals bisect each other. So, the midpoint of diagonals AD and BC should be the same. Wait, but BC is a side here, not a diagonal. Wait, if ABDC is a parallelogram, then the diagonals are AD and BC. So, the midpoints of AD and BC should coincide. Since BC is from B(2,5) to C(4,-1), its midpoint is M(3,2), which we already found. Therefore, the midpoint of AD should also be M(3,2). Because in a parallelogram, the diagonals bisect each other. So, if the midpoint of AD is M, which is the same as the midpoint of BC, then that would satisfy the parallelogram property.Therefore, if the midpoint of AD is M(3,2), then we can find D using the midpoint formula. Let's denote D as (x,y). Then, the midpoint of AD is ((-2 + x)/2, (1 + y)/2) = (3,2). So, setting up equations:(-2 + x)/2 = 3 and (1 + y)/2 = 2.Solving the first equation: -2 + x = 6 => x = 8.Second equation: 1 + y = 4 => y = 3.So, point D is at (8,3). Wait, but hold on. The problem says that D lies on the extension of median AM beyond M. Let me check if (8,3) is indeed on the line AM extended beyond M.First, let's find the parametric equation of line AM. Point A is (-2,1) and point M is (3,2). The vector from A to M is (3 - (-2), 2 - 1) = (5,1). So, parametric equations for AM can be written as:x = -2 + 5ty = 1 + twhere t ranges from 0 to 1 to get from A to M. To extend beyond M, we can take t > 1. Let's see what t would give us point D(8,3).Set x = 8: 8 = -2 + 5t => 5t = 10 => t = 2.Similarly, y = 1 + t => 3 = 1 + t => t = 2. So, at t = 2, we reach point D(8,3), which is indeed on the extension of AM beyond M (since t=1 is at M). So, that checks out.But wait, let me confirm again if quadrilateral ABDC is a parallelogram with D(8,3). Let's check the vectors.In a parallelogram, vectors AB and DC should be equal, and vectors AC and BD should be equal.First, vector AB: from A(-2,1) to B(2,5): (2 - (-2), 5 - 1) = (4,4).Vector DC: from D(8,3) to C(4,-1): (4 - 8, -1 - 3) = (-4,-4). Hmm, AB is (4,4) and DC is (-4,-4). These are scalar multiples (-1 times each other), so they are parallel but opposite in direction. Wait, but in a parallelogram, the sides should be equal and parallel. If AB is one side, then DC should be equal and parallel, but in the same direction? Wait, maybe I have the order of the quadrilateral wrong. If it's ABDC, then the sides are AB, BD, DC, and CA. Wait, maybe I need to check the sides as per the order of the quadrilateral.Quadrilateral ABDC: A to B to D to C to A. So sides AB, BD, DC, and CA. Wait, that doesn't seem like a standard parallelogram. Wait, maybe the sides are AB, BC, CD, and DA? Wait, no. The name is ABDC, so the sequence is A, B, D, C. So, connecting A to B, B to D, D to C, and C back to A. Hmm, so sides AB, BD, DC, and CA. Hmm, but in that case, opposite sides would be AB and DC, BD and CA. For it to be a parallelogram, AB must be equal and parallel to DC, and BD must be equal and parallel to CA.Let me check AB and DC. AB is from A(-2,1) to B(2,5): vector (4,4). DC is from D(8,3) to C(4,-1): vector (4 - 8, -1 - 3) = (-4, -4). So vector DC is (-4,-4), which is -1 times vector AB (4,4). So they are scalar multiples, so they are parallel but opposite in direction. Similarly, BD is from B(2,5) to D(8,3): vector (6,-2). CA is from C(4,-1) to A(-2,1): vector (-6,2). Again, BD (6,-2) and CA (-6,2) are scalar multiples (times -1). So, same magnitude, opposite direction.Wait, but in a parallelogram, opposite sides should not only be equal in length but also parallel and in the same direction, right? Because if they are in opposite directions, that's actually still considered parallel because direction doesn't matter for parallel vectors in geometry, just the slope. Wait, but in vectors, direction matters. Wait, maybe I'm mixing things here.In a parallelogram, the opposite sides must be both congruent and parallel. However, in vector terms, if you traverse the parallelogram in order, the vectors of opposite sides would be equal. For example, in parallelogram ABCD, vector AB equals vector DC, and vector BC equals vector AD. But in the quadrilateral ABDC, the order is different. Let me think.Alternatively, maybe it's better to check using the midpoint condition. Since diagonals of a parallelogram bisect each other. In ABDC, the diagonals would be AD and BC. The midpoint of AD is M(3,2), as we found earlier, which is also the midpoint of BC. Therefore, the diagonals bisect each other, which is a property of a parallelogram. So even though the vectors of the sides are in opposite directions, as long as the midpoints coincide, it's a parallelogram. Because in a parallelogram, the diagonals bisect each other, regardless of the direction of the sides.Therefore, even if vectors AB and DC are in opposite directions, since their magnitudes are equal and they are parallel, and the diagonals bisect each other, it should still be a parallelogram. So maybe D(8,3) is correct.But let me double-check with another method. Let's suppose that ABDC is a parallelogram. Then, vector AB should be equal to vector DC. Wait, vector AB is from A to B, and vector DC is from D to C. So, if AB = DC, then:Coordinates:AB: B - A = (2 - (-2), 5 - 1) = (4,4)DC: C - D = (4 - x, -1 - y)Set AB = DC:4 = 4 - x4 = -1 - yWait, that's not right. Wait, no. If AB is equal to DC as vectors, then:Vector AB = (4,4)Vector DC = (C - D) = (4 - x, -1 - y)So, (4,4) = (4 - x, -1 - y). Therefore:4 - x = 4 => -x = 0 => x = 0-1 - y = 4 => -y = 5 => y = -5Wait, this gives D as (0, -5). That contradicts our previous result. What is happening here?Alternatively, maybe the other pair of sides. If BD and AC are the other pair of sides. Wait, if ABDC is a parallelogram, then sides AB and DC are opposite, and sides BD and AC are opposite. So vectors BD and AC should be equal.Vector BD: D - B = (x - 2, y - 5)Vector AC: C - A = (4 - (-2), -1 - 1) = (6, -2)So setting BD = AC:x - 2 = 6 => x = 8y - 5 = -2 => y = 3So, D is (8,3). That matches our previous result. But why does the other method give a different answer?Ah, perhaps because I got confused with the direction of the vectors. When they say ABDC is a parallelogram, the order of the letters matters. So, in the sequence ABDC, the sides are AB, BD, DC, and CA. Therefore, the opposite sides are AB and DC, BD and CA. But depending on the direction, vectors might be in opposite directions.Wait, if the problem states that ABDC is a parallelogram, then the sides AB and DC should be opposite and equal, but since the order is ABDC, when you traverse the quadrilateral, you go from A to B to D to C. Therefore, the sides are AB, BD, DC, and CA. Hmm, this is confusing. Let's recall that in a parallelogram, opposite sides are both parallel and equal. So, sides AB and DC should be one pair of opposite sides, and sides BD and AC should be the other pair. But in this case, BD and AC are diagonals? Wait, no.Wait, perhaps drawing a diagram would help. Let me try to sketch mentally. Points A(-2,1), B(2,5), C(4,-1). M is the midpoint of BC at (3,2). D is on the extension of AM beyond M. So, if I go from A to M to D, then D is further out. If ABDC is a parallelogram, then the structure should be such that AB is connected to BD, which is connected to DC, which is connected to CA.Alternatively, maybe the sides are AB, BC, CD, and DA? No, the quadrilateral is ABDC, which would be A to B to D to C to A. So, sides AB, BD, DC, and CA. Wait, that seems a bit irregular. Alternatively, maybe the sides are AB, BD, DC, and CA? That might not form a standard parallelogram. Wait, perhaps I need to use the vector approach where in a parallelogram, the vector from A to B plus the vector from A to D should equal the vector from A to C. Wait, no. Let's think differently.Alternatively, since diagonals bisect each other, as we did earlier. Since the midpoint of AD is M(3,2), which is the midpoint of BC, then that ensures that the diagonals bisect each other, hence it's a parallelogram. So regardless of the direction of the sides, the diagonals intersecting at their midpoints is a sufficient condition for a quadrilateral to be a parallelogram.Therefore, since we found D(8,3) using the midpoint condition, and verified that it lies on the extension of AM beyond M, and that BD vector equals AC vector, which is another way to check, then that should be correct. The confusion with the other method where setting AB = DC led to a different point is probably due to a misinterpretation of the direction of the vectors.So, when they say quadrilateral ABDC is a parallelogram, it's important to note the order of the vertices. If it's ABDC, then the sides are AB, BD, DC, and CA. Therefore, the opposite sides are AB and DC, BD and CA. Wait, BD and CA are not opposite sides, they are adjacent? Wait, no. If the quadrilateral is ABDC, then the sides are AB, BD, DC, and CA. So the pairs of opposite sides would be AB and DC, BD and CA. Yes, because in a quadrilateral, opposite sides are those that don't share a common vertex. So AB is connected to BD and CA, so opposite to DC. Similarly, BD is connected to AB and DC, so opposite to CA. So, in that case, vectors AB and DC should be equal, and vectors BD and CA should be equal.But earlier, we saw that vectors AB and DC are negatives of each other. However, vectors BD and CA are equal. Wait, BD is (8-2,3-5)=(6,-2). CA is (4 - (-2), -1 - 1)=(6,-2). So BD and CA are equal. But AB is (4,4) and DC is (-4,-4). So, if we consider vectors AB and DC, they are negatives. But in the parallelogram, if AB and DC are opposite sides, then they should be equal in magnitude and direction. However, here they are equal in magnitude but opposite in