🧠:[<think>Okay, let me try to figure out this problem. So, we have an integer n that's at least 3, and some nonnegative integers x₁, x₂, ..., x_{n-1}. They satisfy two equations: the sum of all x_i is n, and the weighted sum where each x_i is multiplied by its index is 2n - 2. We need to find the minimal value of the function F, which is another weighted sum where each x_k is multiplied by k(2n - k). Hmm, sounds a bit complicated, but let's break it down.First, let me restate the problem in my own words. We have variables x₁ up to x_{n-1}, all nonnegative integers. The total number of "items" (since the sum is n) is n, and the total "weight" (with weights 1, 2, ..., n-1) is 2n - 2. We need to minimize the sum where each x_k is multiplied by k(2n - k). I think this is an optimization problem with constraints. Since all x_i are integers, maybe we can model it as an integer linear programming problem? But since we need an exact solution, maybe there's a combinatorial approach here.Let me first note that k(2n - k) is a quadratic function in terms of k. Let me see what this coefficient looks like for different k. For k=1, it's 1*(2n -1)=2n -1. For k=2, it's 2*(2n -2)=4n -4. For k=3, 3*(2n -3)=6n -9. Hmm, so as k increases, the coefficient increases linearly in n but quadratically in k? Wait, for fixed n, each term k(2n -k) is quadratic in k. So, for a fixed n, the coefficients for x_k first increase and then decrease? Wait, no, because 2n -k is linear. Let me plug in some numbers. Let's say n=5. Then for k=1: 1*(10 -1)=9; k=2: 2*(10-2)=16; k=3: 3*(10 -3)=21; k=4:4*(10 -4)=24. So for n=5, the coefficients are 9,16,21,24. So they increase as k increases. Similarly, for n=4: k=1:7, k=2:12, k=3:15. So again increasing. Wait, maybe for any k from 1 to n-1, since 2n -k is always positive (since k <=n-1, so 2n -k >=2n - (n-1)=n+1, which is positive for n >=3). So, since 2n -k is positive and k is positive, then k(2n -k) is increasing in k? Let's check derivative for real k: d/dk [k(2n -k)] =2n -2k. So derivative is positive when k <n, zero at k=n, negative when k>n. But k is only up to n-1, so derivative is positive for k from 1 to n-1. So yes, the coefficient k(2n -k) is increasing with k when k <n. Therefore, for k from 1 to n-1, the coefficients in F are strictly increasing. Therefore, to minimize F, we want to have as much as possible of the lower k terms, right? Because higher k terms have higher coefficients. So, to minimize F, given the constraints, we should maximize the number of x₁ and minimize the number of higher x_k's.But we also have constraints on the sum of x_i and the weighted sum. So, the problem is similar to a resource allocation problem where we need to distribute n items into different categories (each category k contributes k to the total weight) with the total weight being 2n -2, and we want to minimize the total cost F, which is higher for higher k's. Therefore, the strategy would be to use as many low k's as possible, but subject to the total weight.But since x_i are nonnegative integers, we need integer solutions. Let's consider how this works.Let me try small n first to see a pattern. Let's take n=3. Then we have x₁ + x₂ = 3, and x₁ + 2x₂ = 2*3 -2 =4. So solving these equations:From first equation, x₁ =3 -x₂. Substitute into the second equation: (3 -x₂) +2x₂=4 → 3 +x₂=4 →x₂=1. Therefore x₁=2. Then F=1*(6 -1)*2 +2*(6 -2)*1=5*2 +4*1=10 +4=14. Is this minimal? Let's see if there's another solution. The equations are x₁ +x₂=3, x₁ +2x₂=4. The solution is unique, so F=14. So minimal is 14.Wait, but the problem states n≥3, so n=3 is allowed. Let's check n=4. Then x₁ +x₂ +x₃=4, x₁ +2x₂ +3x₃=2*4 -2=6. Need to minimize F=1*(8 -1)x₁ +2*(8 -2)x₂ +3*(8 -3)x₃=7x₁ +12x₂ +15x₃.We need to solve x₁ +x₂ +x₃=4 and x₁ +2x₂ +3x₃=6. Subtract the first equation from the second: (x₁ +2x₂ +3x₃) - (x₁ +x₂ +x₃)=6 -4 ⇒x₂ +2x₃=2. So x₂=2 -2x₃. Since x₂ and x₃ are nonnegative integers, x₃ can be 0 or 1. If x₃=0, then x₂=2, and x₁=4 -0 -2=2. Then F=7*2 +12*2 +15*0=14 +24=38. If x₃=1, then x₂=0, x₁=4 -1 -0=3. Then F=7*3 +12*0 +15*1=21 +0 +15=36. So the minimal F is 36. So in this case, using x₃=1 (higher k) gives lower F? Wait, that contradicts my previous thought. Wait, but here when x₃=1, which is higher k, but the coefficient for x₃ is 15, which is higher than 12 for x₂ and 7 for x₁. Wait, but in this case, allowing x₃=1 allowed x₂=0, which might have a net effect. Wait, let's calculate. For x₃=1, x₁=3, so more x₁'s. Since x₁ has a lower coefficient (7) compared to x₂ (12). So even though we have to put 1 in x₃ (coefficient 15), but by reducing x₂ from 2 to 0, which reduces 12*2=24 to 0, but increases x₁ from 2 to 3, which adds 7*(3-2)=7, and adds 15*1=15. So total change is -24 +7 +15= -2. Hence, lower total F. So even though we introduced a higher coefficient, we saved more by reducing a middle coefficient. So maybe the minimal is not just about putting as much as possible into the lowest k, but there's a balance.Hmm, so my initial thought was wrong. So perhaps we need a different approach.Wait, let's think in terms of variables. Let me denote the variables x₁, x₂, ..., x_{n-1}. The two equations are:Sum_{k=1}^{n-1} x_k = n,Sum_{k=1}^{n-1} k x_k = 2n - 2.We need to minimize F = Sum_{k=1}^{n-1} k(2n -k) x_k.Let me see if we can relate F to the given equations. Let me expand F:F = Sum_{k=1}^{n-1} [2n k -k²] x_k = 2n Sum_{k=1}^{n-1} k x_k - Sum_{k=1}^{n-1} k² x_k.But from the second equation, Sum_{k=1}^{n-1} k x_k = 2n -2. Therefore, F = 2n*(2n -2) - Sum_{k=1}^{n-1} k² x_k = 4n² -4n - Sum_{k=1}^{n-1} k² x_k.Therefore, to minimize F, we need to maximize Sum_{k=1}^{n-1} k² x_k.So the problem reduces to maximizing the sum of k² x_k, given that Sum x_k =n and Sum k x_k=2n -2, with x_k nonnegative integers.So, instead of minimizing F, we can think of maximizing the sum of k² x_k. That might be easier. So our goal is to maximize Sum k² x_k.So, how can we maximize this sum? Since k² increases with k, to maximize the sum, we should allocate as much as possible to the highest k terms. Because higher k would contribute more to the sum. So, in contrast to before, now we need to maximize the sum, which is achieved by putting as much as possible into the highest k.But subject to Sum x_k =n and Sum k x_k=2n -2.So this is a different perspective. So if we need to maximize the sum of k² x_k, then higher k's are better. So, given the constraints, we need to maximize the number of high k's.But how?Let me consider variables x_{n-1}, the highest variable. Let me see how much we can allocate to x_{n-1}.Each x_{n-1} contributes (n-1) to the total weight and 1 to the total count. So, if we have m copies of x_{n-1}, then the total weight contributed is m(n-1) and the total count is m. The remaining weight needed is 2n -2 -m(n-1), and the remaining count is n -m.So, the remaining variables x₁, ...,x_{n-2} must satisfy:Sum_{k=1}^{n-2} x_k = n -m,Sum_{k=1}^{n-2} k x_k = 2n -2 -m(n -1).We need these to be nonnegative. So, 2n -2 -m(n -1) ≥0, which implies m ≤ (2n -2)/(n -1)=2. So m can be 0,1,2. Since m must be integer.Wait, (2n -2)/(n -1)=2(n -1)/(n -1)=2. So m can be at most 2. Therefore, maximum possible x_{n-1} is 2. Let's check for n=4, which we did earlier. For n=4, m=1 gives x_{n-1}=1, and remaining variables x₁=3, x₂=0, x₃=1. Which was the case where F=36. Indeed, when we set x_{n-1}=1, we achieved a higher Sum k² x_k=1²*3 +2²*0 +3²*1=3 +0 +9=12. If m=2, then x_{n-1}=2, remaining weight=6 -2*3=0, remaining count=4 -2=2. So x₁ +x₂=2, x₁ +2x₂=0. Then x₁=2, x₂=0. Then Sum k² x_k=1²*2 +2²*0 +3²*2=2 +0 +18=20. But wait, but 2n -2 -m(n -1)=6 -6=0. Then the remaining variables have to satisfy Sum x_k=2 and Sum k x_k=0, which implies all x_k=0 except x₁=2. Then Sum k² x_k=1²*2=2. But then total Sum k² x_k=2 +3²*2=2 +18=20. Wait, but in this case, if m=2, then x_{n-1}=2, but then the remaining variables have x₁=2. So Sum k² x_k=1²*2 +3²*2=2 +18=20. However, when m=1, we had Sum k² x_k=12, which is less. Wait, but in reality, when m=1, the remaining variables have x₁=3 and x₃=1? Wait, no, for m=1 in n=4, x₃=1, x₁=3, and x₂=0. So Sum k² x_k=1²*3 +2²*0 +3²*1=3 +0 +9=12. But if m=2, then x₃=2, and remaining variables x₁=2, x₂=0. So Sum k² x_k=1²*2 +3²*2=2 +18=20. Which is higher. But why in the previous calculation, when m=2, the F was higher? Wait, no, in n=4, m=2 is allowed?Wait, in n=4, x₃=2, x₁=2, x₂=0. Then Sum x_k=2 +0 +2=4, Sum k x_k=1*2 +2*0 +3*2=2 +0 +6=8. But 2n -2=6, not 8. So that's invalid. Wait, so m=2 would lead to the remaining variables needing to have Sum x_k=4 -2=2 and Sum k x_k=6 -3*2=0. So x₁=2, x₂=0. Then Sum k x_k=1*2 +2*0=2, which is not 0. Wait, this is a contradiction. So my previous assumption was wrong.Wait, no. Wait, if m=2, then x_{n-1}=2. Then Sum k x_k= (n-1)*m + Sum_{k=1}^{n-2}k x_k=3*2 + Sum_{k=1}^{2}k x_k=6 + (x₁ +2x₂). But the total Sum k x_k needs to be 2n -2=6. Therefore, 6 +x₁ +2x₂=6 ⇒x₁ +2x₂=0. Therefore x₁=0, x₂=0. But Sum x_k= x₁ +x₂ +x₃=0 +0 +2=2≠4. Wait, so m=2 would require Sum x_k=2, but Sum x_k=0 +0 +2=2. So actually, in n=4, if we set x₃=2, then x₁ +x₂=2, but also x₁ +2x₂=0. Therefore, x₁=0, x₂=0, but then x₁ +x₂=0≠2. Contradiction. Therefore, m=2 is not possible for n=4.So my mistake was that I thought m can be up to 2, but for n=4, m=2 leads to an impossible solution. Therefore, m can be at most floor((2n -2)/(n -1))=2, but in reality, when n=4, even m=1 is the maximum allowed. Wait, let's recast.Given the total weight needed is 2n -2. Each x_{n-1} contributes (n -1) to the weight and 1 to the count. So maximum possible m is floor((2n -2)/(n -1))=2, since (2n -2)/(n -1)=2. So m can be 0,1,2.But in n=4, when m=2, the remaining weight is 6 -2*3=0, but remaining count is 4 -2=2. So we need 2 variables with total weight 0. The only way is x₁=2, x₂=0, but then the weight is 2*1 +0*2=2≠0. Therefore, impossible. Therefore, the maximum m is 1 for n=4. Wait, but (2n -2)/(n -1)=2, so why is m=2 impossible? Because even though the weight allows m=2, the count doesn't. So there's a conflict between the remaining count and the remaining weight.Therefore, the actual maximum m is the minimum between floor((2n -2)/(n -1))=2 and the count n. But that's not directly. Wait, perhaps a better way is to consider both constraints. For m to be feasible, we need:After allocating m to x_{n-1}, the remaining weight is 2n -2 -m(n -1) and remaining count is n -m. The remaining variables must satisfy Sum_{k=1}^{n-2}k x_k=remaining weight and Sum_{k=1}^{n-2}x_k=remaining count. The minimal possible weight for the remaining count is remaining count (if all are x₁=1), and the maximal possible weight is (n -2)*remaining count. Therefore, to have a feasible solution, we need:remaining count ≤ remaining weight ≤ (n -2)*remaining count.But remaining weight=2n -2 -m(n -1),remaining count=n -m.So,n -m ≤ 2n -2 -m(n -1) ≤ (n -2)(n -m).Let's compute the left inequality:n -m ≤ 2n -2 -m(n -1)Bring all terms to left:n -m -2n +2 +m(n -1) ≤0-n +2 +m(n -1 -1) ≤0-n +2 +m(n -2) ≤0Similarly, the right inequality:2n -2 -m(n -1) ≤ (n -2)(n -m)Let me compute left inequality first:-n +2 +m(n -2) ≤0=> m(n -2) ≤n -2Since n≥3, n -2 ≥1. So,m ≤(n -2)/(n -2)=1So m ≤1.Therefore, m can be at most 1. Hence, for any n≥3, m≤1.Wait, that's a key insight. Let me verify with n=4. Then m≤1, which matches the earlier result. For n=5:Left inequality: m(n -2) ≤n -2 →m≤1. Therefore, m can be at most 1. Hence, maximum m=1 for all n≥3.Therefore, regardless of n, we can have at most m=1 for x_{n-1}.So, x_{n-1}=1, then remaining variables:Sum x_k=n -1,Sum k x_k=2n -2 - (n -1)=2n -2 -n +1=n -1.So, remaining variables must satisfy Sum x_k= n -1 and Sum k x_k= n -1. Therefore, the remaining variables have both sum and weighted sum equal to n -1. How can that be achieved?The minimal possible weighted sum for Sum x_k= n -1 is n -1 (if all x₁= n -1). So in this case, the weighted sum is exactly n -1, which is required. Therefore, the only solution is x₁= n -1 and all other x_k=0. Therefore, the remaining variables must be x₁= n -1, and x₂=...=x_{n-2}=0.Therefore, the only feasible solution when m=1 is x_{n-1}=1, x₁= n -1, and all others zero. Therefore, the total variables are x₁= n -1, x_{n-1}=1, rest zero.Hence, the maximum Sum k² x_k=1²*(n -1) + (n -1)^2 *1= (n -1) + (n -1)^2= (n -1)(n). Therefore, F=4n² -4n -n(n -1)=4n² -4n -n² +n=3n² -3n.Wait, let's compute F in terms of Sum k² x_k. Since F=4n² -4n -Sum k² x_k, then if Sum k² x_k= n(n -1), then F=4n² -4n -n(n -1)=4n² -4n -n² +n=3n² -3n. Therefore, F=3n(n -1). For n=3, 3*3*2=18, but earlier for n=3, F=14. Wait, contradiction. Therefore, something is wrong here.Wait, in n=3, when m=1, x₂=1, x₁=2. Then Sum k² x_k=1²*2 +2²*1=2 +4=6. Therefore, F=4*9 -4*3 -6=36 -12 -6=18. But earlier, when n=3, we found F=14. So conflict.Wait, wait, for n=3, the equations are x₁ +x₂=3, x₁ +2x₂=4. The solution is x₁=2, x₂=1. Then F=1*(6 -1)*2 +2*(6 -2)*1=5*2 +4*1=10 +4=14. But according to the formula above, Sum k² x_k=1²*2 +2²*1=2 +4=6, then F=4n² -4n -6=4*9 -12 -6=36 -12 -6=18, which is different. So where is the mistake?Wait, let's re-derive F. F was defined as Sum_{k=1}^{n-1}k(2n -k)x_k. Which was expanded as 2n Sum kx_k - Sum k² x_k. Then, since Sum kx_k=2n -2, this becomes 2n(2n -2) - Sum k² x_k=4n² -4n - Sum k² x_k. Therefore, F=4n² -4n - Sum k² x_k. Therefore, if Sum k² x_k=6, then F=36 -12 -6=18. But in reality, for n=3, F=14. Therefore, my expansion is wrong?Wait, let's compute F directly for n=3. F=1*(6 -1)x₁ +2*(6 -2)x₂=5x₁ +8x₂. When x₁=2, x₂=1: 5*2 +8*1=10 +8=18. Wait, but earlier I thought F=14. Wait, no, in the problem statement for n=3, the function is Sum_{k=1}^{n-1}k(2n -k)x_k. For n=3, n-1=2, so it's k=1 and 2. So 1*(6 -1)x₁ +2*(6 -2)x₂=5x₁ +8x₂. With x₁=2, x₂=1: 5*2 +8*1=10 +8=18. But in my initial calculation, I thought F=14, but that was a mistake. Wait, no, hold on. Wait, the problem says "Find the minimal value of F(x₁,x₂,…,x_n) = ∑_{k=1}^{n−1} k(2n−k)x_k." Wait, x_n is not present? Wait, the variables are x₁, x₂,…,x_{n−1}. So for n=3, variables x₁,x₂. So indeed, F=5x₁ +8x₂. For the solution x₁=2, x₂=1: 10 +8=18. So my previous calculation was wrong when I thought F=14. Wait, why did I think that? Maybe I miscalculated.Wait, no. Wait, for n=3, 2n=6. So k=1:1*(6 -1)=5; k=2:2*(6 -2)=8. So x₁=2, x₂=1: 5*2 +8*1=10 +8=18. Therefore, correct value is 18, not 14. So I must have made a mistake earlier when I thought it was 14. Therefore, the formula F=3n(n -1) gives for n=3:3*3*2=18, which matches. For n=4, 3*4*3=36, which matches the earlier result. Therefore, that formula is correct.Therefore, in the case where we set x_{n-1}=1 and x₁= n -1, the Sum k² x_k=1²*(n -1) + (n -1)^2*1=(n -1)+(n -1)^2=n(n -1). Therefore, F=4n² -4n -n(n -1)=4n² -4n -n² +n=3n² -3n=3n(n -1). Therefore, this is the minimal F?But wait, is this the only solution? Suppose there exists another solution where x_{n-1}=0. Then, we need to have Sum x_k=n and Sum k x_k=2n -2. But if x_{n-1}=0, then all variables are x₁,...,x_{n-2}. Then, the maximum possible Sum k x_k is (n -2)*n. But 2n -2 vs (n -2)*n. For n≥3, (n -2)*n= n² -2n. For n=3, it's 3² -6=3, which is greater than 2n -2=4. Wait, n=3, x_{n-1}=x₂=0, then Sum k x_k= x₁ +2x₂= x₁. But x₁ +x₂=3, so x₁=3, Sum k x_k=3. But 2n -2=4, so impossible. Therefore, for n=3, we cannot set x_{n-1}=0. For n=4, if x₃=0, then Sum k x_k= x₁ +2x₂ ≤3x₁ +3x₂=3*4=12. But 2n -2=6. So possible. Wait, but we need to reach 6. Let's see. For n=4, x₃=0. Then x₁ +x₂ +x₃=4 ⇒x₁ +x₂=4. x₁ +2x₂=6. Subtract: x₂=2. Then x₁=2. Therefore, x₁=2, x₂=2, x₃=0. Then Sum k² x_k=1²*2 +2²*2=2 +8=10. Therefore, F=4*16 -16 -10=64 -16 -10=38. Which was our first solution for n=4. So if we set x_{n-1}=0, we can still get a solution, but with higher F (38 vs 36). So, in this case, even though we can set x_{n-1}=0, the F is higher. Therefore, the minimal F occurs when we set x_{n-1}=1 and x₁= n -1.But wait, in n=4, we had another solution where x₃=1, x₁=3, x₂=0. Which gave Sum k² x_k=1²*3 +3²*1=3 +9=12. Hence, F=4*16 -16 -12=64 -16 -12=36, which is less than 38. Therefore, how was this solution obtained? Because in the previous logic, we thought that m=1 gives x_{n-1}=1, and x₁=n -1. But in n=4, x_{n-1}=1, x₁=3, but there is another variable x₂=0. Wait, but according to the previous logic, after setting m=1, the remaining variables must satisfy Sum x_k= n -1=3, Sum k x_k= n -1=3. Which would require x₁=3, x₂=0, x₃=0 (but x₃ is already set to 1). Wait, no, in the case where x_{n-1}=1, then remaining variables are x₁ to x_{n-2}. For n=4, remaining variables are x₁, x₂. So, Sum x₁ +x₂=3, Sum x₁ +2x₂=3. Then solving gives x₁=3, x₂=0. Hence, the solution x₁=3, x₂=0, x₃=1. So indeed, Sum k² x_k=1²*3 +3²*1=3 +9=12. Then F=4n² -4n -12=64 -16 -12=36. Which is correct.But in n=3, when we set x_{n-1}=1 (x₂=1), then remaining variables x₁=2. Therefore, Sum k² x_k=1²*2 +2²*1=2 +4=6. Then F=4*9 -12 -6=36 -12 -6=18, which is correct.So, the general solution is x_{n-1}=1, x₁= n -1, rest zero. Hence, Sum k² x_k=1²*(n -1) + (n -1)^2*1= n(n -1). Therefore, F=4n² -4n -n(n -1)=3n² -3n. This seems to hold.But wait, is this the only possible solution? Let me check for n=5.For n=5, we need x₁ +x₂ +x₃ +x₄=5,x₁ +2x₂ +3x₃ +4x₄=8.If we set x₄=1, then remaining:x₁ +x₂ +x₃=4,x₁ +2x₂ +3x₃=8 -4=4.Subtract: x₂ +2x₃=0. Therefore, x₂=0, x₃=0. Hence, x₁=4. So solution x₄=1, x₁=4. Then Sum k² x_k=1²*4 +4²*1=4 +16=20. Then F=4*25 -20 -20=100 -20 -20=60. Alternatively, if we don't set x₄=1, perhaps another solution.Suppose x₄=0. Then x₁ +x₂ +x₃=5,x₁ +2x₂ +3x₃=8.Subtract: x₂ +2x₃=3. So possible solutions:x₃=0: x₂=3, x₁=2. Sum k² x_k=1²*2 +2²*3=2 +12=14. F=4*25 -20 -14=100 -20 -14=66.x₃=1: x₂=1, x₁=3. Sum k² x_k=1*3 +2*1 +3*1=3 +2 +3=8. F=100 -20 -8=72.x₃=2: x₂= -1 invalid.So, maximum Sum k² x_k is 14 when x₄=0, x₂=3, x₁=2. Hence, F=66. Which is higher than the case when x₄=1, F=60. Alternatively, is there a solution with x₄=2?x₄=2: Then x₁ +x₂ +x₃=3,Sum k x_k=8 -4*2=0. Therefore, x₁=3, x₂=x₃=0. Sum k² x_k=1*3 +4²*2=3 +32=35. Then F=4*25 -20 -35=100 -20 -35=45. Wait, this is better? But wait, if x₄=2, then Sum x_k=3 +0 +0 +2=5. Sum k x_k=1*3 +4*2=3 +8=11≠8. So invalid.Wait, conflict. So if x₄=2, remaining Sum k x_k=8 -4*2=0. So x₁ +2x₂ +3x₃=0. Hence, x₁=x₂=x₃=0. But x₁ +x₂ +x₃=5 -2=3≠0. Contradiction. Therefore, x₄=2 is invalid. Similarly, x₄=1 gives valid solution. Hence, minimal F=60. Which corresponds to 3n(n -1)=3*5*4=60. Correct.Therefore, the formula seems to hold. Therefore, the minimal F is 3n(n -1).But wait, let's check another n=6.For n=6, the minimal F should be 3*6*5=90.Let's verify. Set x₅=1, then remaining variables x₁=5, rest zero. Sum k² x_k=1²*5 +5²*1=5 +25=30. Then F=4*36 -24 -30=144 -24 -30=90. Correct.Alternatively, if we set x₅=0, then x₁ +x₂ +x₃ +x₄=6,x₁ +2x₂ +3x₃ +4x₄=10.Subtract: x₂ +2x₃ +3x₄=4.Possible solutions:Maximizing Sum k² x_k would require putting as much as possible into higher k. Let's try x₄=1: Then x₂ +2x₃=1. So x₃=0, x₂=1. Then x₁=6 -1 -0 -1=4. Sum k² x_k=1²*4 +2²*1 +4²*1=4 +4 +16=24. Then F=4*36 -24 -24=144 -24 -24=96. Which is higher than 90.x₄=0: Then x₂ +2x₃=4. To maximize Sum k² x_k, set x₃=2, x₂=0, x₁=6 -0 -2=4. Sum k² x_k=1*4 +3²*2=4 +18=22. F=144 -24 -22=98.x₄=1, x₃=1, x₂=1: x₂ +2x₃=1 +2=3, remaining x₁=6 -1 -1 -1=3. Sum k² x_k=1*3 +2*1 +3*1 +4*1=3 +2 +3 +4=12. F=144 -24 -12=108.So, the minimal F when x₅=1 is 90, which is indeed less than other solutions. Therefore, the formula holds.Hence, in general, the minimal F is 3n(n -1). This is achieved by setting x_{n-1}=1 and x₁=n -1, with all other variables zero. This satisfies both equations:Sum x_k= (n -1) +1= n,Sum k x_k=1*(n -1) + (n -1)*1= n -1 +n -1=2n -2.Therefore, the minimal value is 3n(n -1).Final AnswerThe minimal value of ( F ) is boxed{3n(n - 1)}.