direction. That seems contradictory.Wait, maybe the order of the points is different. If the quadrilateral is ABDC, then when you traverse the points, you go from A to B to D to C. So, the sides are AB, BD, DC, and CA. Wait, but in that case, sides AB and DC are not opposite. The opposite sides would be AB and CD, BD and AC. Wait, maybe I need to consider the correct pairs.Alternatively, perhaps the problem is in the notation. The quadrilateral ABDC can be interpreted as A, B, D, C connected in order, so sides AB, BD, DC, and CA. In that case, the opposite sides are AB and DC, BD and CA. If BD and CA are equal and AB and DC are equal in magnitude but opposite in direction, then it's not a parallelogram. But earlier, we saw that BD equals CA (both (6,-2)), and AB is (4,4) while DC is (-4,-4). So AB and DC are not equal vectors, but BD and CA are equal. Hmm, that would mean only one pair of sides is equal and parallel, which is not sufficient for a parallelogram.But we know from the midpoint condition that it's a parallelogram because the diagonals bisect each other. Therefore, perhaps the confusion arises from different definitions. Let's check the properties again.In a parallelogram, both pairs of opposite sides are parallel and equal. The diagonals bisect each other. If the diagonals bisect each other, then it's a parallelogram. So, even if one pair of opposite sides are negatives of each other, as long as the diagonals bisect each other, it's a parallelogram.Wait, but if vectors AB and DC are negatives, then they are still parallel and equal in magnitude, just opposite in direction. So, in terms of geometry, direction doesn't matter for parallelism; only the slope matters. So, even if they point in opposite directions, they are still parallel. Therefore, AB and DC are parallel and equal in length, and BD and AC are equal and parallel. Therefore, ABDC is a parallelogram.Therefore, D(8,3) is correct.But to resolve the confusion with the vector directions: when we traverse the quadrilateral ABDC, the side AB is from A to B, and DC is from D to C. If AB is (4,4), then DC should be the same vector. However, DC is (C - D) = (4 - 8, -1 - 3) = (-4, -4). So DC is (-4, -4), which is -1*(4,4). So, they are scalar multiples, hence parallel, but opposite in direction. However, in the context of a parallelogram, when you traverse the quadrilateral, the sides need to be congruent and parallel, but the direction depends on the order of traversal.If you traverse AB and then DC, but since DC is in the opposite direction, does that matter? Wait, in a parallelogram, the opposite sides should be both congruent and parallel, regardless of direction. So, even if their vectors are negatives, as long as they have the same magnitude and slope, they are considered congruent and parallel. Therefore, it's still a parallelogram.So, perhaps the key here is that in vector terms, the vectors AB and DC are negatives, but as geometric sides, they are still congruent and parallel. Therefore, the quadrilateral ABDC is a parallelogram.Therefore, after thorough checking, D(8,3) seems to be the correct answer.But let me check another way. Let's compute the coordinates using the properties of a parallelogram.In a parallelogram, the vector from A to B should be the same as the vector from D to C. Wait, vector AB is (4,4). So vector DC should be (4,4). Therefore, point D must be such that when you go from D to C, you move (4,4). Therefore, D = C - vector DC = (4, -1) - (4,4) = (0, -5). But this gives D(0,-5), which conflicts with the previous result.Wait, this is the same issue as before. If we consider vector AB = vector DC, then vector DC should be (4,4). Therefore, D = C - vector DC = (4 - 4, -1 - 4) = (0, -5). But this point (0, -5) is not on the extension of median AM. So that contradicts the problem's condition that D is on the extension of AM beyond M.Therefore, there's a conflict here. On one hand, using the midpoint condition gives D(8,3) on AM beyond M, and checking via vectors BD and AC being equal also gives D(8,3). However, using the vector AB = DC gives D(0,-5), which is not on AM. Therefore, which one is correct?The problem states that quadrilateral ABDC is a parallelogram. Depending on the order of the letters, the parallelogram can be formed differently. If ABDC is a parallelogram, the sides could be AB, BD, DC, and CA, but if it's intended to be a standard notation where the sides are AB, BC, CD, DA, then the order would be ABCD. But in this case, it's ABDC. So, the order is important.Alternatively, maybe the problem is mistyped, and it should be ABCD instead of ABDC. But assuming the problem is correct, we have to work with ABDC.Alternatively, perhaps the problem considers ABDC as a quadrilateral where sides AB and DC are opposite, and sides AC and BD are the other pair. Wait, but in that case, the sides AC and BD are diagonals. Hmm.Alternatively, maybe the problem is referring to a parallelogram with vertices A, B, D, C, so in order, meaning AB is adjacent to B and D, so the sides are AB, BD, DC, and CA. But in that case, it's a non-convex quadrilateral? Maybe not.Alternatively, perhaps the correct way is to use the midpoint method. Since the problem states that D is on the extension of median AM beyond M, and quadrilateral ABDC is a parallelogram, then using the midpoint condition (diagonals bisect each other) is the correct approach here, which gives D(8,3).Moreover, if we check the coordinates of D(8,3), and verify if ABDC is a parallelogram:- Calculate vectors AB, BD, DC, and CA.AB: B - A = (2 - (-2), 5 - 1) = (4,4)BD: D - B = (8 - 2, 3 - 5) = (6, -2)DC: C - D = (4 - 8, -1 - 3) = (-4, -4)CA: A - C = (-2 - 4, 1 - (-1)) = (-6, 2)Now, in a parallelogram, opposite sides should be equal and parallel. So AB and DC should be equal and parallel. AB is (4,4), DC is (-4,-4). These are scalar multiples (multiplied by -1), so they are parallel and have the same magnitude. Similarly, BD is (6,-2), and CA is (-6,2). Again, BD and CA are scalar multiples (multiplied by -1), so they are parallel and equal in magnitude. Therefore, even though the vectors are in opposite directions, they are considered parallel in the geometric sense (same slope) and equal in length. Therefore, quadrilateral ABDC is indeed a parallelogram.Therefore, even though using the vector equality AB = DC leads to a different point, the problem specifies that D is on the extension of AM beyond M, which is only satisfied by D(8,3). The other point D(0,-5) is not on AM, hence invalid. Therefore, the correct answer must be D(8,3).Another way to confirm is by using coordinates. If we plot the points:- A(-2,1), B(2,5), C(4,-1), M(3,2), D(8,3)Check if ABDC is a parallelogram:- AB: from (-2,1) to (2,5)- BD: from (2,5) to (8,3)- DC: from (8,3) to (4,-1)- CA: from (4,-1) to (-2,1)Calculating the slopes:- Slope of AB: (5 - 1)/(2 - (-2)) = 4/4 = 1- Slope of DC: (-1 - 3)/(4 - 8) = (-4)/(-4) = 1- Slope of BD: (3 - 5)/(8 - 2) = (-2)/6 = -1/3- Slope of CA: (1 - (-1))/(-2 - 4) = 2/(-6) = -1/3So, slopes of AB and DC are equal (1), and slopes of BD and CA are equal (-1/3). Therefore, opposite sides are parallel, confirming it's a parallelogram.Therefore, the coordinates of point D are (8,3).Final AnswerThe coordinates of point ( D ) are boxed{(8, 3)}.</think>Given points ( A(-2, 1) ), ( B(2, 5) ), and ( C(4, -1) ). Point ( D ) lies on the extension of median ( AM ) beyond point ( M ), and quadrilateral ( ABDC ) is a parallelogram. We need to find the coordinates of point ( D ).1. Find the midpoint ( M ) of ( BC ): - Using the midpoint formula: [ M = left( frac{2 + 4}{2}, frac{5 + (-1)}{2} right) = (3, 2) ]2. Determine the coordinates of ( D ): - Since ( D ) lies on the extension of median ( AM ) beyond ( M ), we use the midpoint condition for the diagonals of the parallelogram. The midpoint of ( AD ) must be the same as the midpoint of ( BC ), which is ( M(3, 2) ). - Let ( D = (x, y) ). The midpoint of ( AD ) is: [ left( frac{-2 + x}{2}, frac{1 + y}{2} right) = (3, 2) ] - Solving the equations: [ frac{-2 + x}{2} = 3 implies x = 8 ] [ frac{1 + y}{2} = 2 implies y = 3 ] - Therefore, the coordinates of ( D ) are ( (8, 3) ).3. Verification: - Check if ( D(8, 3) ) lies on the line ( AM ). The parametric equations for line ( AM ) are: [ x = -2 + 5t, quad y = 1 + t ] - For ( t = 2 ), we get ( x = 8 ) and ( y = 3 ), confirming ( D ) is on the extension of ( AM ). - Check vectors for the parallelogram: - Vectors ( AB = (4, 4) ) and ( DC = (-4, -4) ) are parallel and equal in magnitude. - Vectors ( BD = (6, -2) ) and ( CA = (-6, 2) ) are parallel and equal in magnitude. - Slopes of opposite sides are equal, confirming parallelism.Thus, the coordinates of point ( D ) are (boxed{(8, 3)}).]

🔑:## First method.The coordinates of point Mleft(x_{0} ; y_{0}right) are the arithmetic means of the corresponding coordinates of the endpoints of segment B C, i.e.,x_{0}=frac{2+4}{2}=3, y_{0}=frac{5-1}{2}=2Since the diagonals of a parallelogram are bisected by the point of intersection, Mleft(x_{0} ; y_{0}right) is the midpoint of the segment with endpoints at points A(-2 ; 1) and Dleft(x_{1} ; y_{1}right). Therefore,x_{0}=frac{-2+x_{1}}{2}=3, y_{0}=frac{1+y_{1}}{2}=2From this, we find that x_{1}=8, y_{1}=3.## Second method.Let x_{1}, y_{1} be the coordinates of point D. If A B C D is a parallelogram, then overrightarrow{B D}=overrightarrow{A C}, and sinceoverrightarrow{B D}=overrightarrow{left(x_{1}-2 ; y_{1}-5right)}, overrightarrow{A C}=overrightarrow{(4-(-2) ;-1-1)}=overrightarrow{(6 ;-2)}thenx_{1}-2=6, y_{1}-5=-2 text {. }From this, we find that x_{1}=8, y_{1}=3.## Answer(8 ; 3).