</think>Given an integer ( n geq 3 ) and nonnegative integers ( x_1, x_2, ldots, x_{n-1} ) satisfying the equations:[begin{cases}x_1 + x_2 + cdots + x_{n-1} = n, x_1 + 2x_2 + cdots + (n-1)x_{n-1} = 2n - 2,end{cases}]we need to find the minimal value of ( F(x_1, x_2, ldots, x_{n-1}) = sum_{k=1}^{n-1} k(2n - k)x_k ).To solve this, we first express ( F ) in terms of the given constraints. Expanding ( F ):[F = sum_{k=1}^{n-1} k(2n - k)x_k = 2n sum_{k=1}^{n-1} kx_k - sum_{k=1}^{n-1} k^2 x_k.]Using the second constraint ( sum_{k=1}^{n-1} kx_k = 2n - 2 ), we substitute this into the expression for ( F ):[F = 2n(2n - 2) - sum_{k=1}^{n-1} k^2 x_k = 4n^2 - 4n - sum_{k=1}^{n-1} k^2 x_k.]To minimize ( F ), we need to maximize ( sum_{k=1}^{n-1} k^2 x_k ). By analyzing the constraints and the structure of the problem, we find that the optimal solution occurs when we allocate as much as possible to the highest ( k ) term while satisfying both constraints. This leads to setting ( x_{n-1} = 1 ) and ( x_1 = n - 1 ), with all other ( x_k = 0 ).For this solution, the sum ( sum_{k=1}^{n-1} k^2 x_k ) is calculated as:[1^2 cdot (n - 1) + (n - 1)^2 cdot 1 = (n - 1) + (n - 1)^2 = (n - 1)n.]Substituting this back into the expression for ( F ):[F = 4n^2 - 4n - n(n - 1) = 4n^2 - 4n - n^2 + n = 3n^2 - 3n = 3n(n - 1).]Thus, the minimal value of ( F ) is (boxed{3n(n - 1)}).,<think>Okay, let's try to tackle this problem. So, we have an integer n ≥ 3, and we need to find the minimal value of the function F(x₁, x₂, ..., x_{n-1}) = Σₖ=1^{n-1} k(2n - k)x_k. The constraints are that the sum of x₁ + x₂ + ... + x_{n-1} equals n, and the weighted sum x₁ + 2x₂ + ... + (n-1)x_{n-1} equals 2n - 2. All the x_k are nonnegative integers. Hmm. So, it's an optimization problem where we need to minimize F given those two linear constraints. Let me recall, in linear programming, the optimal solution usually lies at a vertex of the feasible region, but here the variables are integers, so it's more like integer linear programming. But maybe since the problem is structured, there's a combinatorial way to approach it without getting into the complexities of integer programming.First, let's write down the function F more clearly. Each term in the sum is k(2n - k)x_k. Let's see, expanding that, it's (2n k - k²)x_k. So, F = Σ [2n k x_k - k² x_k] = 2n Σ k x_k - Σ k² x_k. Wait, but from the constraints, we have Σ k x_k = 2n - 2. So, substituting that into F, we get F = 2n(2n - 2) - Σ k² x_k. Therefore, minimizing F is equivalent to maximizing Σ k² x_k. So, if I can find the maximum value of Σ k² x_k given the constraints, then subtracting that from 2n(2n - 2) will give me the minimal F.Alright, so the problem reduces to maximizing Σ k² x_k with the same constraints: Σ x_k = n and Σ k x_k = 2n - 2. That seems more manageable. Now, how can we maximize the sum of k² x_k? Since the coefficients k² increase with k, to maximize the sum, we want to allocate as much as possible to the variables with higher k. But subject to the constraints. Let's think.We have two constraints: the total number of variables (sum x_k) is n, and the total weighted sum (sum k x_k) is 2n - 2. Let me denote the variables as x₁, x₂, ..., x_{n-1}. To maximize Σ k² x_k, we need to maximize the number of higher k terms. So, ideally, we would want as many x_{n-1} as possible, since (n-1)² is the largest coefficient. But we have to satisfy the constraints. Let's see.Suppose we try to set x_{n-1} as large as possible. Let's denote x_{n-1} = t. Then, the remaining variables x₁, x₂, ..., x_{n-2} must satisfy:Sum_{k=1}^{n-2} x_k = n - tSum_{k=1}^{n-2} k x_k = (2n - 2) - (n - 1)tSo, we need to solve for integers x₁, ..., x_{n-2} such that their sum is n - t and their weighted sum is (2n - 2) - (n - 1)t. But we need to ensure that the weighted sum is achievable. Let's check the feasibility. The minimal possible weighted sum given that sum x = n - t is achieved when all the x's are as small as possible. Since the minimal weighted sum would be if we set all x₁ = n - t, and x₂=...=x_{n-2}=0, which gives weighted sum 1*(n - t). The maximal weighted sum (if possible) would be if we set x_{n-2} = n - t, giving (n - 2)*(n - t). But we need the required weighted sum, which is (2n - 2) - (n - 1)t, to be between 1*(n - t) and (n - 2)*(n - t). So:(n - t) ≤ (2n - 2) - (n - 1)t ≤ (n - 2)(n - t)Let's solve these inequalities for t.First inequality:(n - t) ≤ (2n - 2) - (n - 1)tMultiply out:n - t ≤ 2n - 2 - (n - 1)tBring all terms to left:n - t - 2n + 2 + (n - 1)t ≤ 0(-n + 2) + (n - 2)t ≤ 0(n - 2)t ≤ n - 2Since n ≥ 3, n - 2 ≥ 1. So, divide both sides:t ≤ 1Second inequality:(2n - 2) - (n - 1)t ≤ (n - 2)(n - t)Expand RHS:(n - 2)(n - t) = n(n - 2) - (n - 2)tSo, inequality:2n - 2 - (n - 1)t ≤ n(n - 2) - (n - 2)tBring all terms to left:2n - 2 - (n - 1)t - n(n - 2) + (n - 2)t ≤ 0Simplify term by term:2n - 2 - n(n - 2) + [ - (n - 1)t + (n - 2)t ] = 2n - 2 - n² + 2n + (-n + 1 + n - 2)t = (2n + 2n) - 2 - n² + (-1)t = 4n - 2 - n² - t ≤ 0So,-n² + 4n - 2 - t ≤ 0=> t ≥ -n² + 4n - 2But t is a nonnegative integer. The right-hand side is -n² + 4n - 2. Let's see for n ≥ 3:For n = 3: -9 + 12 - 2 = 1For n = 4: -16 + 16 - 2 = -2So, starting from n=3, this inequality gives t ≥ 1 for n=3, and t ≥ negative numbers for higher n, which is always true since t is nonnegative. So, combining both inequalities, from the first inequality, t ≤ 1, and from the second inequality, for n=3, t ≥1. Therefore, for n=3, t must be exactly 1. For n ≥4, the second inequality is automatically satisfied since t ≥0, and first inequality says t ≤1. So, t can be 0 or 1.Wait, but for n=3, the constraints for the remaining variables would be sum x₁ + x₂ = 3 - t and 1x₁ + 2x₂ = 6 - 2 - 2t = 4 - 2t. But for n=3, if t=1, then sum x₁ + x₂ = 2, and 1x₁ + 2x₂ = 4 - 2*1 = 2. So, solving x₁ + x₂ = 2 and x₁ + 2x₂ = 2. Subtract: x₂ = 0, so x₁ = 2. So, possible. Then Σ k² x_k = 1²*2 + 2²*0 + 3²*1 = 2 + 0 + 9 = 11. Then F = 2*3*(6 - 2) - 11 = 6*4 -11=24 -11=13. Wait, but let's check if that's minimal. Alternatively, if t=0 for n=3, then sum x₁ +x₂ =3, and x₁ +2x₂=6-0=6. Then solving x₁ + x₂=3 and x₁ +2x₂=6. Subtract: x₂=3, so x₁=0. Then Σk²x_k=1²*0 +2²*3 +3²*0=0 +12 +0=12. Then F=24 -12=12, which is lower. Wait, but earlier conclusion was for n=3, t must be 1, but if t=0 is allowed, maybe the first inequality wasn't correctly derived.Wait, let's go back. For n=3, when solving the first inequality:Original inequality:(n - t) ≤ (2n - 2) - (n - 1)tFor n=3:(3 - t) ≤ (6 - 2) - 2t => 3 - t ≤ 4 - 2t => 3 - t -4 + 2t ≤0 => (-1 + t) ≤0 => t ≤1.So t can be 0 or 1. However, for n=3, the second inequality:t ≥ -n² +4n -2 = -9 +12 -2=1. So t ≥1.Therefore, t must be exactly 1 when n=3. So, how come when we set t=0, we get a feasible solution? Let's check the feasibility. If t=0, then sum x₁ +x₂=3, and sum x₁ +2x₂=2n -2=4. Wait, for n=3, 2n -2=4. So, x₁ +2x₂=4, x₁ +x₂=3. Subtract: x₂=1. Then x₁=2. So that's feasible. Then Σk²x_k=1²*2 +2²*1=2 +4=6. Wait, but earlier calculation when t=0 for n=3 gave Σk²x_k=12. Wait, no, maybe I messed up variables. Wait, when n=3, n-1=2, so variables are x₁ and x₂. If t is x_{n-1}=x₂. So, if t=0, then x₂=0, but in that case, solving x₁ +x₂=3 and x₁ +2x₂=4 would require x₂=1, x₁=2. So, actually, x₂=1, which is t=1? Wait, no. Wait, the variables are x₁, x₂ for n=3. If t is x_{n-1}=x₂. So, if we set x₂= t, then sum x₁=3 - t, and sum x₁ +2x₂=4. Then x₁=3 - t, so 3 - t +2t=4 => 3 + t=4 => t=1. Therefore, x₂=1, x₁=2. Therefore, when n=3, t must be 1. So, if we set t=0, we can't satisfy the equations. Wait, but when I tried to set t=0, I had x₁ +x₂=3 and x₁ +2x₂=4. Solving gives x₂=1, which is t=1. Therefore, x₂ can't be 0. So, actually, the minimal t is 1. Therefore, for n=3, t=1 is the only possible value. Therefore, for n=3, x₂=1, x₁=2, leading to Σk²x_k=1²*2 +2²*1=2+4=6. Then F=2*3*(4) -6=24 -6=18. Wait, hold on, previously, I thought 2n(2n -2)=6*4=24, and Σk²x_k=6, so F=24 -6=18. Wait, but in the beginning, when I considered t=0, I thought Σk²x_k=12, but that was a miscalculation. So, actually, for n=3, the maximum Σk²x_k is 6, so F=24 -6=18. But let's check with the variables: x₁=2, x₂=1. Then F=1*(6 -1)*2 + 2*(6 -2)*1= (5*2)+(4*2)=10 +8=18. Correct. So that seems right.But maybe there's a different allocation. Wait, for n=3, the variables must satisfy x₁ +x₂=3 and x₁ +2x₂=4. So, x₂=1, x₁=2. There's no other solution. Therefore, that's the only possible solution. So, the maximum Σk²x_k=6, leading to F=18. Therefore, for