🧠:[<think>Okay, let's try to figure out this mowing problem. Hmm, so there's a team of mowers who have to mow two meadows. One meadow is twice as large as the other. Let me break down the information step by step.First, the team worked on the larger meadow for half a day. Then, they split into two groups. One group continued working on the larger meadow and finished it by evening. The other group worked on the smaller meadow but didn't finish it, and the remaining work on the smaller meadow was done by one mower the next day. We need to find out how many mowers were in the original team.Let me denote the number of mowers in the team as ( n ). So, the total number of mowers is ( n ).The meadows: Let's call the area of the smaller meadow ( A ). Then, the larger meadow is ( 2A ).Now, let's think about the time they spent. The team worked on the larger meadow for half a day. Then, they split into two halves. Wait, "split in half" – does that mean each half has ( n/2 ) mowers? That seems logical. So, after splitting, one half (( n/2 ) mowers) stays on the larger meadow, and the other half (( n/2 ) mowers) goes to the smaller meadow.They started working on the larger meadow in the morning. They spent half a day (let's assume a day is considered as, say, 12 hours, but maybe we can just use half a day as a unit of time). So, the entire team worked on the larger meadow for half a day. Then, for the remaining half day, half the team worked on the larger meadow, and the other half worked on the smaller meadow.But wait, the problem says "by evening." So, maybe they split after working half a day on the larger meadow, and then each group worked another half day? So total time for the larger meadow would be half a day with the entire team and half a day with half the team. And the smaller meadow was worked on for half a day with half the team, but not finished, and the remaining part was done by one mower the next day.We need to relate the work done to the areas. Let's assume that each mower works at a constant rate. Let's define the rate of one mower as ( r ) (area per unit time). So, the work done is rate multiplied by time multiplied by number of mowers.First, the larger meadow. The total work done on the larger meadow is done in two parts: first, the entire team (( n ) mowers) working for half a day, and then half the team (( n/2 ) mowers) working for another half a day. The sum of these two should equal the area of the larger meadow, which is ( 2A ).So, translating that into an equation:First part: ( n times r times frac{1}{2} ) daySecond part: ( frac{n}{2} times r times frac{1}{2} ) dayTotal work on larger meadow:( n r times frac{1}{2} + frac{n}{2} r times frac{1}{2} = 2A )Similarly, for the smaller meadow. The team (half of them, so ( n/2 ) mowers) worked on the smaller meadow for half a day, but didn't finish it. The remaining work was done by one mower the next day. Let's assume the next day is another full day? Or is it another half day? Wait, the problem says "the remaining work was completed by one mower on the following day." Since they were working until evening on the first day, so the following day would be a full day's work? But the problem doesn't specify, so maybe we need to assume that "day" is the same unit as before. If the first day had a morning and afternoon (each half day), then the following day would be another full day? Wait, maybe the time units are all in half days. Let's clarify.Wait, in the first day, the team worked on the larger meadow for half a day (morning), then split and worked another half day (afternoon). So total first day is morning and afternoon, each half day. Then, the next day, the remaining work on the smaller meadow was completed by one mower. So, how much time did that one mower take? If the next day is considered a full day, that would be 1 day, but if it's only half a day, then it's half a day. But the problem says "the remaining work was completed by one mower on the following day." It doesn't specify how long it took, just that the work was done by one mower the next day. So maybe we need to assume that the next day's work is a full day. Wait, but if we are working in half-day units, maybe the remaining work was done in one half-day? Hmm, this is unclear.Wait, actually, since the problem mentions "by evening" for the first day, meaning that they worked the entire day until evening. So maybe the first day is one full day, but they split the work into two halves? Wait, let me read the problem again:"A mowing team had to mow two meadows, one twice as large as the other. For half a day, the team mowed the larger meadow. After that, they split in half: one half stayed on the large meadow and finished it by evening, while the other half mowed the smaller meadow but did not finish it. How many mowers were in the team, if it is known that the remaining work was completed by one mower on the following day?"So, "for half a day, the team mowed the larger meadow." Then, "after that," meaning after the half day, they split in half. Then, "one half stayed on the large meadow and finished it by evening." So, if they started in the morning, worked half a day, then split, and then worked until evening. So, "evening" would be the end of the day. So, total time spent on the larger meadow is half a day plus another half day (after splitting). So, total of one day on the larger meadow. Similarly, the smaller meadow was worked on for half a day (after splitting) by half the team, and the remaining work was done by one mower the next day.So, the time units here are half days. Let's define a half day as a unit of time. So, the entire team worked on the larger meadow for 1 unit of time (half a day). Then, after splitting, half the team worked on the larger meadow for another 1 unit of time (another half day), totaling 2 units of time (1 full day) on the larger meadow. Similarly, the smaller meadow was worked on by half the team for 1 unit of time (half a day), and then the remaining work was done by 1 mower for 1 unit of time (the next half day? Wait, but the problem says "on the following day," which would be another full day? Wait, maybe the next day is a full day, but if they were working half days, then maybe the remaining work took 1 mower a full day (2 units of time). Hmm, this is a bit ambiguous.Wait, perhaps we need to make the following assumption: The initial work on the larger meadow was for half a day (let's say morning). Then, the team splits, and the two halves work for the rest of the day (afternoon) on their respective meadows. So, the mowing on both meadows after the split is done in the afternoon (another half day). Then, "the remaining work was completed by one mower on the following day." So, the remaining work on the smaller meadow was done by one mower the next day, which would be another full day? Or another half day?This is critical because the time taken affects the amount of work done. Let me re-examine the problem:"For half a day, the team mowed the larger meadow. After that, they split in half: one half stayed on the large meadow and finished it by evening, while the other half mowed the smaller meadow but did not finish it. How many mowers were in the team, if it is known that the remaining work was completed by one mower on the following day?"Key points:1. The team mowed the larger meadow for half a day.2. Then, they split into two halves.3. One half finished the larger meadow by evening.4. The other half mowed the smaller meadow but didn't finish.5. The remaining work on the smaller meadow was done by one mower the next day.Assuming a "day" is composed of a morning and afternoon, each being half a day. So, the first half day (morning) was spent by the entire team on the larger meadow. Then, in the afternoon (the second half of the day), the team split into two halves. One half worked on the larger meadow, finishing it by evening (i.e., by the end of the afternoon). The other half worked on the smaller meadow during the afternoon but didn't finish it, and the remaining work was done by one mower the next day. So, the next day would be another full day, or another half day? The problem says "the following day," so perhaps a full day. However, since the initial work was done in half days, maybe the "following day" is also considered a half day? Hmm, the problem is a bit ambiguous here.But let's think: If the remaining work on the smaller meadow was done by one mower on the following day, then how much time did that take? If the following day is a full day, then the work done is 1 mower * 1 day. If it's a half day, then 1 mower * 0.5 day. But the problem doesn't specify, so maybe we need to assume that "the following day" is the same as the previous working time, which was half a day? Wait, but in the problem statement, they split after half a day and worked until evening, which would be another half day. So, perhaps each "day" has two half days: morning and afternoon. Then, "the following day" would be the next morning and afternoon. But the problem says "the remaining work was completed by one mower on the following day," which could mean that the one mower worked the entire following day (two half days) or just one half day. This is unclear.However, given that when they worked on the meadows after splitting, they only worked for half a day (afternoon), so maybe the remaining work on the smaller meadow was a half day's work for one mower. Wait, but the problem says "the remaining work was completed by one mower on the following day." If "completed on the following day," it might mean that the one mower took the entire following day to do it. So, full day. But if they were working in half-day increments, maybe the following day is another half day. Hmm.This is a critical ambiguity. Let's consider both possibilities.First assumption: The remaining work on the smaller meadow took one mower a full day (i.e., two half days) to complete.Second assumption: The remaining work took one mower a half day to complete.But in the problem, when they split the team, they worked "by evening," which would be the end of the day. So, the initial half day (morning) plus another half day (afternoon) makes a full day. Then, the next day would be another full day.But if the remaining work was done "on the following day," that might mean a full day's work. However, the problem says "the remaining work was completed by one mower on the following day." So, it's possible that the one mower worked the entire next day to finish the remaining work.Alternatively, if the following day is considered the same as the previous working periods (half days), then maybe it took one half day. But the problem doesn't specify. Since the problem is likely mathematical and expects an integer answer, maybe the remaining work was equivalent to one mower working one full day (i.e., two half days). Let's proceed with that assumption and see if it leads to an integer answer. If not, we can reconsider.Let me try to formalize this.Let’s define the following:- Let ( n ) = total number of mowers in the team.- Let the area of the smaller meadow be ( A ), so the larger meadow is ( 2A ).- Let the work rate of one mower be ( r ) (area per half day). Since the problem uses half days as units, let's define the rate per half day.So, work done is calculated as (number of mowers) × (time in half days) × (rate per half day). But since the rate is per half day, time is 1 unit for each half day.So, the work done on the larger meadow:- First half day: all ( n ) mowers work on it. So, work done = ( n times 1 times r = nr ).- Second half day: half the team (( n/2 ) mowers) works on it. Work done = ( (n/2) times 1 times r = (n/2)r ).Total work done on larger meadow: ( nr + (n/2)r = (3n/2)r ).But the larger meadow's area is ( 2A ), so:( (3n/2)r = 2A ). Let's note this as equation (1).For the smaller meadow:Work done by half the team (( n/2 ) mowers) in the second half day: ( (n/2) times 1 times r = (n/2)r ).The remaining work on the smaller meadow is ( A - (n/2)r ).This remaining work was completed by one mower on the following day. If we assume "the following day" is a full day (i.e., two half days), then the work done by one mower is ( 1 times 2 times r = 2r ). Therefore:( A - (n/2)r = 2r ).Alternatively, if the following day is considered one half day, then the work done is ( 1 times 1 times r = r ), so:( A - (n/2)r = r ).We need to see which assumption gives us an integer solution for ( n ).First, let's assume the remaining work took a full day (two half days):Equation (1): ( (3n/2)r = 2A )Equation (2): ( A - (n/2)r = 2r )From equation (2): ( A = (n/2)r + 2r = r(n/2 + 2) )Plug this into equation (1):( (3n/2)r = 2 times r(n/2 + 2) )Divide both sides by ( r ):( 3n/2 = 2(n/2 + 2) )Simplify the right-hand side:( 3n/2 = n + 4 )Multiply both sides by 2:( 3n = 2n + 8 )Subtract ( 2n ):( n = 8 )Okay, that gives us ( n = 8 ). Let's check if this makes sense.If ( n = 8 ), then from equation (2):( A = r(8/2 + 2) = r(4 + 2) = 6r )So, smaller meadow is ( 6r ), larger meadow is ( 12r ).From equation (1):( (3*8)/2 * r = 12r ). Left side is ( 12r ), which matches the larger meadow area. Good.So, remaining work on smaller meadow is ( 6r - (8/2)r = 6r - 4r = 2r ). Then, one mower working two half days (a full day) would do ( 2r ), which matches. So, this works.Alternatively, if we had assumed the remaining work took one half day, then equation (2) would be:( A - (n/2)r = r )So, ( A = (n/2 + 1)r )Plug into equation (1):( (3n/2)r = 2(n/2 + 1)r )Divide by r:( 3n/2 = 2(n/2 + 1) )Simplify RHS:( 3n/2 = n + 2 )Multiply both sides by 2:( 3n = 2n + 4 )Subtract 2n:( n = 4 )Check if this works.If ( n = 4 ), then ( A = (4/2 + 1)r = 3r )Larger meadow is ( 6r ). From equation (1):( (3*4)/2 * r = 6r ). Correct.Remaining work on smaller meadow: ( 3r - (4/2)r = 3r - 2r = r ). Then, one mower working one half day (since the remaining work was done on the following day, which if it's a half day) would do ( 1 * r ). So that works too.But the problem states "the remaining work was completed by one mower on the following day." If "the following day" is considered a full day, then the first assumption (8 mowers) is correct. If "the following day" is considered a half day, then 4 mowers. However, in the problem statement, when they worked until evening, they split after half a day and worked another half day. So, each half day is a unit. Then, "the following day" might refer to the next day, which would be a full day (two half days). But in some contexts, a day is considered as a single unit. Since the problem is likely translated from Russian (original context), and in such problems, often a "day" refers to a full day's work, which could be different.But in the problem, the initial mowing on the larger meadow was "half a day," then after splitting, they worked "by evening," which would mean they worked the remaining half day. So, the total time spent on the larger meadow was 1 full day (half day by entire team, half day by half team). The smaller meadow was worked on for half a day by half the team, and then the remaining work was done by one mower the next day. If the next day is considered a full day (i.e., two half days), then the answer is 8. If it's considered a half day, then the answer is 4.But in the problem statement, it says "the remaining work was completed by one mower on the following day." Since they worked until evening on the first day, the following day would start the next morning. So, the one mower would have the entire next day (i.e., a full day) to work on the remaining part. Therefore, considering a full day as two half days, then the answer would be 8. However, in some interpretations, a "day" is a single unit, so if they used half days as units, then "the following day" is a half day. But the problem mentions "for half a day" initially, then "by evening," so splitting the day into two halves.Given that, if the remaining work was completed by one mower on the following day, which is a full day (two half days), then the answer is 8. However, if "the following day" is considered one half day, answer is 4. But since the problem is likely expecting an integer answer and 8 is more plausible if we consider the following day as a full day, then 8 is the answer.Wait, but let me check once again with both possibilities.Case 1: Following day is a full day (two half days):Remaining work: 2r. So, one mower works two half days: 2r. So, 2r = 2r, which works with n=8.Case 2: Following day is a half day:Remaining work: r. So, one mower works one half day: r. Which works with n=4.But in the problem statement, it says "the remaining work was completed by one mower on the following day." The phrase "on the following day" often implies that the work was done within that day, not necessarily the entire day. So, if the remaining work was 2r, then one mower would take two half days (a full day) to do it. If the remaining work was r, then one mower would take one half day.But the problem states that "the remaining work was completed by one mower on the following day," which could mean that the entire following day was used, regardless of the amount of work. However, since the problem is in a math context, it's likely that the following day is the same unit as before. Since before, the work was split into half days, perhaps the following day is also a half day. Therefore, the answer is 4. But the problem may be expecting 8.Wait, let's see. If n=8:- Larger meadow area: 12r- Work done on larger meadow: 8*1*r + 4*1*r = 12r. Correct.- Smaller meadow area: 6r.- Work done on smaller meadow: 4*1*r = 4r. Remaining work: 2r. Then, one mower works next day (two half days): 2r. So, correct.If n=4:- Larger meadow area: 6r- Work done: 4*1*r + 2*1*r = 6r. Correct.- Smaller meadow area: 3r.- Work done: 2*1*r = 2r. Remaining work: 1r. One mower works next day (half day): 1r. Correct.But the problem says "the remaining work was completed by one mower on the following day." If "completed on the following day" means that the one mower took the entire following day to finish it, which would be a full day (two half days), then 8 is the answer. However, if "completed on the following day" just means that the mower did it the next day, without specifying how long it took, but since the previous work was in half days, maybe it's one half day. But in that case, the answer is 4.Since both 4 and 8 are possible depending on interpretation, but in most problems of this type, they usually assume that "day" is a full day. So, given the problem mentions "half a day" initially, then splits into another half day, and then the remaining work is done "on the following day," which is a separate day. So, the following day is a full day. Therefore, the remaining work would take a full day, so 8 mowers is the answer.Hence, I think the answer is 8.Final AnswerThe number of mowers in the team was boxed{8}.</think>Let ( n ) be the number of mowers in the team. The larger meadow is twice the size of the smaller meadow, denoted as ( 2A ) and ( A ) respectively.1. Work done on the larger meadow: - The entire team worked for half a day: ( n times frac{1}{2} ) day. - After splitting, half the team (( frac{n}{2} )) worked for another half day: ( frac{n}{2} times frac{1}{2} ) day. - Total work done on the larger meadow: [ n times frac{1}{2} + frac{n}{2} times frac{1}{2} = frac{n}{2} + frac{n}{4} = frac{3n}{4} ] - This equals the area of the larger meadow ( 2A ): [ frac{3n}{4} = 2A quad Rightarrow quad 3n = 8A quad Rightarrow quad A = frac{3n}{8} ]2. Work done on the smaller meadow: - Half the team (( frac{n}{2} )) worked for half a day: ( frac{n}{2} times frac{1}{2} ) day. - Work done: ( frac{n}{4} ) - Remaining work: ( A - frac{n}{4} ) - The remaining work was completed by one mower the next day (assumed to be a full day, i.e., 1 day): [ A - frac{n}{4} = 1 times 1 quad Rightarrow quad A = frac{n}{4} + 1 ]3. Equating the two expressions for ( A ): - From the larger meadow: ( A = frac{3n}{8} ) - From the smaller meadow: ( A = frac{n}{4} + 1 ) - Setting them equal: [ frac{3n}{8} = frac{n}{4} + 1 ] - Solving for ( n ): [ frac{3n}{8} = frac{2n}{8} + frac{8}{8} quad Rightarrow quad 3n = 2n + 8 quad Rightarrow quad n = 8 ]Thus, the number of mowers in the team was (boxed{8}).,<think>Okay, let's see. There's this mowing team that had to mow two meadows. One meadow is twice as large as the other. So, let's call the smaller meadow size S, then the larger one would be 2S. The problem says the team worked on the larger meadow for half a day. Then, after that, they split into two halves. One half stayed on the large meadow and finished it by evening. The other half went to mow the smaller meadow but didn't finish it. The remaining work on the smaller meadow was done by one mower the next day. We need to find how many mowers were in the team.Hmm. Let's break this down step by step. Let's denote the number of mowers as N. First, the team worked on the larger meadow (which is 2S) for half a day. So, the entire team of N mowers worked for half a day. Then, they split into two groups: N/2 stayed on the large meadow, and N/2 went to the smaller meadow. The group that stayed on the large meadow finished it by evening. Since the first half day was already spent by N mowers, the remaining work on the large meadow would have been done by N/2 mowers for the other half day. Similarly, the other half of the team, N/2 mowers, worked on the smaller meadow for half a day but didn't finish it. The remaining work was done by one mower the next day, which I assume is a full day. So, we need to model the amount of work done on each meadow. Let's think in terms of man-days. A man-day is the amount of work one mower can do in a day. Let’s let the work rate be such that one mower can mow 1 unit per day. Then, a man-day is 1 unit of work. So, first, the larger meadow is 2S. The total work done on the larger meadow is the work done by N mowers for half a day plus the work done by N/2 mowers for another half day. Half a day is 0.5 days. So, work done on larger meadow: N * 0.5 + (N/2) * 0.5. Let's compute that:N * 0.5 + (N/2) * 0.5 = 0.5N + 0.25N = 0.75N. But this work equals 2S. So, 0.75N = 2S. Therefore, S = (0.75N)/2 = 0.375N. Now, let's look at the smaller meadow. The smaller meadow is S. The work done on it is by N/2 mowers for half a day. Then, the remaining work was completed by one mower the next day, which is one man-day. So, work done on smaller meadow: (N/2) * 0.5 + 1 * 1 = 0.25N + 1. But this work should equal S. So:0.25N + 1 = S. But from earlier, we have S = 0.375N. So, substituting:0.25N + 1 = 0.375N. Subtract 0.25N from both sides:1 = 0.125N. Therefore, N = 1 / 0.125 = 8. So, the number of mowers in the team is 8. Wait, let me verify this again. If there are 8 mowers, then the work on the larger meadow:First half day: 8 mowers * 0.5 day = 4 man-days.Then they split into two groups of 4. The group on the larger meadow works another half day: 4 * 0.5 = 2 man-days.Total for larger meadow: 4 + 2 = 6 man-days. Since the larger meadow is twice the smaller, that means the smaller meadow would be 3 man-days. On the smaller meadow: the other half of the team (4 mowers) works half a day: 4 * 0.5 = 2 man-days. Then the remaining work is 3 - 2 = 1 man-day, which is done by one mower the next day. That matches. Therefore, the answer is 8. Wait, let me check if the equations are consistent. So, according to the equations, S = 0.375N. If N=8, then S=0.375*8=3. So, the smaller meadow is 3 man-days. The larger meadow is 6 man-days. First, large meadow: 8 mowers for 0.5 days is 4 man-days. Then 4 mowers for 0.5 days is 2 man-days. Total 6. Correct. Small meadow: 4 mowers for 0.5 days is 2 man-days. Remaining 1 man-day done by 1 mower. Correct. Yes, that all checks out. Therefore, the number of mowers is 8.Final AnswerThe number of mowers in the team was boxed{8}.</think>Let's denote the number of mowers as ( N ). The larger meadow is twice the size of the smaller one, so we denote the smaller meadow as ( S ) and the larger meadow as ( 2S ).1. Work on the larger meadow: - The entire team of ( N ) mowers worked for half a day, contributing ( N times 0.5 ) man-days. - After splitting, half the team (( frac{N}{2} )) worked for another half day, contributing ( frac{N}{2} times 0.5 ) man-days. - Total work done on the larger meadow: ( 0.5N + 0.25N = 0.75N ) man-days. - Since the larger meadow is ( 2S ), we have ( 0.75N = 2S ). Solving for ( S ), we get ( S = 0.375N ).2. Work on the smaller meadow: - Half the team (( frac{N}{2} )) worked for half a day, contributing ( frac{N}{2} times 0.5 = 0.25N ) man-days. - The remaining work was completed by one mower in one full day, contributing ( 1 ) man-day. - Total work done on the smaller meadow: ( 0.25N + 1 ) man-days. - Since the smaller meadow is ( S ), we have ( 0.25N + 1 = S ).3. Equating the two expressions for ( S ): - From the larger meadow: ( S = 0.375N ). - From the smaller meadow: ( 0.25N + 1 = 0.375N ). - Solving for ( N ): ( 1 = 0.125N ) leads to ( N = frac{1}{0.125} = 8 ).Verification:- Larger meadow: ( 8 ) mowers for half a day (( 4 ) man-days) and ( 4 ) mowers for another half day (( 2 ) man-days) totals ( 6 ) man-days, which is ( 2S ). Thus, ( S = 3 ) man-days.- Smaller meadow: ( 4 ) mowers for half a day (( 2 ) man-days) plus ( 1 ) man-day by one mower totals ( 3 ) man-days, which matches ( S ).Thus, the number of mowers in the team was (boxed{8}).]