n=3, minimal F is 18.But the problem says n ≥3. Let's check for n=4. Maybe we can see a pattern.For n=4, the constraints are x₁ +x₂ +x₃=4 and x₁ +2x₂ +3x₃=2*4 -2=6. We need to maximize Σ k²x_k =1²x₁ +2²x₂ +3²x₃.To maximize this sum, we want to maximize the higher k terms, so set x₃ as large as possible. Let t=x₃. Then x₁ +x₂=4 -t and x₁ +2x₂=6 -3t. Let's solve for x₁ and x₂.From x₁ +x₂=4 -t and x₁ +2x₂=6 -3t. Subtracting the first equation from the second: x₂=6 -3t - (4 -t)=2 -2t. Then x₁=4 -t -x₂=4 -t - (2 -2t)=2 +t.But x₁ and x₂ must be nonnegative integers. Therefore:x₂=2 -2t ≥0 ⇒ 2 -2t ≥0 ⇒ t ≤1.Also, x₁=2 + t ≥0, which is always true since t is nonnegative. Since t is integer, t can be 0 or 1.If t=1: Then x₂=2 -2*1=0, x₁=2 +1=3. So variables x₁=3, x₂=0, x₃=1. Then Σk²x_k=1*3 +4*0 +9*1=3 +0 +9=12. Then F=2*4*(2*4 -2) -12=8*6 -12=48 -12=36.If t=0: Then x₂=2 -0=2, x₁=2 +0=2. Variables x₁=2, x₂=2, x₃=0. Σk²x_k=1*2 +4*2 +9*0=2 +8 +0=10. Then F=8*6 -10=48 -10=38.Therefore, maximum Σk²x_k is 12 when t=1, leading to minimal F=36. So, for n=4, the minimal F is 36.Wait, but let's check if there are other possible allocations. Suppose we set x₃=1, which gives the maximum coefficient. If x₃=1, then x₁=3, x₂=0. Alternatively, if x₃=0, then x₁=2, x₂=2. But maybe another combination? For example, x₂=1, x₃=1. Then x₁=4 -1 -1=2. Then check if x₁ +2x₂ +3x₃=2 +2 +3=7, which is more than 6. Not allowed. So, no. Therefore, only possible t=0 or 1. So, indeed, the maximum Σk²x_k is 12, leading to F=36.So, for n=3, minimal F=18; n=4, F=36. Let's check for n=5 to see a pattern.For n=5, constraints: x₁ +x₂ +x₃ +x₄=5 and x₁ +2x₂ +3x₃ +4x₄=2*5 -2=8. Maximize Σk²x_k=1x₁ +4x₂ +9x₃ +16x₄.To maximize, set x₄ as large as possible. Let t=x₄. Then x₁ +x₂ +x₃=5 -t and x₁ +2x₂ +3x₃=8 -4t.We need to solve for x₁, x₂, x₃. Let's express in terms of variables. Let’s denote x₁ = a, x₂ = b, x₃ = c. Then:a + b + c =5 - ta + 2b +3c =8 -4tSubtract the first equation from the second:( a + 2b +3c ) - ( a + b + c ) = (8 -4t) - (5 -t)Which gives:b + 2c = 3 -3tHence, b =3 -3t -2cBut since b ≥0, 3 -3t -2c ≥0. Also, a =5 - t -b -c=5 - t - (3 -3t -2c) -c=5 -t -3 +3t +2c -c=2 +2t +cSo, a=2 +2t +c. Since a ≥0, this is always true as long as c is nonnegative.Now, since all variables a, b, c must be nonnegative integers, we can express the constraints as:From b =3 -3t -2c ≥0:3 -3t -2c ≥0 ⇒ 3 -3t ≥2c ⇒ c ≤ (3 -3t)/2Since c must be a nonnegative integer, we can find possible values of t and c.But t is x₄, so t is nonnegative integer. Let's check possible t.To maximize Σk²x_k, we need to set t as large as possible. Let's try t=1.If t=1:Then b =3 -3(1) -2c=0 -2cSo, 0 -2c ≥0 ⇒ -2c ≥0 ⇒ c ≤0. Therefore, c=0.Then b=0 -2*0=0.a=2 +2*1 +0=4.Check variables: a=4, b=0, c=0, t=1. Sum:4+0+0+1=5. Weighted sum:4*1 +0*2 +0*3 +1*4=4 +0 +0 +4=8. Correct.Σk²x_k=1*4 +4*0 +9*0 +16*1=4 +0 +0 +16=20. Then F=2*5*(10 -2) -20=10*8 -20=80 -20=60.If t=2:Then b=3 -3*2 -2c=3 -6 -2c=-3 -2c ≥0. Which is impossible since -3 -2c is negative. So t=2 is invalid.If t=0:Then b=3 -0 -2c=3 -2c ≥0 ⇒ c ≤1.5 ⇒ c=0 or 1.If c=0: Then b=3, a=2 +0 +0=2. So variables a=2, b=3, c=0, t=0. Check sum:2+3+0+0=5. Weighted sum:2*1 +3*2 +0*3 +0*4=2 +6 +0 +0=8. Correct.Σk²x_k=1*2 +4*3 +9*0 +16*0=2 +12 +0 +0=14.If c=1: Then b=3 -2*1=1, a=2 +0 +1=3. Variables a=3, b=1, c=1, t=0. Check sum:3 +1 +1 +0=5. Weighted sum:3*1 +1*2 +1*3 +0*4=3 +2 +3 +0=8. Correct.Σk²x_k=1*3 +4*1 +9*1 +16*0=3 +4 +9 +0=16.So, for t=0, the maximum Σk²x_k is 16 when c=1. So compare with t=1, which gives Σk²x_k=20. Therefore, t=1 is better, leading to F=60.Therefore, for n=5, minimal F=60.So, for n=3, F=18; n=4, F=36; n=5, F=60. Let's see if there's a pattern here.18 = 3*636 = 4*960 =5*12Hmm, 6,9,12... which are 3*2, 3*3, 3*4. So, 3(n-1). For n=3:3*2=6, n=3: F=3*6=18; n=4:4*9=36=4*(3*3)=12*3; Wait, maybe not. Alternatively, 18=3*6=3*(3*2), 36=4*9=4*(3*3), 60=5*12=5*(3*4). So, F= n*3*(n-1). Let's check:For n=3: 3*3*2=18 ✔️n=4:4*3*3=36 ✔️n=5:5*3*4=60 ✔️So, the pattern seems F=3n(n-1). Let me test this hypothesis for n=6.For n=6, constraints: x₁ +x₂ +x₃ +x₄ +x₅=6 and x₁ +2x₂ +3x₃ +4x₄ +5x₅=10. Maximize Σk²x_k=1x₁ +4x₂ +9x₃ +16x₄ +25x₅.Maximizing Σk²x_k, set x₅ as large as possible. Let t=x₅. Then:x₁ +x₂ +x₃ +x₄=6 -tx₁ +2x₂ +3x₃ +4x₄=10 -5tWe need to solve:a + b + c + d =6 -t (where a=x₁, b=x₂, c=x₃, d=x₄)a +2b +3c +4d=10 -5tSubtract first equation from the second:b +2c +3d=4 -4tThus, b=4 -4t -2c -3dSince b ≥0, 4 -4t -2c -3d ≥0.Also, a=6 -t -b -c -d=6 -t - (4 -4t -2c -3d) -c -d=6 -t -4 +4t +2c +3d -c -d=2 +3t +c +2dSo, a=2 +3t +c +2d ≥0. Which is always true if t,c,d nonnegative.To maximize Σk²x_k=1a +4b +9c +16d +25t, we need to set t as large as possible. Let's check possible t.Start with t=2:Then, b=4 -8 -2c -3d= -4 -2c -3d. Since b must be nonnegative, this is impossible. So t=2 invalid.t=1:Then, b=4 -4 -2c -3d=0 -2c -3d. So, b= -2c -3d. But b must be ≥0, so -2c -3d ≥0 ⇒ c=d=0. Then b=0. Then a=2 +3*1 +0 +0=5. So variables: a=5, b=0, c=0, d=0, t=1. Check sum:5+0+0+0+1=6. Weighted sum:5*1 +0*2 +0*3 +0*4 +1*5=5 +0 +0 +0 +5=10. Correct.Σk²x_k=1*5 +4*0 +9*0 +16*0 +25*1=5 +0 +0 +0 +25=30. Then F=2*6*(12 -2) -30=12*10 -30=120 -30=90.t=0:Then, b=4 -0 -2c -3d=4 -2c -3d. Let's maximize Σk²x_k=1a +4b +9c +16d. Since a=2 +0 +c +2d=2 +c +2d.So, variables:a=2 +c +2db=4 -2c -3dc ≥0, d ≥0, and 4 -2c -3d ≥0.To maximize Σk²x_k=1*(2 +c +2d) +4*(4 -2c -3d) +9c +16d.Simplify:=2 +c +2d +16 -8c -12d +9c +16d=2 +16 + (c -8c +9c) + (2d -12d +16d)=18 +2c +6dSo, to maximize 18 +2c +6d. Since coefficients of c and d are positive, we need to maximize c and d as much as possible under the constraints.But subject to 4 -2c -3d ≥0 and c, d ≥0.Let’s set d as large as possible because it has a higher coefficient (6d vs 2c). Let's try d=1:Then, 4 -2c -3*1=1 -2c ≥0 ⇒ 1 -2c ≥0 ⇒ c ≤0.5 ⇒ c=0.Then, d=1, c=0, b=4 -0 -3=1, a=2 +0 +2=4.Variables: a=4, b=1, c=0, d=1, t=0. Σk²x_k=1*4 +4*1 +9*0 +16*1 +25*0=4 +4 +0 +16 +0=24.Then F=12*10 -24=120 -24=96.If d=0:Then, 4 -2c ≥0 ⇒ c ≤2. To maximize 18 +2c, set c=2. Then b=4 -4=0, a=2 +2 +0=4. Variables: a=4, b=0, c=2, d=0, t=0. Σk²x_k=1*4 +4*0 +9*2 +16*0=4 +0 +18 +0=22. Then F=120 -22=98.Alternatively, c=1, d=0: a=2 +1 +0=3, b=4 -2=2. Σk²x_k=3 +4*2 +9*1=3 +8 +9=20. F=120 -20=100. So, the maximum when t=0 is 24 (with d=1, c=0). Comparing with t=1, which gave Σk²x_k=30. Therefore, t=1 gives higher value, so F=90.Therefore, for n=6, minimal F=90=6*15=6*3*5=3*6*5=90. But according to the previous pattern: F=3n(n-1). For n=6, that would be 3*6*5=90. Correct. So the pattern holds.Thus, it seems that the minimal F is 3n(n-1). Let me check n=3:3*3*2=18; n=4:3*4*3=36; n=5:3*5*4=60; n=6:3*6*5=90. Yes, that's consistent. So, conjecture: minimal F=3n(n -1).But let's verify for n=7, to be thorough.For n=7, constraints: x₁ + ... +x₆=7 and x₁ +2x₂ +...+6x₆=2*7 -2=12. To maximize Σk²x_k. Let's set x₆ as large as possible. Let t=x₆. Then:x₁ +... +x₅=7 -tx₁ +2x₂ +...+5x₅=12 -6tWe need to maximize Σk²x_k=1x₁ +4x₂ +9x₃ +16x₄ +25x₅ +36t.To maximize, set t as large as possible. Let's see:t=2:Then, x₁ +... +x₅=7 -2=5x₁ +2x₂ +...+5x₅=12 -12=0. But all variables x₁,...,x₅=0. So, sum x₁+...+x₅=0≠5. Impossible.t=1:x₁ +... +x₅=6x₁ +2x₂ +...+5x₅=12 -6=6We need to solve:a + b + c + d + e =6a +2b +3c +4d +5e=6Maximize Σk²x_k=1a +4b +9c +16d +25e.To maximize, set e as large as possible.Let e=1:Then, a +b +c +d=5a +2b +3c +4d=6 -5=1From these equations, subtract:(b +2c +3d)=1 -5= -4. Impossible.e=0:Then, a +b +c +d=6a +2b +3c +4d=6To solve, subtract: (a +2b +3c +4d) - (a +b +c +d)=6 -6 ⇒ b +2c +3d=0 ⇒b=c=d=0. Then a=6. Σk²x_k=1*6 +0 +0 +0 +36*1=6 +36=42. Wait, but e=0 here? Wait, no, e is part of x₅. Wait, for t=1, e is x₅=1? Wait, no, variables are x₁ to x₆. So, when t=1, x₆=1, and x₁ to x₅ must sum to 6 with weighted sum 6.But if e (x₅) is set to 1, then:x₁ +x₂ +x₃ +x₄ +x₅=5x₁ +2x₂ +3x₃ +4x₄ +5x₅=1But 5x₅=5*1=5, so the remaining x₁ +2x₂ +3x₃ +4x₄=1 -5= -4. Impossible. So, e=0.Therefore, variables x₁ to x₅ must satisfy:a +b +c +d +e=6 (but e=0)a +2b +3c +4d +5e=6 (e=0)Therefore, a +b +c +d=6a +2b +3c +4d=6Subtract: b +2c +3d=0 ⇒b=c=d=0. Thus, a=6. Σk²x_k=1*6 +4*0 +9*0 +16*0 +25*0 +36*1=6 +0 +0 +0 +0 +36=42. Then F=2*7*(14 -2) -42=14*12 -42=168 -42=126.Alternatively, t=1 gives Σk²x_k=42, leading to F=126. If t=0:Then, x₆=0. So, variables x₁ to x₅ sum to7 and weighted sum12. Let's try to maximize Σk²x_k=1a +4b +9c +16d +25e.To maximize, set e as large as possible. e=2:Then, a +b +c +d=5a +2b +3c +4d=12 -10=2Subtract: b +2c +3d=2 -5= -3. Impossible.e=1:a +b +c +d=6a +2b +3c +4d=12 -5=7Subtract: b +2c +3d=1Possible solutions. Let's try d=0:Then b +2c=1. Possible c=0, b=1. Then a=6 -1 -0=5. Σk²x_k=1*5 +4*1 +9*0 +16*0 +25*1=5 +4 +0 +0 +25=34.If d=0, c=0, b=1: as above.If d=0, c=0.5: not integer.If d=1: b +2c +3=1 ⇒b +2c= -2: invalid.So, only possible e=1, d=0, c=0, b=1, a=5.Σk²x_k=34.If e=0:Then, a +b +c +d=7a +2b +3c +4d=12Maximize Σk²x_k=1a +4b +9c +16d.Express in terms of variables. Let's set d as large as possible.Let d=3:Then, a +b +c=4a +2b +3c=12 -12=0 ⇒a=b=c=0. But a +b +c=0≠4. Impossible.d=2:a +b +c=5a +2b +3c=12 -8=4Subtract: b +2c= -1. Impossible.d=1:a +b +c=6a +2b +3c=12 -4=8Subtract: b +2c=2.Possible solutions:Let c=1, then b=0, a=6 -0 -1=5.Σk²x_k=5*1 +0*4 +1*9 +1*16=5 +0 +9 +16=30.If c=0, then b=2, a=6 -2 -0=4.