🔑:7. If the whole brigade mowed a large meadow for half a day and half of the brigade mowed for another half a day, it is clear that the brigade mowed frac{2}{3} of the meadow in half a day, and half of the brigade mowed frac{1}{3} of the meadow in half a day. Since the second meadow is half the size of the first, after working on it with half of the brigade for half a day, the uncut area is frac{1}{2}-frac{1}{3}=frac{1}{6}. This area was mowed by one mower in one day. Therefore, one mower mows frac{1}{6} of the meadow in one day, and the whole brigade mows frac{4}{3}. Answer: there were 8 mowers in the brigade.

🧠:[<think>Okay, let me try to figure out this problem. So, I need to calculate the cube root of x divided by (2015 + 2016), where x is the harmonic mean of the numbers a and b. First, let me recall what harmonic mean is. The harmonic mean of two numbers a and b is given by c such that 1/c is the average of 1/a and 1/b. So, formula-wise, that would be 1/c = (1/a + 1/b)/2. Therefore, c = 2/(1/a + 1/b). So, x here is this harmonic mean c.Given that a and b are these two fractions:a = (2016 + 2015)/(2016² + 2016*2015 + 2015²)andb = (2016 - 2015)/(2016² - 2016*2015 + 2015²)So, first, I need to compute a and b, then find their harmonic mean x, then compute x divided by (2015 + 2016), and finally take the cube root of that result.Hmm. Let me take this step by step.First, let's compute the numerators and denominators for a and b.Let's note that 2016 and 2015 are consecutive integers, right? 2016 = 2015 + 1. So maybe there's a way to simplify these expressions using some algebraic identities. Let's see.First, compute the numerator of a: 2016 + 2015. That's 2015 + 2016 = 4031.Denominator of a: 2016² + 2016*2015 + 2015². Hmm, that looks like the formula for a³ - b³, which is (a - b)(a² + ab + b²). Wait, but here we have a² + ab + b². Since the denominator is a² + ab + b², if I can relate this to a³ - b³, maybe?Wait, but the numerator is a + b. Let me write a and b as variables to see if there's a relationship. Let me let m = 2016 and n = 2015. Then a = (m + n)/(m² + mn + n²) and b = (m - n)/(m² - mn + n²). Maybe substituting variables would make it clearer.So, let m = 2016, n = 2015. Then:a = (m + n)/(m² + mn + n²)b = (m - n)/(m² - mn + n²)Hmm. Let me see. Maybe these expressions can be related to the sum and difference of cubes? Because m³ - n³ = (m - n)(m² + mn + n²) and m³ + n³ = (m + n)(m² - mn + n²). Wait, that's right! Let me verify:m³ + n³ = (m + n)(m² - mn + n²)m³ - n³ = (m - n)(m² + mn + n²)Yes, that's correct. So, then, the denominators of a and b are the same as the second factors in those cube formulas. Therefore, the denominators can be expressed as (m³ - n³)/(m - n) for a, and (m³ + n³)/(m + n) for b. So:Denominator of a: m² + mn + n² = (m³ - n³)/(m - n)Denominator of b: m² - mn + n² = (m³ + n³)/(m + n)Therefore, substituting back into a and b:a = (m + n)/[(m³ - n³)/(m - n)] = (m + n)(m - n)/(m³ - n³) = (m² - n²)/(m³ - n³)Similarly, b = (m - n)/[(m³ + n³)/(m + n)] = (m - n)(m + n)/(m³ + n³) = (m² - n²)/(m³ + n³)Wait, but hold on:Wait, for a: denominator is (m³ - n³)/(m - n), so invert and multiply:a = (m + n) * (m - n)/(m³ - n³) = (m² - n²)/(m³ - n³)Similarly, for b: denominator is (m³ + n³)/(m + n), so invert and multiply:b = (m - n)*(m + n)/(m³ + n³) = (m² - n²)/(m³ + n³)Wait, so both a and b have the same numerator, m² - n², but different denominators: a has denominator m³ - n³, and b has denominator m³ + n³.So, a = (m² - n²)/(m³ - n³) and b = (m² - n²)/(m³ + n³)Therefore, 1/a = (m³ - n³)/(m² - n²)Similarly, 1/b = (m³ + n³)/(m² - n²)So, harmonic mean x is 2/(1/a + 1/b) = 2/[ (m³ - n³ + m³ + n³)/(m² - n²) ) ] = 2/[ (2m³)/(m² - n²) ) ] = 2 * (m² - n²)/(2m³) ) = (m² - n²)/m³Therefore, x = (m² - n²)/m³Wait, that seems straightforward. Let me check that again.Starting with harmonic mean:1/c = (1/a + 1/b)/2So, 1/c = [ (m³ - n³)/(m² - n²) + (m³ + n³)/(m² - n²) ] / 2Since both 1/a and 1/b have the same denominator (m² - n²), we can factor that out:1/c = [ ( (m³ - n³) + (m³ + n³) ) / (m² - n²) ] / 2Combine the numerators:(m³ - n³ + m³ + n³) = 2m³Thus:1/c = (2m³ / (m² - n²)) / 2 = m³ / (m² - n²)Therefore, c = (m² - n²)/m³So, x, which is the harmonic mean, is (m² - n²)/m³Therefore, x = (m² - n²)/m³So now, substituting back m = 2016 and n = 2015.But let's compute m² - n² first. Since m² - n² = (m - n)(m + n). Since m and n are consecutive integers, m - n = 1. So, m² - n² = 1*(m + n) = m + n = 2016 + 2015 = 4031Therefore, x = (m + n)/m³ = 4031 / (2016³)Therefore, x is 4031 divided by 2016 cubed.Now, the problem asks for the cube root of x divided by (2015 + 2016). Let's write that out:Cube root of [x / (2015 + 2016)] = cube root [ (4031 / 2016³) / 4031 ]Wait, because 2015 + 2016 is 4031, same as m + n.So, x is 4031 / 2016³, and we divide that by 4031:x / 4031 = (4031 / 2016³) / 4031 = 1 / 2016³Therefore, the cube root of 1 / 2016³ is 1/2016So, the answer should be 1/2016, and that's the cube root.Wait, let me verify that step again.x = 4031 / 2016³Then x / (2015 + 2016) = x / 4031 = (4031 / 2016³) / 4031 = (1 / 2016³) * (4031 / 4031) = 1 / 2016³Then, the cube root of 1 / 2016³ is indeed (1 / 2016³)^(1/3) = 1 / 2016Therefore, the final answer is 1/2016. So, boxed, that's boxed{dfrac{1}{2016}}But wait, let me double-check to make sure I didn't skip a step or make a mistake in simplifying.First, m = 2016, n = 2015. So, m - n = 1, m + n = 4031.a = (m + n)/(m² + mn + n²) = 4031/(m² + mn + n²). But we found that m² + mn + n² = (m³ - n³)/(m - n) = (2016³ - 2015³)/1. So, a = 4031 / (2016³ - 2015³). Similarly, b was (m - n)/(m² - mn + n²) = 1/( (m³ + n³)/(m + n) ) = (m + n)/(m³ + n³) = 4031/(2016³ + 2015³)Wait, but earlier, when we simplified a and b, we ended up with a = (m² - n²)/(m³ - n³) and b = (m² - n²)/(m³ + n³). But since m² - n² = 4031, this gives a = 4031/(m³ - n³) and b = 4031/(m³ + n³). Therefore, 1/a = (m³ - n³)/4031 and 1/b = (m³ + n³)/4031. Then, the harmonic mean is 2/(1/a + 1/b) = 2/[ (m³ - n³ + m³ + n³)/4031 ] = 2/[ 2m³ / 4031 ] = (2 * 4031) / (2m³) = 4031/m³. So, x = 4031/m³. Then, x divided by (m + n) is 4031/m³ divided by 4031 = 1/m³. Then, cube root of 1/m³ is 1/m. Since m is 2016, the result is 1/2016. So yes, that checks out.Therefore, the answer is indeed 1/2016. That seems correct.Just to recap, by recognizing the denominators of a and b as parts of the sum and difference of cubes formulas, we could express a and b in terms that made calculating their reciprocals straightforward. Then, adding the reciprocals for the harmonic mean led to cancellation of terms, simplifying the expression significantly. Finally, dividing by the sum of 2015 and 2016 (which was already present in the numerator) canceled out the remaining terms, leaving a simple cube root of 1 over 2016 cubed, which is 1 over 2016.So, I think that's solid. No steps seem to be missing, and all the algebra checks out. So, the answer is 1/2016.Final Answerboxed{dfrac{1}{2016}}</think>To calculate (sqrt[3]{frac{x}{2015+2016}}), where (x) is the harmonic mean of the numbers (a) and (b):1. Identify the harmonic mean formula: The harmonic mean (c) of two numbers (a) and (b) is given by (frac{1}{c} = frac{1}{2}left(frac{1}{a} + frac{1}{b}right)). Thus, (x = frac{2}{frac{1}{a} + frac{1}{b}}).2. Express (a) and (b) in terms of (m = 2016) and (n = 2015): - (a = frac{m + n}{m^2 + mn + n^2}) - (b = frac{m - n}{m^2 - mn + n^2})3. Simplify using sum and difference of cubes: - (m^2 + mn + n^2 = frac{m^3 - n^3}{m - n}) - (m^2 - mn + n^2 = frac{m^3 + n^3}{m + n})4. Rewrite (a) and (b): - (a = frac{(m + n)(m - n)}{m^3 - n^3} = frac{m^2 - n^2}{m^3 - n^3}) - (b = frac{(m - n)(m + n)}{m^3 + n^3} = frac{m^2 - n^2}{m^3 + n^3})5. Calculate reciprocals: - (frac{1}{a} = frac{m^3 - n^3}{m^2 - n^2}) - (frac{1}{b} = frac{m^3 + n^3}{m^2 - n^2})6. Find the harmonic mean (x): - (frac{1}{x} = frac{1}{2}left(frac{m^3 - n^3 + m^3 + n^3}{m^2 - n^2}right) = frac{1}{2} cdot frac{2m^3}{m^2 - n^2} = frac{m^3}{m^2 - n^2}) - (x = frac{m^2 - n^2}{m^3})7. Simplify (x): - (m^2 - n^2 = (m - n)(m + n) = 1 cdot 4031 = 4031) - (x = frac{4031}{m^3} = frac{4031}{2016^3})8. Compute (frac{x}{2015 + 2016}): - (frac{x}{4031} = frac{4031 / 2016^3}{4031} = frac{1}{2016^3})9. Take the cube root: - (sqrt[3]{frac{1}{2016^3}} = frac{1}{2016})Thus, the final answer is (boxed{dfrac{1}{2016}}).,<think>Okay, let's tackle this problem step by step. The task is to calculate the cube root of x divided by (2015 + 2016), where x is the harmonic mean of two numbers a and b. The numbers a and b are given as fractions involving 2015 and 2016. First, let me recall what the harmonic mean is. The harmonic mean c of two positive numbers a and b is defined such that 1/c is the average of 1/a and 1/b. So, mathematically, that would be:1/c = (1/a + 1/b)/2Therefore, c = 2/(1/a + 1/b). So, to find x, which is the harmonic mean of a and b, I need to compute x = 2/(1/a + 1/b). Then, once I have x, I need to compute the cube root of x divided by (2015 + 2016). Given that, let's first write down the expressions for a and b:a = (2016 + 2015)/(2016² + 2016*2015 + 2015²)b = (2016 - 2015)/(2016² - 2016*2015 + 2015²)So, a and b are both fractions. Let me note that 2016 and 2015 are consecutive integers, with 2016 being the larger one. Let's compute 2016 + 2015 and 2016 - 2015 first to simplify a and b.2016 + 2015 = 40312016 - 2015 = 1Therefore, the numerators of a and b are 4031 and 1, respectively.So now, a = 4031 / (2016² + 2016*2015 + 2015²)and b = 1 / (2016² - 2016*2015 + 2015²)Next, let's compute the denominators of a and b. Let's start with the denominator of a:Denominator of a: 2016² + 2016*2015 + 2015²Similarly, denominator of b: 2016² - 2016*2015 + 2015²Hmm, these denominators look similar to the formula for the sum and difference of cubes. Recall that:For any two numbers m and n,m³ + n³ = (m + n)(m² - mn + n²)m³ - n³ = (m - n)(m² + mn + n²)But in our denominators, the terms are m² ± mn + n², which are the factors of m³ ∓ n³. Wait, so if we consider m = 2016 and n = 2015, then:Denominator of a: 2016² + 2016*2015 + 2015² = (2016³ - 2015³)/(2016 - 2015) Similarly, denominator of b: 2016² - 2016*2015 + 2015² = (2016³ + 2015³)/(2016 + 2015)Wait, let me verify that.Starting with m³ - n³ = (m - n)(m² + mn + n²). Therefore, m² + mn + n² = (m³ - n³)/(m - n)Similarly, m³ + n³ = (m + n)(m² - mn + n²). Therefore, m² - mn + n² = (m³ + n³)/(m + n)Yes, that's correct. Therefore, the denominator of a is (2016³ - 2015³)/(2016 - 2015), but since m - n is 1 here (2016 - 2015 = 1), then denominator of a is 2016³ - 2015³ divided by 1, which is just 2016³ - 2015³.Similarly, denominator of b is (2016³ + 2015³)/(2016 + 2015). Since 2016 + 2015 = 4031, denominator of b is (2016³ + 2015³)/4031.Therefore, substituting back into a and b:a = 4031 / (2016³ - 2015³)b = 1 / [(2016³ + 2015³)/4031] = 4031 / (2016³ + 2015³)So, a = 4031 / (2016³ - 2015³)and b = 4031 / (2016³ + 2015³)Therefore, now we can write 1/a and 1/b:1/a = (2016³ - 2015³)/40311/b = (2016³ + 2015³)/4031So, the harmonic mean x is 2/(1/a + 1/b) = 2 / [(2016³ - 2015³ + 2016³ + 2015³)/4031] = 2 / [2*2016³ / 4031] = 2 * [4031/(2*2016³)] = 4031 / 2016³Wait, let's check that again. Let's compute 1/a + 1/b:1/a + 1/b = (2016³ - 2015³)/4031 + (2016³ + 2015³)/4031Combine the numerators:[(2016³ - 2015³) + (2016³ + 2015³)] / 4031 = [2*2016³] / 4031Therefore, 1/a + 1/b = 2*2016³ / 4031Hence, the harmonic mean x = 2 / (2*2016³ / 4031) ) = 2 * (4031 / (2*2016³)) ) = 4031 / 2016³Therefore, x = 4031 / 2016³Wait, that's interesting. So x is 4031 divided by 2016 cubed.Now, the problem asks for the cube root of x divided by (2015 + 2016). Since 2015 + 2016 is 4031, as we computed earlier.Therefore, we need to compute:cube_root( (x) / 4031 ) = cube_root( (4031 / 2016³) / 4031 ) = cube_root(1 / 2016³) = 1 / 2016Because the cube root of 1/2016³ is 1/2016.Wait, let's verify that step.If we have x = 4031 / 2016³, then x / 4031 = (4031 / 2016³) / 4031 = (1 / 2016³)Then, the cube root of 1 / 2016³ is indeed 1 / 2016, since (1/2016)^3 = 1 / 2016³.Therefore, the result is 1/2016.So, summarizing all steps:1. Calculated a and b, found they were 4031/(2016³ - 2015³) and 4031/(2016³ + 2015³).2. Took reciprocals of a and b, which summed to 2*2016³/4031.3. Therefore, harmonic mean x = 4031 / 2016³.4. Divided x by 4031, getting 1/2016³, took the cube root to get 1/2016.Therefore, the answer is 1/2016.But let me check if I made any miscalculations in the steps above.Starting with a and b:a = (2016 + 2015)/(2016² + 2016*2015 + 2015²) = 4031 / (2016² + 2016*2015 + 2015²)Denominator: 2016² + 2016*2015 + 2015². As per the formula for m² + mn + n² = (m³ - n³)/(m - n). Here, m = 2016, n = 2015. So denominator is (2016³ - 2015³)/ (2016 - 2015) = (2016³ - 2015³)/1. So denominator is 2016³ - 2015³, hence a = 4031 / (2016³ - 2015³). Correct.Similarly for b:b = (2016 - 2015)/(2016² - 2016*2015 + 2015²) = 1 / (2016² - 2016*2015 + 2015²)Denominator: 2016² - 2016*2015 + 2015². Using the formula m² - mn + n² = (m³ + n³)/(m + n). Therefore, denominator is (2016³ + 2015³)/ (2016 + 2015) = (2016³ + 2015³)/4031. Thus, b = 1 / [(2016³ + 2015³)/4031] = 4031 / (2016³ + 2015³). Correct.Therefore, 1/a = (2016³ - 2015³)/4031 and 1/b = (2016³ + 2015³)/4031. Adding them gives:(2016³ - 2015³ + 2016³ + 2015³)/4031 = (2*2016³)/4031. Therefore, harmonic mean x = 2 / [(2*2016³)/4031] = (2 * 4031) / (2*2016³) ) = 4031 / 2016³. Correct.Then, x divided by 4031 is (4031 / 2016³)/4031 = 1 / 2016³. The cube root of that is 1 / 2016. Correct.So, the answer should be 1/2016. Therefore, boxed, it is boxed{dfrac{1}{2016}}.Final Answerboxed{dfrac{1}{2016}}</think>To calculate (sqrt[3]{frac{x}{2015+2016}}), where (x) is the harmonic mean of the numbers (a) and (b):1. Calculate the harmonic mean (x): - The harmonic mean (c) of two numbers (a) and (b) is given by (frac{1}{c} = frac{1}{2}left(frac{1}{a} + frac{1}{b}right)). - Given (a = frac{4031}{2016^3 - 2015^3}) and (b = frac{4031}{2016^3 + 2015^3}): - (frac{1}{a} = frac{2016^3 - 2015^3}{4031}) - (frac{1}{b} = frac{2016^3 + 2015^3}{4031}) - Sum of reciprocals: (frac{1}{a} + frac{1}{b} = frac{(2016^3 - 2015^3) + (2016^3 + 2015^3)}{4031} = frac{2 cdot 2016^3}{4031}) - Harmonic mean (x): (x = frac{2}{frac{2 cdot 2016^3}{4031}} = frac{4031}{2016^3})2. Calculate the cube root: - The expression to evaluate is (sqrt[3]{frac{x}{4031}}) (since (2015 + 2016 = 4031)): - (frac{x}{4031} = frac{frac{4031}{2016^3}}{4031} = frac{1}{2016^3}) - Cube root: (sqrt[3]{frac{1}{2016^3}} = frac{1}{2016})Thus, the final answer is (boxed{dfrac{1}{2016}}).]

🔑:Solution:Let's calculatefrac{1}{a}+frac{1}{b}=frac{2016^{2}+2016 cdot 2015+2015^{2}}{2016+2015}+frac{2016^{2}-2016 cdot 2015+2015^{2}}{2016-2015}==frac{(2016-2015)left(2016^{2}+2016 cdot 2015+2015^{2}right)+(2016+2015)left(2016^{2}-2016 cdot 2015+2015^{2}right)}{(2016+2015)(2016-2015)}=frac{2016^{3}-2015^{3}+2016^{3}+2015^{3}}{(2016+2015)(2016-2015)}=frac{2 cdot 2016^{3}}{4031}, thenfrac{1}{c}=frac{1}{2} cdot frac{2 cdot 2016^{3}}{4031}=frac{2016^{3}}{4031}, hence x=frac{4031}{2016^{3}}Finally, sqrt[3]{frac{x}{2015+2016}}=sqrt[3]{frac{4031}{(2015+2016) cdot 2016^{3}}}=frac{1}{2016}Answer: frac{1}{2016}.