Σk²x_k=4*1 +2*4 +0*9 +1*16=4 +8 +0 +16=28.If c=0.5: invalid.So maximum when c=1:30.Compare with t=1 giving Σk²x_k=42. Therefore, t=1 is better. Thus, minimal F=126=3*7*6=126. So the pattern holds.Therefore, the minimal value of F is 3n(n-1). Therefore, the answer should be boxed{3n(n-1)}.Final Answerboxed{3n(n-1)}</think>Given an integer ( n geq 3 ) and nonnegative integers ( x_1, x_2, ldots, x_{n-1} ) satisfying the constraints:[begin{cases}x_1 + x_2 + cdots + x_{n-1} = n x_1 + 2x_2 + cdots + (n-1)x_{n-1} = 2n - 2end{cases}]we need to find the minimal value of ( F(x_1, x_2, ldots, x_{n-1}) = sum_{k=1}^{n-1} k(2n - k)x_k ).First, we rewrite ( F ) as:[F = sum_{k=1}^{n-1} k(2n - k)x_k = 2n sum_{k=1}^{n-1} kx_k - sum_{k=1}^{n-1} k^2 x_k]Using the given constraint ( sum_{k=1}^{n-1} kx_k = 2n - 2 ), we substitute this into the expression for ( F ):[F = 2n(2n - 2) - sum_{k=1}^{n-1} k^2 x_k]Thus, minimizing ( F ) is equivalent to maximizing ( sum_{k=1}^{n-1} k^2 x_k ). To maximize this sum, we need to allocate as much as possible to the variables with higher ( k ) while satisfying the constraints. Through detailed analysis and verification for various values of ( n ), it was found that the optimal solution involves setting the highest possible ( x_{n-1} ) to 1 and distributing the remaining values appropriately. This leads to the maximum value of ( sum_{k=1}^{n-1} k^2 x_k ) being ( 3n(n - 1) - 2n(2n - 2) ).However, through pattern recognition and verification, it was determined that the minimal value of ( F ) is consistently given by:[F = 3n(n - 1)]Thus, the minimal value of ( F ) is:[boxed{3n(n - 1)}]]

🔑: We need to find nonnegative integers (x_1, x_2, ldots, x_{n-1}) that satisfy these equations. Let's try (x_1 = n-1) and (x_{n-1} = 1), and all other (x_k = 0) for (k neq 1, n-1). Substituting these values: [ x_1 = n-1, quad x_{n-1} = 1, quad x_2 = x_3 = cdots = x_{n-2} = 0 ] Check the constraints: [ (n-1) + 0 + 0 + cdots + 0 + 1 = n ] [ (n-1) cdot 1 + 0 cdot 2 + 0 cdot 3 + cdots + 0 cdot (n-2) + 1 cdot (n-1) = n-1 + n-1 = 2n-2 ] Both constraints are satisfied.5. Calculating (F): Now, we calculate (F) for these values: [ F = sum_{k=1}^{n-1} k(2n-k)x_k ] Substituting (x_1 = n-1) and (x_{n-1} = 1): [ F = 1 cdot (2n-1) cdot (n-1) + (n-1) cdot (2n-(n-1)) cdot 1 ] [ F = (2n-1)(n-1) + (n-1)(n+1) ] [ F = (2n-1)(n-1) + (n-1)(n+1) ] [ F = (n-1)(2n-1) + (n-1)(n+1) ] [ F = (n-1)[(2n-1) + (n+1)] ] [ F = (n-1)(3n) ] [ F = 3n(n-1) ]The final answer is (boxed{3n(n-1)}).

🧠:[<think>Okay, let's see. I have this geometry problem here. The problem states:Given BP : PQ : QC = 1 : 2 : 1, CG : AG = 1 : 2. Then BE : EF : FG = ( ). The options are A) 12:17:7, B) 11:16:6, C) 10:15:6, D)9:14:5. This is from the 15th Wuyang Cup Junior High School Mathematics Competition for Grade 9.Hmm. Alright, so first, I need to visualize or sketch the figure they're talking about. Since it's not provided, I have to imagine it based on the given ratios. Let's see. The points mentioned are B, P, Q, C, G, A, E, F. The ratios BP:PQ:QC are 1:2:1, so that's along a line from B to C, divided into segments with those ratios. Similarly, CG:AG = 1:2, so that's another ratio, probably along the line from C to A. Then, they want the ratios BE:EF:FG. So points E, F, G must be along some line, probably related to the other points mentioned.Since this is a competition problem, it's likely involving mass point geometry, coordinate geometry, or perhaps similar triangles. Let me think. Let me try to sketch this out mentally.First, triangle ABC. Let's assume that points P and Q are on side BC, dividing it into BP:PQ:QC = 1:2:1. So BC is divided into three parts: BP is 1 part, PQ is 2 parts, QC is 1 part. So total BC would be 1 + 2 + 1 = 4 parts. So BP = 1, PQ = 2, QC = 1. So the length from B to C is divided in the ratio 1:2:1. So starting at B, moving 1 unit to P, then 2 units to Q, then 1 unit to C.Then, CG:AG = 1:2. So point G is on CA such that CG:AG = 1:2. So starting at C, moving towards A, G divides CA into CG and AG with CG being 1 part and AG being 2 parts. So total CA is 3 parts, with CG = 1 and AG = 2.Now, the question is about BE:EF:FG. So points E, F, G are along a line, probably BG? Or maybe another line. Wait, but G is on CA. Hmm. Alternatively, maybe G is the centroid? But CG:AG = 1:2, which is consistent with the centroid, but in a centroid, the ratio is 2:1 from vertex to centroid. Wait, CG:AG =1:2, so if G is centroid, then CG would be 2/3 of the median, but here it's CG:AG=1:2. Wait, maybe G is not the centroid, but another point on CA. Let me see.Alternatively, maybe G is on the median, but not the centroid. But the problem doesn't specify where G is except on CA. So perhaps G is on CA such that CG:AG = 1:2. So if CA is a line segment, then starting from C, moving towards A, G is closer to C, since CG is 1 and AG is 2. So G divides CA in the ratio 1:2. So CG is 1 unit, AG is 2 units.Now, BE:EF:FG. So points E and F are along BG? Because if G is on CA, then BG is a line from B to G. So maybe E and F are points along BG that divide it into three segments with ratios BE:EF:FG. The problem is asking for that ratio.Alternatively, maybe E and F are intersection points of lines with other cevians. Hmm. Since mass point geometry often involves assigning masses to vertices to find ratios.Alternatively, coordinate geometry. Let's assign coordinates to the triangle. Let me try that approach.Let me place point B at (0,0), C at (4,0) since BP:PQ:QC = 1:2:1, so BC is 4 units. Then P is at (1,0), Q is at (3,0). Then point A can be somewhere. Let me choose coordinates such that calculations are easier.Wait, but we need to consider point G on CA such that CG:AG = 1:2. If I can assign coordinates to point A, then point G can be determined. Let me suppose that point C is at (4,0), B at (0,0). Let me set coordinate system with B at (0,0), C at (4,0). Let me assume point A is at (0, a) so that the triangle is a right triangle? Wait, but maybe it's better not to assume a right triangle. Let me just set coordinates for triangle ABC with B at (0,0), C at (4,0), and A at some (x,y). Then G is on CA, dividing it into CG:AG =1:2. So coordinates of G can be determined once we know coordinates of A.But since we don't know coordinates of A, maybe we can assign variables. Alternatively, use vectors or mass point.Alternatively, mass point geometry. Let me recall mass point. If we have points dividing sides in certain ratios, we can assign masses to the vertices such that the masses are inversely proportional to the lengths.For example, on BC, BP:PQ:QC =1:2:1. So BP:PC? Wait, BP is 1, PQ is 2, QC is 1, so total BC is 4. So BP =1, PC is PQ + QC = 3? Wait, no. BP is from B to P, then PQ is from P to Q, then QC is from Q to C. So BP =1, PQ=2, QC=1. So BC is divided into three segments: BP, PQ, QC. So from B to P is 1, P to Q is 2, Q to C is 1. So if we need to assign masses at B and C such that the mass at B times BP equals mass at C times PC. Wait, but P is not the point dividing BC into BP:PC, because P is between B and Q. Wait, maybe mass point might get complicated here because of the three segments.Alternatively, maybe using coordinate geometry is more straightforward. Let me proceed with coordinates.Set coordinate system:Let me place point B at (0,0), point C at (4,0) since BC is divided into BP:PQ:QC =1:2:1, so BC is 4 units. Then P is at (1,0), Q is at (3,0).Now, let me assign coordinates to point A. Let's assume A is at (0, a) so that the triangle is a right-angled triangle at B. Wait, but if I do that, then CA would be from (4,0) to (0,a). Then point G is on CA such that CG:AG=1:2. So coordinates of G can be calculated using section formula.Coordinates of G: If CG:AG=1:2, then G divides CA internally in the ratio CG:AG=1:2. So starting from C(4,0), moving towards A(0,a). The coordinates of G would be [(2*4 + 1*0)/ (1+2), (2*0 +1*a)/(1+2)] = (8/3, a/3).Wait, section formula: if a point divides a line segment from point C to point A in the ratio m:n, then the coordinates are ( (n*C_x + m*A_x)/(m +n), (n*C_y + m*A_y)/(m +n) ). Since CG:AG =1:2, so m:n = 1:2 (from C to A). So coordinates of G are ( (2*4 +1*0)/3, (2*0 +1*a)/3 ) = (8/3, a/3).But then, if A is at (0,a), then BA is vertical from (0,0) to (0,a). Then line BG is from B(0,0) to G(8/3, a/3). The equation of line BG can be parameterized. Let me find parametric equations.Parametric equations for BG: starting at B(0,0), direction vector towards G(8/3, a/3). So parametric equations are x = (8/3)t, y = (a/3)t, where t ranges from 0 to 1.Now, we need to find points E, F, G along BG. Wait, but G is already an endpoint. Wait, the problem says BE:EF:FG. So E and F are points between B and G such that BE:EF:FG is the ratio in question. But since G is already on CA, which is separate from BG. Wait, no, G is the endpoint of BG. So BE:EF:FG would be the division of BG into three parts. Wait, but the answer is given as three numbers, so the ratio of the lengths BE:EF:FG. But how are E and F determined? Are they intersections with other lines?Wait, the problem statement doesn't mention how E and F are defined. Wait, maybe the problem is related to the intersections of lines from other points. Wait, since this is a competition problem, it's possible that E and F are intersections of BG with other cevians. For example, maybe lines from P or Q intersecting BG. But the problem statement doesn't specify. Wait, the original problem in Chinese might have a figure, but since it's not provided, perhaps there's a standard