🧠:[<think>Okay, let's see. I need to find the limit as x approaches 2 of [cos x / cos 2] raised to the power of 1/(x - 2). Hmm. Alright, so when x is approaching 2, the exponent 1/(x - 2) is going to either go to positive or negative infinity depending on the direction from which x approaches 2. But since the problem doesn't specify left or right, I guess it's just the general limit. First, let me check what the base of the exponent is when x approaches 2. The base is (cos x / cos 2). If x approaches 2, then cos x approaches cos 2, right? Because cosine is continuous. So the base approaches cos 2 / cos 2 = 1. And the exponent approaches infinity. So this is a 1^infinity indeterminate form, which means I need to use some technique to evaluate it. I remember that when dealing with limits of the form [f(x)]^g(x) where f(x) approaches 1 and g(x) approaches infinity, the standard method is to use the exponential function and natural logarithm. Specifically, the limit can be rewritten as e^{limit of g(x) * (f(x) - 1) as x approaches a}. Wait, is that right? Or is it that we take the natural log of the expression, compute the limit, and then exponentiate? Let me recall. Yes, the standard approach is: Let L be the original limit. Then take natural log of both sides: ln L = lim_{x->2} [ (1/(x - 2)) * ln(cos x / cos 2) ) ]. Then compute this limit, and then exponentiate the result to get L. So first, compute the limit of (ln(cos x / cos 2)) / (x - 2) as x approaches 2. Let me write that out: lim_{x->2} [ln(cos x) - ln(cos 2)] / (x - 2). That looks like the definition of the derivative of ln(cos x) at x = 2. Because the derivative of ln(cos a) with respect to a is -sin a / cos a = -tan a. So if I recognize this limit as the derivative of ln(cos x) at x = 2, then the limit becomes -tan 2. Therefore, ln L = -tan 2, so L = e^{-tan 2}. Therefore, the answer should be e^{-tan 2}. Wait, let me verify step by step. Let me set t = x - 2, so as x approaches 2, t approaches 0. Then the limit becomes lim_{t->0} [ (cos(2 + t)/cos 2)^{1/t} ]. Let's see, that substitution might make it easier. So, cos(2 + t) can be expanded using the cosine addition formula: cos(2 + t) = cos 2 cos t - sin 2 sin t. Therefore, cos(2 + t)/cos 2 = [cos 2 cos t - sin 2 sin t]/cos 2 = cos t - tan 2 sin t. So the expression becomes [cos t - tan 2 sin t]^{1/t}. Hmm, as t approaches 0, cos t ~ 1 - t^2/2, and sin t ~ t. So substituting the approximations, we have [1 - t^2/2 - tan 2 * t]^{1/t}. Hmm, then taking the logarithm, we get (1/t) * ln(1 - tan 2 * t - t^2/2). Now, for small t, ln(1 + a) ~ a - a^2/2 + ... So here, a = - tan 2 * t - t^2/2. So ln(1 + a) ~ a - a^2/2. Therefore, the logarithm becomes approximately (- tan 2 * t - t^2/2) - [(- tan 2 * t - t^2/2)^2]/2. Then divide by t: [ - tan 2 * t - t^2/2 - ( (tan^2 2 * t^2 + tan 2 * t^3 + t^4 /4 ) / 2 ) ] / t. Simplifying, the first term is - tan 2, then - t/2, then subtract (tan^2 2 * t + tan 2 * t^2 / 2 + t^3 /8). As t approaches 0, all the terms with t go to zero, so the limit of the logarithm is - tan 2. Therefore, again, the original limit is e^{-tan 2}. So that seems consistent. Alternatively, using the derivative approach, since the limit is the derivative of ln(cos x) at x = 2, which is (d/dx) ln(cos x) = - tan x. Evaluated at x = 2, it's - tan 2. Therefore, ln L = - tan 2, so L = e^{- tan 2}. Therefore, the answer should be e^{-tan 2}. Let me check if there's another way to do this. Maybe using L’Hospital’s Rule? Let's see. If we consider lim_{x->2} [ln(cos x) - ln(cos 2)] / (x - 2). Let me set h = x - 2, so as h approaches 0. Then the limit becomes [ln(cos(2 + h)) - ln(cos 2)] / h. Then, as h approaches 0, the numerator can be approximated by the derivative. But if we want to apply L’Hospital’s Rule, we can consider the form 0/0. Let's check: numerator as h approaches 0 is ln(cos(2 + h)) - ln(cos 2) which approaches ln(cos 2) - ln(cos 2) = 0, denominator h approaches 0. So it's 0/0, so we can apply L’Hospital. Taking derivative of numerator with respect to h: derivative of ln(cos(2 + h)) is ( - sin(2 + h ) / cos(2 + h ) ) * 1, derivative of denominator is 1. Therefore, the limit becomes [ - sin(2 + h ) / cos(2 + h ) ] / 1 as h approaches 0. Which evaluates to - sin(2)/cos(2) = - tan 2. Therefore, same result. Thus, regardless of the method, the limit of the logarithm is - tan 2, so the original limit is e^{- tan 2}. Therefore, the answer is e^{-tan 2}. Let me check if there's any possible mistake here. For example, could there be an error in recognizing the form? Let me confirm the initial form: when x approaches 2, the base (cos x / cos 2) approaches 1, and the exponent 1/(x - 2) approaches infinity. So indeed, it's 1^infinity, which is indeterminate. So the steps taken are correct. Another way to think about it is using the standard limit formula: lim_{t->0} (1 + a t + b t^2 + ...)^{1/t} = e^{a}, if higher order terms are negligible. In our substitution, t = x - 2 approaching 0, so (cos(2 + t)/cos 2) = 1 + [ - tan 2 * t - (t^2)/2 ] + ... So the coefficient of t is - tan 2, so the limit would be e^{- tan 2}, which matches. Hence, I'm confident that the answer is e^{- tan 2}.Final AnswerThe limit is boxed{e^{-tan 2}}.</think>To find the limit of the function (lim _{x rightarrow 2}left(frac{cos x}{cos 2}right)^{frac{1}{x-2}}), we recognize it as a (1^infty) indeterminate form. We use the natural logarithm to transform the limit:[L = lim _{x rightarrow 2}left(frac{cos x}{cos 2}right)^{frac{1}{x-2}}]Taking the natural logarithm of both sides, we get:[ln L = lim _{x rightarrow 2} frac{1}{x-2} ln left(frac{cos x}{cos 2}right)]This simplifies to:[ln L = lim _{x rightarrow 2} frac{ln(cos x) - ln(cos 2)}{x - 2}]Recognizing this as the definition of the derivative of (ln(cos x)) at (x = 2), we compute the derivative:[frac{d}{dx} ln(cos x) = -tan x]Evaluating this derivative at (x = 2) gives:[ln L = -tan 2]Exponentiating both sides to solve for (L), we find:[L = e^{-tan 2}]Thus, the limit is (boxed{e^{-tan 2}}).,<think>Okay, so I need to find the limit as x approaches 2 of (cos x / cos 2) raised to the power of 1/(x - 2). Hmm, let me think. Limits with exponents can sometimes be tricky, especially when they're in this form. Maybe I can rewrite it using exponentials and natural logarithms? Because I remember that a^b is the same as e^{b ln a}, right? So maybe taking the natural log first, finding that limit, and then exponentiating the result. That sounds like a plan.Let me set L as the original limit. Then,L = lim_{x→2} (cos x / cos 2)^{1/(x - 2)}Taking the natural logarithm of both sides:ln L = lim_{x→2} [ (1/(x - 2)) * ln (cos x / cos 2) ]So now, I need to compute this limit, which is the limit of [ln(cos x) - ln(cos 2)] divided by (x - 2) as x approaches 2. Wait a second, that looks like the definition of the derivative of ln(cos x) at x = 2. Because the derivative of a function f(x) at a point a is lim_{x→a} [f(x) - f(a)] / (x - a). Here, f(x) = ln(cos x) and a = 2. So this limit is f'(2). Therefore, ln L = f'(2).Alright, so let's compute the derivative of ln(cos x). The derivative of ln(u) is (u')/u, so here u = cos x, so derivative is (-sin x)/cos x = -tan x. Therefore, f'(x) = -tan x. Thus, f'(2) = -tan 2. Therefore, ln L = -tan 2. Then, exponentiating both sides, L = e^{-tan 2}.Wait, let me double-check. The derivative of ln(cos x) is indeed -tan x. Yes, because d/dx ln(cos x) = (1/cos x)(-sin x) = -sin x / cos x = -tan x. So that's correct. Then substituting x = 2, so f'(2) = -tan 2, so ln L = -tan 2, so L = e^{-tan 2}.Alternatively, maybe I can use L’Hospital’s Rule on the limit expression [ln(cos x) - ln(cos 2)] / (x - 2). Let me see. Let me set t = x - 2, so as x approaches 2, t approaches 0. Then the limit becomes [ln(cos(2 + t)) - ln(cos 2)] / t. Then as t approaches 0, this is the derivative of ln(cos(2 + t)) at t = 0, which is the same as before. So the derivative is -tan(2 + t) evaluated at t = 0, so -tan 2. Same result.Alternatively, maybe expand cos x around x = 2 using Taylor series? Let's try that approach for verification.The Taylor series expansion of cos x around x = 2. Let me recall that cos(a + h) = cos a cos h - sin a sin h. So if we let h = x - 2, then x = 2 + h, and as x approaches 2, h approaches 0. So cos x = cos(2 + h) = cos 2 cos h - sin 2 sin h.So cos x / cos 2 = [cos 2 cos h - sin 2 sin h] / cos 2 = cos h - (sin 2 / cos 2) sin h = cos h - tan 2 sin h.Then, the expression (cos x / cos 2)^{1/(x - 2)} becomes [cos h - tan 2 sin h]^{1/h} because h = x - 2.So we need to compute lim_{h→0} [cos h - tan 2 sin h]^{1/h}.Hmm, maybe take the logarithm again:ln L = lim_{h→0} (1/h) * ln(cos h - tan 2 sin h)Let me expand cos h and sin h for small h:cos h ≈ 1 - h²/2 + h^4/24 - ...sin h ≈ h - h^3/6 + h^5/120 - ...So