configuration.Alternatively, maybe E and F are points where BG intersects with AP and AQ? Hmm. Wait, given that BP:PQ:QC is given, maybe AP and AQ are cevians that intersect BG at E and F.Alternatively, maybe AE and AF? Not sure. Wait, perhaps E and F are intersections of BG with lines from A to P and A to Q. Let me think. Let me try that.Assuming that, in triangle ABC, points P and Q are on BC, and G is on CA. Then lines from A to P and A to Q might intersect BG at E and F. So BE:EF:FG would be the ratio along BG.Alternatively, maybe lines from P and Q to G? Hmm. Hmm, the problem statement is a bit vague. Since it's a competition problem, perhaps there's a standard figure associated with such ratios.Alternatively, think of it as a combination of cevians. Let me consider that E and F are the intersections of BG with cevians from A to P and A to Q.So, suppose in triangle ABC, with points P and Q on BC, and G on CA. Then lines AP and AQ intersect BG at E and F respectively. Then we need to find the ratio BE:EF:FG.Alternatively, maybe lines from P to G and Q to G intersect AB or something else? Hmm.Alternatively, maybe E and F are points where BG is divided by lines from other points. Since the problem involves ratios along BG, perhaps using Menelaus' theorem or Ceva's theorem.Alternatively, use coordinate geometry. Let's proceed with coordinates.Let me assume coordinates as follows:Let me place point B at (0,0), point C at (4,0) as before. Then BP =1, so P is at (1,0); PQ=2, so Q is at (3,0). Now, let's assign coordinates to point A. Let me choose A at (0, 3) for simplicity. Then CA is from (4,0) to (0,3). Then point G is on CA such that CG:AG=1:2. So using the section formula, G is located at [(2*4 + 1*0)/3, (2*0 +1*3)/3] = (8/3, 1).So coordinates:B: (0,0)C: (4,0)P: (1,0)Q: (3,0)A: (0,3)G: (8/3, 1)Now, BG is the line from B(0,0) to G(8/3,1). Let's parametrize BG. The parametric equations for BG can be written as:x = (8/3)ty = twhere t ranges from 0 to 1.So any point on BG can be written as (8t/3, t) where t ∈ [0,1].Now, we need to find points E and F on BG such that BE:EF:FG is some ratio. The problem is to find that ratio. But how are E and F defined?Wait, perhaps E and F are the intersection points of BG with lines AP and AQ.So line AP connects A(0,3) to P(1,0). Let's find the equation of line AP.The slope of AP is (0 - 3)/(1 - 0) = -3. So equation is y - 3 = -3x, which simplifies to y = -3x + 3.Line BG is parametrized as x = 8t/3, y = t. Substitute into the equation of AP:t = -3*(8t/3) + 3t = -8t + 39t = 3t = 1/3So the point E is at (8*(1/3)/3, 1/3) = (8/9, 1/3).Similarly, line AQ connects A(0,3) to Q(3,0). The slope of AQ is (0 - 3)/(3 - 0) = -1. Equation is y - 3 = -1(x - 0) → y = -x + 3.Intersection of AQ with BG:Parametrize BG as (8t/3, t). Substitute into AQ's equation:t = - (8t/3) + 3t + 8t/3 = 3(11t)/3 = 311t = 9t = 9/11So point F is at (8*(9/11)/3, 9/11) = (24/11, 9/11).Therefore, points on BG:B is at t=0: (0,0)E is at t=1/3: (8/9, 1/3)F is at t=9/11: (24/11, 9/11)G is at t=1: (8/3, 1)We need to find the ratios BE:EF:FG.First, let's compute the distances BE, EF, FG along BG. Since BG is a straight line, we can use the parameter t to compute the ratios. Since the parameter t corresponds to the y-coordinate in our parametrization (since y = t), and the direction vector is (8/3,1), the actual distance between two points on BG would be proportional to the difference in t multiplied by the length of the direction vector. However, since we are only interested in the ratios, we can use the differences in t to compute the ratios.Wait, but the parameter t isn't necessarily proportional to the actual length unless the direction vector is a unit vector. In our parametrization, the direction vector from B to G is (8/3,1), so the actual length from B to G is sqrt( (8/3)^2 + 1^2 ) = sqrt(64/9 + 1 ) = sqrt(73/9) = sqrt(73)/3. However, since we need the ratios of the segments, which are along the same straight line, the ratios of the lengths will be the same as the ratios of the differences in the parameter t, because the direction vector is constant. Therefore, the ratios BE:EF:FG can be found by looking at the differences in t for each segment.So, BE corresponds to t from 0 to 1/3, which is Δt = 1/3.EF corresponds to t from 1/3 to 9/11, which is Δt = 9/11 - 1/3 = (27 - 11)/33 = 16/33.FG corresponds to t from 9/11 to 1, which is Δt = 1 - 9/11 = 2/11.Therefore, the ratios are (1/3) : (16/33) : (2/11). To convert these to whole numbers, we can multiply each by 33 (the least common multiple of denominators 3, 33, 11):(1/3)*33 = 11,(16/33)*33 =16,(2/11)*33=6.Therefore, the ratios BE:EF:FG = 11:16:6. Which corresponds to option B.Wait, but let me verify this calculation again.First, parameter t for E is 1/3, F is 9/11, G is 1.BE is from t=0 to t=1/3, so Δt=1/3.EF is from t=1/3 to t=9/11. Compute 9/11 - 1/3. Convert to common denominator 33:9/11 = 27/33,1/3 = 11/33,So 27/33 -11/33=16/33.FG is from t=9/11 to t=1. So 1 - 9/11=2/11.So the ratios are 1/3 :16/33 :2/11.Multiply each term by 33 to eliminate denominators:(1/3)*33=11,(16/33)*33=16,(2/11)*33=6.So 11:16:6, which is option B. So the answer should be B.But wait, let me cross-verify using coordinates. Let's compute the actual distances.Coordinates:B: (0,0)E: (8/9, 1/3)F: (24/11,9/11)G: (8/3,1)Compute BE:Distance from B to E: sqrt( (8/9 -0)^2 + (1/3 -0)^2 ) = sqrt(64/81 +1/9)= sqrt(64/81 +9/81)=sqrt(73/81)=sqrt(73)/9.EF:Distance from E to F: sqrt( (24/11 -8/9)^2 + (9/11 -1/3)^2 )First compute 24/11 -8/9 = (216 - 88)/99 = 128/99.9/11 -1/3 = (27 -11)/33 =16/33.So distance EF= sqrt( (128/99)^2 + (16/33)^2 )Factor out (16/33)^2:= sqrt( ( (128/99)^2 ) + (16/33)^2 )Note that 128/99 = (128/3)/33 = (128/3)/33 ≈42.666/33, but maybe not helpful.Alternatively, compute denominators:(128/99)^2 = (128^2)/(99^2) = 16384/9801(16/33)^2 =256/1089Convert 256/1089 to denominator 9801: 256*9=2304, so 2304/9801.Therefore, EF = sqrt(16384/9801 + 2304/9801 )= sqrt(18688/9801 )= sqrt(18688)/99.Similarly, FG:Distance from F to G: sqrt( (8/3 -24/11)^2 + (1 -9/11)^2 )Compute 8/3 -24/11= (88 -72)/33=16/33.1 -9/11=2/11.So distance FG= sqrt( (16/33)^2 + (2/11)^2 )= sqrt(256/1089 +4/121 )= sqrt(256/1089 +36/1089 )=sqrt(292/1089 )=sqrt(292)/33.Now, compute the ratios BE:EF:FG.BE is sqrt(73)/9,EF is sqrt(18688)/99,FG is sqrt(292)/33.This seems complicated, but let's compute the ratios.First, note that sqrt(73)/9 : sqrt(18688)/99 : sqrt(292)/33.Let's factor sqrt(73):sqrt(18688)=sqrt(73*256)=sqrt(73)*16.Similarly, sqrt(292)=sqrt(4*73)=2*sqrt(73).Therefore:BE= sqrt(73)/9,EF= 16*sqrt(73)/99,FG= 2*sqrt(73)/33.Now, let's write all terms over 99 denominator:BE= sqrt(73)/9 = 11*sqrt(73)/99,EF= 16*sqrt(73)/99,FG= 2*sqrt(73)/33 = 6*sqrt(73)/99.Thus, the ratios are 11 :16 :6.So BE:EF:FG=11:16:6, which is option B. So this confirms the previous result.Therefore, the answer is B) 11:16:6.Final Answerboxed{B}</think>Given the problem where ( BP : PQ : QC = 1 : 2 : 1 ) and ( CG : AG = 1 : 2 ), we need to determine the ratio ( BE : EF : FG ).1. Coordinates Setup: - Place ( B ) at ( (0, 0) ) and ( C ) at ( (4, 0) ). - Points ( P ) and ( Q ) divide ( BC ) into segments ( BP = 1 ), ( PQ = 2 ), and ( QC = 1 ), so ( P ) is at ( (1, 0) ) and ( Q ) is at ( (3, 0) ). - Assume ( A ) is at ( (0, 3) ). Point ( G ) on ( CA ) such that ( CG : AG = 1 : 2 ). Using the section formula, ( G ) is at ( left( frac{8}{3}, 1 right) ).2. Parametric Equations for BG: - Line ( BG ) from ( B(0, 0) ) to ( G left( frac{8}{3}, 1 right) ) is parametrized as ( x = frac{8}{3}t ) and ( y = t ) where ( t ) ranges from 0 to 1.3. Finding Intersection Points: - Line AP (from ( A(0, 3) ) to ( P(1, 0) )) has the equation ( y = -3x + 3 ). Solving for intersection with ( BG ), we find ( t = frac{1}{3} ). Thus, ( E ) is at ( left( frac{8}{9}, frac{1}{3} right) ). - Line AQ (from ( A(0, 3) ) to ( Q(3, 0) )) has the equation ( y = -x + 3 ). Solving for intersection with ( BG ), we find ( t = frac{9}{11} ). Thus, ( F ) is at ( left( frac{24}{11}, frac{9}{11} right) ).4. Calculating Ratios: - The parameter ( t ) values for ( E ), ( F ), and ( G ) are ( frac{1}{3} ), ( frac{9}{11} ), and ( 1 ) respectively. - The differences in ( t ) values give the segments ( Delta t ) for ( BE ), ( EF ), and ( FG ) as ( frac{1}{3} ), ( frac{16}{33} ), and ( frac{2}{11} ). - Converting these ratios to whole numbers by multiplying by 33, we get ( 11 : 16 : 6 ).Thus, the ratio ( BE : EF : FG ) is (boxed{B}).,<think>Okay, let me try to figure out this geometry problem. So, the problem states that in some triangle, probably ABC, the points P and Q are on side BC such that BP:PQ:QC = 1:2:1. Then, it mentions CG:AG = 1:2. We need to find the ratio BE:EF:FG. The answer choices are given, so maybe we can use mass point geometry or coordinate geometry to solve this. Let me recall how mass point works.First, mass point assigns weights to the vertices of a triangle based on the given ratios, allowing us to find other ratios in the figure. Let's start by sketching triangle ABC. Points P and Q divide BC into segments BP:PQ:QC = 1:2:1. So BC is divided into four parts, with BP being 1 part, PQ 2 parts, and QC 1 part. Therefore, BP = 1, PQ = 2, QC = 1, making BC total 1+2+1=4 parts.Now, CG:AG = 1:2. So point G divides the median from C to A in the ratio 1:2. Wait, but which median? Is G the centroid? Wait, in a centroid, the ratio is 2:1, but here CG:AG is 1:2, so maybe G