cos h - tan 2 sin h ≈ (1 - h²/2) - tan 2*(h - h^3/6)= 1 - h²/2 - tan 2 * h + (tan 2) * h^3/6But since h is approaching 0, maybe higher-order terms beyond h can be neglected? Wait, but we need to keep terms up to h, perhaps?Wait, let's write the expansion up to h^2:cos h ≈ 1 - h²/2sin h ≈ hTherefore,cos h - tan 2 sin h ≈ 1 - h²/2 - tan 2 * hSo ln(cos h - tan 2 sin h) ≈ ln(1 - tan 2 * h - h²/2)Now, use the expansion ln(1 + ε) ≈ ε - ε²/2 + ε^3/3 - ... where ε = -tan 2 * h - h²/2So,ln(1 - tan 2 h - h²/2) ≈ (-tan 2 h - h²/2) - [(-tan 2 h - h²/2)^2]/2 + ...Let's compute up to the first order in h, maybe:The leading term is -tan 2 h - h²/2. Then the next term is - [ (tan^2 2 h^2 + tan 2 h * h² + h^4 /4 ) ] / 2, but h^4 is negligible. So approximately:- [ tan^2 2 h^2 + ... ] / 2. So up to h^2 terms.But since we are dividing by h (from 1/h), let's see:ln(cos h - tan 2 sin h) ≈ (-tan 2 h - h²/2) - (tan^2 2 h^2)/2 + ...Therefore,(1/h) * ln(...) ≈ [ -tan 2 h - h²/2 - (tan^2 2 h^2)/2 ] / h= -tan 2 - h/2 - (tan^2 2 h)/2 + ... as h approaches 0.Therefore, as h approaches 0, this tends to -tan 2. So again, ln L = -tan 2, so L = e^{-tan 2}. So same result.Therefore, regardless of the method, the limit is e^{-tan 2}.Wait, but just to make sure there's no mistake in the substitution. Let me check again the substitution step.Original limit: x approaching 2, so let h = x - 2, h approaches 0.So the expression becomes [cos(2 + h)/cos 2]^{1/h}Which is [cos 2 cos h - sin 2 sin h)/cos 2]^{1/h}Which is [cos h - tan 2 sin h]^{1/h}Taking ln:(1/h) * ln(cos h - tan 2 sin h). As h approaches 0.Expanding the numerator:cos h - tan 2 sin h ≈ 1 - h²/2 - tan 2 hSo ln(1 - tan 2 h - h²/2) ≈ -tan 2 h - h²/2 - ( (-tan 2 h)^2 ) /2Wait, let me do that again. The expansion of ln(1 + ε) where ε is small is ε - ε²/2 + ε³/3 - ... So here, ε = - tan 2 h - h²/2. Therefore,ln(1 + ε) ≈ ε - ε²/2So substituting:ε = - tan 2 h - h²/2Then,ln(1 + ε) ≈ (- tan 2 h - h²/2) - [(- tan 2 h - h²/2)^2]/2Compute the square term:(- tan 2 h - h²/2)^2 = tan² 2 h² + 2 * tan 2 h * (h²/2) + (h²/2)^2= tan² 2 h² + tan 2 h^3 + h^4 /4So, when h approaches 0, the h^3 and h^4 terms are negligible compared to h². So approximately, the square term is tan² 2 h².Therefore,ln(1 + ε) ≈ (- tan 2 h - h²/2) - (tan² 2 h²)/2Now, multiplying by (1/h):[ (- tan 2 h - h²/2 ) - (tan² 2 h²)/2 ] / h= [ - tan 2 h / h - h²/(2h) - (tan² 2 h²)/(2h) ]= - tan 2 - h/2 - (tan² 2 h)/2Now, taking the limit as h approaches 0:The first term is - tan 2, the second term goes to 0, the third term also goes to 0. So the limit is - tan 2. Hence, ln L = - tan 2, so L = e^{- tan 2}Therefore, this confirms the previous result. So the answer is e^{-tan 2}But wait, just to make sure, maybe I can plug in some numerical value close to 2 and check? Let's see. Let's take x = 2 + 0.001, so h = 0.001Compute cos(2.001)/cos(2) first:But computing cos(2) and cos(2.001) numerically. Let me compute in calculator terms:First, note that 2 radians is approximately 114.59 degrees. Let me use calculator for approximate values.cos(2) ≈ cos(114.59°) ≈ -0.4161cos(2.001) ≈ cos(2 + 0.001). Let's compute this using the Taylor expansion:cos(2 + 0.001) ≈ cos(2) - sin(2)*0.001 - 0.5 * cos(2)*(0.001)^2Compute each term:cos(2) ≈ -0.4161sin(2) ≈ 0.9093So:cos(2.001) ≈ -0.4161 - 0.9093 * 0.001 - 0.5*(-0.4161)*(0.000001)≈ -0.4161 - 0.0009093 + 0.000000208≈ -0.4161 - 0.0009093 ≈ -0.4170093 + 0.000000208 ≈ -0.4170091So cos(2.001)/cos(2) ≈ (-0.4170091)/(-0.4161) ≈ 1.00218So then (1.00218)^{1/0.001} = (1.00218)^{1000}But 1.00218^1000 ≈ e^{1000 * ln(1.00218)} ≈ e^{1000 * 0.00218 - ... } since ln(1+x) ≈ x - x²/2 + ..., so ln(1.00218) ≈ 0.00218 - (0.00218)^2 / 2 ≈ 0.00218 - 0.00000237 ≈ 0.0021776Thus, 1000 * 0.0021776 ≈ 2.1776So e^{2.1776} ≈ 8.82But according to our result, the limit is e^{-tan 2}Compute tan 2: tan(2 radians) ≈ tan(114.59°) ≈ -2.185So e^{-tan 2} = e^{2.185} ≈ 8.89Hmm, the actual value when h=0.001 gives approximately 8.82, and e^{2.185} is approximately 8.89. Close, considering that h=0.001 is not extremely small. Let's try with h=0.0001.x = 2.0001, h=0.0001cos(2.0001) ≈ cos(2) - sin(2)*0.0001 - 0.5*cos(2)*(0.0001)^2≈ -0.4161 - 0.9093*0.0001 - 0.5*(-0.4161)*(0.00000001)≈ -0.4161 - 0.00009093 + 0.00000000208≈ -0.41619093 + 0.00000000208 ≈ -0.41619093So cos(2.0001)/cos(2) ≈ (-0.41619093)/(-0.4161) ≈ 1.000218So (1.000218)^{1/0.0001} = (1.000218)^{10000}ln(1.000218) ≈ 0.000218 - (0.000218)^2 / 2 ≈ 0.000218 - 0.0000000237 ≈ 0.0002179763So 10000 * 0.0002179763 ≈ 2.179763e^{2.179763} ≈ 8.84Which is closer to e^{-tan 2} ≈ e^{2.185} ≈ 8.89. The discrepancy is because even h=0.0001 is not infinitesimal, but as h approaches 0, the value approaches e^{2.185}. So this numerical check supports the analytical result.Therefore, I think the answer is indeed e^{-tan 2}, which is the same as e^{|tan 2|} but since tan 2 is negative (since 2 radians is in the second quadrant where tangent is negative), but wait, tan(2) is actually negative? Wait, 2 radians is approximately 114.59 degrees, which is in the second quadrant (90° to 180°), where sine is positive, cosine is negative, so tan is sine/cosine, so negative. Therefore, tan 2 ≈ -2.185, so -tan 2 ≈ 2.185, so e^{-tan 2} = e^{2.185}. So yes, the numerical approximations are approaching around e^{2.185}, which is approximately 8.89, and our calculation gives e^{-tan 2} which is e^{2.185}, so that's correct.Therefore, the final answer is e^{-tan 2}.Final AnswerThe limit is equal to boxed{e^{-tan 2}}.</think>To find the limit of the function [lim _{x rightarrow 2}left(frac{cos x}{cos 2}right)^{frac{1}{x-2}},]we start by taking the natural logarithm of the expression to simplify the limit. Let ( L ) be the original limit. Then,[ln L = lim _{x rightarrow 2} frac{1}{x-2} ln left( frac{cos x}{cos 2} right).]This can be rewritten as:[ln L = lim _{x rightarrow 2} frac{ln(cos x) - ln(cos 2)}{x - 2}.]Recognizing this as the definition of the derivative of ( ln(cos x) ) at ( x = 2 ), we compute the derivative. The derivative of ( ln(cos x) ) is ( -tan x ). Therefore, evaluating this derivative at ( x = 2 ):[ln L = -tan 2.]Exponentiating both sides to solve for ( L ):[L = e^{-tan 2}.]To verify, we can use a Taylor series expansion around ( x = 2 ). Let ( h = x - 2 ), then as ( x rightarrow 2 ), ( h rightarrow 0 ). Expanding ( cos(2 + h) ) around ( h = 0 ):[cos(2 + h) approx cos 2 cos h - sin 2 sin h.]Thus,[frac{cos(2 + h)}{cos 2} approx cos h - tan 2 sin h.]For small ( h ), this approximates to:[cos h - tan 2 sin h approx 1 - frac{h^2}{2} - tan 2 cdot h.]Taking the natural logarithm and dividing by ( h ):[frac{1}{h} ln left( 1 - tan 2 cdot h - frac{h^2}{2} right) approx -tan 2 - frac{h}{2} - frac{tan^2 2 cdot h}{2}.]As ( h rightarrow 0 ), the higher-order terms vanish, confirming the limit is ( -tan 2 ). Therefore, exponentiating the result gives the final answer:[boxed{e^{-tan 2}}.]]

🔑:## Solutionlim _{x rightarrow 2}left(frac{cos x}{cos 2}right)^{frac{1}{x-2}}=lim _{x rightarrow 2}left(e^{ln left(frac{cos x}{cos 2}right)}right)^{frac{1}{x-2}}==lim _{x rightarrow 2} e^{frac{1}{x-2} cdot ln left(frac{cos x}{cos 2}right)}=exp left{lim _{x rightarrow 2} frac{1}{x-2} cdot ln left(frac{cos x}{cos 2}right)right}=Substitution:x=y+2 Rightarrow y=x-2x rightarrow 2 Rightarrow y rightarrow 0We get:begin{aligned}& =exp left{lim _{y rightarrow 0} frac{1}{(y+2)-2} cdot ln left(frac{cos (y+2)}{cos 2}right)right}= & =exp left{lim _{y rightarrow 0} frac{1}{y} cdot ln left(frac{cos y cdot cos 2-sin y cdot sin 2}{cos 2}right)right}= & =exp left{lim _{y rightarrow 0} frac{1}{y} cdot ln (cos y-sin y cdot tan 2)right}= & =exp left{lim _{y rightarrow 0} frac{1}{y} cdot ln left(1-2 sin ^{2} frac{y}{2}-2 sin frac{y}{2} cdot cos frac{y}{2} cdot tan 2right)right}= & =exp left{lim _{y rightarrow 0} frac{1}{y} cdot ln left(1-2 sin frac{y}{2}left(sin frac{y}{2}+cos frac{y}{2} cdot tan 2right)right)right}=end{aligned}Using the substitution of equivalent infinitesimals:ln left(1-2 sin frac{y}{2}left(sin frac{y}{2}+cos frac{y}{2} cdot tan 2right)right) sim-2 sin frac{y}{2}left(sin frac{y}{2}+cos frac{y}{2} cdot tan 2right), as y rightarrow 0left(-2 sin frac{y}{2}left(sin frac{y}{2}+cos frac{y}{2} cdot tan 2right) rightarrow 0right)We get:=exp left{lim _{y rightarrow 0} frac{1}{y} cdot(-2) sin frac{y}{2}left(sin frac{y}{2}+cos frac{y}{2} cdot tan 2right)right}=Using the substitution of equivalent infinitesimals:We get:=exp left{lim _{y rightarrow 0} frac{1}{y} cdot(-2) cdot frac{y}{2} cdotleft(sin frac{y}{2}+cos frac{y}{2} cdot tan 2right)right}=begin{aligned}& =exp left{lim _{y rightarrow 0}-sin frac{y}{2}-cos frac{y}{2} cdot tan 2right}= & =exp left{-sin frac{0}{2}-cos frac{0}{2} cdot tan 2right}=exp {-0-1 cdot tan 2}= & =exp {-tan 2}=e^{-tan 2}end{aligned}## Problem Kuznetsov Limits 19-4