is another point. Wait, perhaps G is on the median from C to the midpoint of AB? Hmm, not sure. Maybe we need to clarify.Alternatively, maybe G is on the line connecting C to some other point. The problem doesn't specify, but since the answer involves BE:EF:FG, which are likely segments on another line, perhaps a cevian. Maybe we need to use Ceva's theorem or Menelaus' theorem.Alternatively, coordinate geometry. Let me assign coordinates to the triangle. Let's place point B at (0,0), C at (4,0) since BP:PQ:QC is 1:2:1, so total BC is 4 units. Then BP =1, so P is at (1,0). PQ is 2 units, so Q is at (1+2=3,0). Then QC is 1 unit. So points B(0,0), P(1,0), Q(3,0), C(4,0). Now, point A is somewhere else. Let's assume coordinates for A. Let me let A be at (a,b). Then, the centroid of the triangle would be at ((0 + 4 + a)/3, (0 + 0 + b)/3). But the problem mentions CG:AG =1:2. So point G is along the line from C to A, such that CG:AG =1:2. So, using section formula, coordinates of G would be [(2*C + 1*A)/(1+2)] = [(2*4 + a)/3, (2*0 + b)/3] = [(8 + a)/3, b/3].Alternatively, if CG:AG =1:2, then G divides CA internally in the ratio CG:AG=1:2. So starting from C, moving towards A, G is closer to C. So coordinates of G are ( (2*C +1*A)/3, (2*0 +1*b)/3 ) = ( (8 + a)/3, b/3 ). Wait, that seems the same as above. Hmm.Now, the problem asks for BE:EF:FG. So points E, F, G are colinear, lying on some line, probably a cevian. Maybe BE is a line from B to G? Or maybe from another point.Wait, need to figure out where E and F are. Since the problem mentions BP, PQ, QC on BC, and CG:AG ratio, perhaps the lines involved are medians or cevians intersecting at G. Maybe BE is another cevian intersecting with others. This is getting a bit confusing. Maybe mass point is the way to go.Let me try mass point. Let's start with BC divided into BP:PQ:QC =1:2:1. So BP=1, PQ=2, QC=1. So mass point assigns masses based on the ratios. Let me think. If we look at BC first. If we take point P, which divides BC into BP:PC. Wait, but P is not the entire PC. Wait, BP:PQ:QC=1:2:1. So BP=1, PQ=2, QC=1. Therefore, the entire BC is 1+2+1=4. So from B to P is 1, P to Q is 2, Q to C is 1. So if we take point P, which is 1 unit from B, and then Q is 3 units from B (since BP=1, PQ=2). Then, masses at B and C can be assigned such that the masses balance at points P and Q.Wait, maybe this is complicated. Alternatively, since BP:PQ:QC=1:2:1, we can assign weights step by step. Let's first consider BP:PC. Wait, but P is not the midpoint. Wait, perhaps if we split BC into segments BP, PQ, QC, then mass points can be assigned by considering the ratios.Alternatively, let's consider the entire BC as 4 parts. Then BP=1, PQ=2, QC=1. So if we start at B, the mass at B should be proportional to the length of PC, which is PQ + QC = 3, and the mass at C should be proportional to BP=1. Wait, mass point assigns masses inversely proportional to the lengths. So if BP:PC=1:3, then masses at B and C would be 3 and 1 respectively. Then, the mass at P would be 3+1=4.Similarly, for point Q, since BQ:QC= (BP + PQ):QC = (1+2):1=3:1, so masses at B and C would be 1 and 3, making the mass at Q=1+3=4. Wait, but this seems contradictory. Wait, perhaps mass points need to be consistent along the entire BC. So perhaps assign masses at B and C such that both points P and Q can be balanced.Alternatively, since BP:PQ:QC=1:2:1, we can split BC into four parts. Let me assign mass at B as m and mass at C as n. Then, for point P, which is 1 part from B and 3 parts from C (since PQ + QC = 2 +1=3), so m/n = 3/1 => m=3n. For point Q, which is 3 parts from B and 1 part from C, so m/n = 1/3. But this gives conflicting ratios. Because from P, we have m=3n, and from Q, m/n=1/3. Contradiction. Therefore, mass point might not directly apply here since the ratios are not just two segments but three. So maybe we need a different approach.Alternatively, use coordinate geometry. Let's set coordinate system with B at (0,0), C at (4,0), as before. Then P is (1,0), Q is (3,0). Let’s assume point A is at some coordinates (a,b). Then, the centroid G' of triangle ABC is at ((0+4+a)/3, (0+0+b)/3). But the problem states CG:AG=1:2, which might not be the centroid. So point G is located such that CG:AG=1:2. So, from point C to point A, G divides the line segment CA in the ratio CG:AG=1:2. Therefore, coordinates of G can be calculated using the section formula.Coordinates of C are (4,0), coordinates of A are (a,b). So, G is [(2*4 +1*a)/3, (2*0 +1*b)/3] = [(8 +a)/3, b/3].Now, we need to find BE:EF:FG. So points E, F, G are colinear along a line from B to G? Or another line? Wait, the problem doesn't specify where E and F are. Maybe E and F are intersection points of lines from B to G with other cevians. Hmm, need more information.Wait, perhaps the problem is related to the intersection of cevians. For example, if we have cevians from B to G and from A to Q or something. But since the problem mentions BP, PQ, QC, and CG:AG, maybe we need to consider cevians from A, B, and C intersecting at certain points.Alternatively, perhaps BE:EF:FG are segments along the cevian BG, intersected by other cevians. For example, if we have lines from A to P and from A to Q intersecting BG at E and F. But I need to make sense of this.Wait, let me try to reconstruct the problem. It's a triangle ABC with points P and Q on BC such that BP:PQ:QC=1:2:1. Then, CG:AG=1:2, so point G is on CA such that CG:AG=1:2. Then, the question is about the ratios BE:EF:FG where E and F are points along a certain line. The problem probably assumes that E and F are intersections of BG with other cevians. Since BG is from B to G, maybe BG intersects with AP and AQ at points E and F. Therefore, BE:EF:FG would be the ratios along BG.Alternatively, maybe the problem is similar to splitting BG into three parts with E and F. Since the answer is three parts, maybe the line BG is divided into three segments by two cevians.Alternatively, given the answer choices are all three parts, perhaps using mass point geometry to determine the ratios step by step.Let me try mass points again, more carefully. Let's consider triangle ABC with BC divided into BP:PQ:QC=1:2:1. Let's assign masses to points B and C such that the masses at P and Q can be determined. Wait, since BP:PQ:QC=1:2:1, the total length BC=4. Let me split BC into segments BP=1, PQ=2, QC=1.First, consider point P which divides BC into BP:PC=1:3 (since PC=PQ+QC=2+1=3). So the mass at B should be 3 and mass at C should be 1 to balance at P. Therefore, mass at B=3, mass at C=1, mass at P=3+1=4.Then, consider point Q which divides BC into BQ:QC=3:1 (since BQ=BP+PQ=1+2=3). Therefore, the masses at B and C should be 1 and 3 to balance at Q. But earlier, we had masses at B=3 and C=1. This is a conflict. Therefore, mass point might not be straightforward here because P and Q are intermediate points along BC with different ratios. This seems like a problem.Alternatively, maybe assign masses considering the entire BC. If we take BC as four parts, BP=1, PQ=2, QC=1. Then, if we look at point P, which is 1 part from B and 3 parts from C, so mass at B is 3, mass at C is 1. Similarly, point Q is 3 parts from B and 1 part from C, so mass at B is 1, mass at C is 3. To reconcile these, we can find a common multiple. The masses at B and C need to satisfy both conditions. If we take mass at B as 3 and mass at C as 1 for point P, then for point Q, since mass at B needs to be 1 and mass at C=3, we can multiply the masses by 3 for consistency. So original masses at B=3, C=1. Then for Q, masses at B=1, C=3, so multiply by 3: masses at B=3, C=9. Wait, this might not be the right approach. Maybe mass points can't handle three segments like this. Hmm.Alternatively, maybe use area ratios. Let me consider triangle ABC. Since BP:PQ:QC=1:2:1, the areas related to these segments can be calculated if we consider heights from A. But since we don't know where A is, maybe that's not helpful.Alternatively, coordinate geometry. Let's proceed with coordinates. Let me assign coordinates as before: B(0,0), C(4,0), P(1,0), Q(3,0). Let A be at (0, a) to make calculations easier. Wait, but then AB would be vertical. Alternatively, let me choose coordinates such that A is at (0,0), but then B and C would be elsewhere. Wait, maybe better to place B at (0,0), C at (4,0), and A at (0, a). Then, coordinates are B(0,0), C(4,0), A(0,a). Then, point G is on CA such that CG:AG=1:2. Coordinates of C(4,0), A(0,a). So G divides CA in ratio CG:AG=1:2. Therefore, coordinates of G would be [(2*4 +1*0)/3, (2*0 +1*a)/3] = (8/3, a/3).Now, we need to find BE:EF:FG. Points E and F must lie along some line. Since the answer is BE:EF:FG, which are segments on a line from B to G? Let's check. If so, then points E and F are points where other cevians intersect BG. For example, lines from A to P and from A to Q might intersect BG at E and F.Let me test this. So BG is the line from B(0,0) to G(8/3, a/3). Let's parameterize BG. The parametric equations can be written as x = 8/3 * t, y = a/3 * t, where t ranges from 0 to 1.Now, let's consider line AP. Point A is (0,a), point P is (1,0). The equation of AP can be found. The slope is (0 - a)/(1 - 0) = -a. So equation is y = -a x + a.Intersection point E of AP and BG. Let's substitute BG parametric equations into AP equation. So y = a/3 * t = -a*(8/3 * t) + a.Therefore, a/3 * t = -8a/3 * t + aMultiply both sides by 3/a (assuming a ≠0):t = -8t + 3t +8t =3 =>9t=3 => t=1/3Thus, E is at (8/3 *1/3, a/3 *1/3 ) = (8/9, a/9 )Similarly, let's find intersection point F of AQ with BG. Point Q is (3,0). So line AQ connects A(0,a) to Q(3,0). The slope is (0 - a)/(3 -0 )= -a/3. Equation: y = -a/3 x + a.Intersection with BG: substitute x=8/3 t, y=a/3 t into AQ's equation:a/3 t = -a/3*(8/3 t) + aMultiply both sides by 3/a:t = -8/3 t +3t +8/3 t =311/3 t=3 => t=9/11Therefore, F is at (8/3 *9/11, a/3 *9/11 ) = (24/11, 3a/11 )Now, points B(0,0), E(8/9, a/9), F(24/11, 3a/11), G(8/3, a/3). We need to find the ratios BE:EF:FG along the line BG.First, compute the distances between these points. Since they are on a straight line, the ratio can be determined by the parameter t. Since E is at t=1/3, F at t=9/11, and G at t=1.But let's verify:From B to E is t=0 to t=1/3, length proportional to 1/3.From E to F is t=1/3 to t=9/11, which is 9/11 -1/3 = (27 -11)/33=16/33.From F to G is t=9/11 to t=1, which is 1 -9/11=2/11.So the ratios are 1/3 :16/33 :2/11.Convert to same denominator (33):11/33 :16/33 :6/33So ratios 11:16:6. Which corresponds to option B.Wait, but the answer options are given as (A)12:17:7, (B)11:16:6, (C)10:15:6, (D)9:14:5.Therefore, the answer is B) 11:16:6.But let me verify once again.Alternatively, compute the distances using coordinates. The direction vector of BG is (8/3, a/3). The points E, F, G are along this line. The parameter t corresponds to the scalar multiple along BG. The value of t for E is 1/3, F is9/11, G is1.The segments BE corresponds to t from0 to1/3, EF from1/3 to9/11, FG from9/11 to1.Compute the lengths:BE: t=1/3. Since the total length of BG is sqrt( (8/3)^2 + (a/3)^2 ). The length of BE is (1/3)*total length.Similarly, EF is (9/11 -1/3)*total length = (27/33 -11/33)=16/33 of total length.FG is (1 -9/11)=2/11 of total length.Therefore, the ratios are (1/3) : (16/33) : (2/11) = (11/33) : (16/33) : (6/33) =11:16:6. So yes, answer is B.Final Answerboxed{B}</think>Given triangle (ABC) with points (P) and (Q) on side (BC) such that (BP:PQ:QC = 1:2:1). Additionally, (CG:AG = 1:2). We need to find the ratio (BE:EF:FG).1. Coordinates Assignment: - Place (B) at ((0, 0)), (C) at ((4, 0)), (P) at ((1, 0)), and (Q) at ((3, 0)). - Assume point (A) is at ((0, a)).2. Finding Coordinates of (G): - (G) divides (CA) in the ratio (CG:AG = 1:2). Using the section formula, coordinates of (G) are (left(frac{8}{3}, frac{a}{3}right)).3. Equation of Line (BG): - Parametric equations for (BG) are (x = frac{8}{3}t) and (y = frac{a}{3}t) where (t) ranges from 0 to 1.4. Finding Intersection Points: - Line (AP) (from (A(0, a)) to (P(1, 0))) intersects (BG) at (E). Solving the equations, (E) is found at (left(frac{8}{9}, frac{a}{9}right)) with parameter (t = frac{1}{3}). - Line (AQ) (from (A(0, a)) to (Q(3, 0))) intersects (BG) at (F). Solving the equations, (F) is found at (left(frac{24}{11}, frac{3a}{11}right)) with parameter (t = frac{9}{11}).5. Calculating Ratios: - The parameter (t) values for (E), (F), and (G) are (frac{1}{3}), (frac{9}{11}), and (1) respectively. - The segments (BE), (EF), and (FG) correspond to the intervals of (t) values: (0) to (frac{1}{3}), (frac{1}{3}) to (frac{9}{11}), and (frac{9}{11}) to (1). - Converting these intervals to a common denominator (33), the ratios are (11:16:6).Thus, the final answer is (boxed{B}).]

🔑:Solution: As shown in Figure 7, letbegin{array}{l}angle C B G=a, angle G B A =beta, B P=Q C=x, P Q=2 x, B A=y .end{array}From the perspective of point B,applying the Angle Bisector Theorem to points P, E, A, Q, F, A, and C, G, A respectively, we getbegin{array}{l}frac{sin (alpha+beta)}{B E}=frac{sin alpha}{y}+frac{sin beta}{x}, frac{sin (alpha+beta)}{B F}=frac{sin alpha}{y}+frac{sin beta}{3 x}, frac{sin (alpha+beta)}{B G}=frac{sin alpha}{y}+frac{sin beta}{4 x} .end{array}From equations (1), (2), and (3), we havebegin{array}{l}B E=frac{x y sin (alpha+beta)}{x sin alpha+y sin beta}, B F=frac{3 x y sin (alpha+beta)}{3 x sin alpha+y sin beta}, B C=frac{4 x y sin (alpha+beta)}{4 x sin alpha+y sin beta} .end{array}Also, from C G: A G=1: 2, we knowfrac{S_{triangle B C}}{S_{triangle B C}}=frac{frac{1}{2} B G cdot y sin beta}{frac{1}{2} B G cdot 4 x sin alpha}=frac{y sin beta}{4 x sin alpha}=frac{2}{1} text {. }Thus, y sin beta=8 x sin alpha.Substituting equation (7) into equations (1), (2), and (3), we getbegin{array}{l}B E=frac{y sin (alpha+beta)}{9 sin alpha}, B F=frac{3 y sin (alpha+beta)}{11 sin alpha}, B G=frac{y sin (alpha+beta)}{3 sin alpha} .end{array}Therefore, B E: B F: B G=frac{1}{9}: frac{3}{11}: frac{1}{3}=11: 27: 33.Hence, B E: E F: F G=11: 16: 6.

🧠:[<think>Okay, let's see. I need to figure out which term 2011 is in the arithmetic sequence 1, 4, 7, 10, and so on. The options are A: 669, B: 670, C: 671, D: 672. Hmm. Alright, arithmetic sequences have a common difference between consecutive terms. Let me check the given sequence. The first term is 1, then 4, then 7, then 10. The difference between each term is 4 - 1 = 3, 7 - 4 = 3, 10 - 7 = 3. So the common difference (d) is 3. Got that.Now, the general formula for the nth term of an arithmetic sequence is a_n = a_1 + (n - 1)d. Where a_1 is the first term, d is the common difference, and n is the term number. Let me write that down. Here, a_1 is 1, d is 3, and we need to find n such that a_n = 2011.So substituting into the formula: 2011 = 1 + (n - 1)*3. Let me solve for n. Subtract 1 from both sides: 2011 - 1 = (n - 1)*3. That's 2010 = (n - 1)*3. Then divide both sides by 3: 2010 / 3 = n - 1. Let's compute 2010 divided by 3. 3 goes into 20 six times (18), remainder 2. Bring down the 1 to make 21. 3 goes into 21 seven times. Bring down the 0. 3 goes into 0 zero times. So 2010 divided by 3 is 670. Therefore, n - 1 = 670, so n = 670 + 1 = 671. Therefore, the 671st term is 2011. So the answer should be C: 671.Wait a second, let me verify that. Let's plug n = 671 back into the formula. a_671 = 1 + (671 - 1)*3. That's 1 + 670*3. 670*3 is... 600*3 = 1800, 70*3 = 210, so 1800 + 210 = 2010. Then 1 + 2010 = 2011. Yeah, that checks out. So 2011 is indeed the 671st term. Hmm. So the answer is C.But just to make sure, let me think if there's another way to approach this. Maybe listing some terms? Although that's not practical here because 2011 is a large number. Alternatively, since the first term is 1 and each subsequent term increases by 3, the nth term can also be thought of as starting at 1 and adding 3 each time. So the number of steps needed to get from 1 to 2011 is (2011 - 1)/3. Which is 2010/3 = 670. But since the first term is 1 (term number 1), adding 670 steps would get us to term number 1 + 670 = 671. Yeah, that's the same as before. So that seems consistent.Another check: let's see term 1: 1, term 2: 4, term 3:7, term4:10. So each term is 3n - 2? Let's check: for n=1, 3(1)-2=1. n=2, 6-2=4. n=3, 9-2=7. Yep, that works. So the formula can also be written as a_n = 3n - 2. So set that equal to 2011: 3n -2 =2011. Then 3n=2013, so n=2013/3=671. Exactly. So again, n=671. So answer C. That's three different ways of looking at it, all leading to the same answer. So I feel confident that C is the correct choice.Final Answerboxed{C}</think>To determine which term 2011 is in the arithmetic sequence 1, 4, 7, 10, ..., we use the formula for the nth term of an arithmetic sequence: (a_n = a_1 + (n - 1)d). Here, the first term (a_1 = 1) and the common difference (d = 3). We need to find (n) such that (a_n = 2011).1. Start with the formula for the nth term: [ 2011 = 1 + (n - 1) cdot 3 ]2. Subtract 1 from both sides: [ 2010 = (n - 1) cdot 3 ]3. Divide both sides by 3: [ 670 = n - 1 ]4. Add 1 to both sides: [ n = 671 ]To verify, we can substitute (n = 671) back into the formula:[a_{671} = 1 + (671 - 1) cdot 3 = 1 + 670 cdot 3 = 1 + 2010 = 2011]Another verification using the formula (a_n = 3n - 2) (derived from the sequence):[3n - 2 = 2011 implies 3n = 2013 implies n = 671]Thus, 2011 is the 671st term of the sequence. The answer is (boxed{C}).,<think>Okay, let's see. The problem is asking which term 2011 is in the arithmetic sequence 1, 4, 7, 10, and so on. The options are A: 669, B: 670, C: 671, D: 672. Hmm. I need to figure out the position of 2011 in this sequence. First, let me recall what an arithmetic sequence is. It's a sequence where each term after the first is obtained by adding a constant difference. In this case, looking at the given numbers: 1, 4, 7, 10... The difference between each term is 4 - 1 = 3, 7 - 4 = 3, 10 - 7 = 3. So the common difference, d, is 3.The general formula for the nth term of an arithmetic sequence is a_n = a_1 + (n - 1)d. Here, a_1 is the first term, which is 1, d is 3. So the formula becomes a_n = 1 + (n - 1)*3.We need to find n such that a_n = 2011. Let me set up the equation:1 + (n - 1)*3 = 2011.Now, solving for n. Subtract 1 from both sides:(n - 1)*3 = 2010.Then divide both sides by 3:n - 1 = 2010 / 3.Calculating 2010 divided by 3. Let me do that. 3 goes into 2010 how many times? Well, 3*700 = 2100, which is too big. So 3*600 = 1800. 2010 - 1800 = 210. Then 3*70 = 210. So 600 + 70 = 670. Therefore, 2010 / 3 = 670. Hence, n - 1 = 670, so n = 671. Wait, that gives n as 671. Let me check the options. Option C is 671. So the answer should be C. But let me verify once again to make sure I didn't make a mistake.Let me plug n = 671 back into the formula. a_671 = 1 + (671 - 1)*3 = 1 + 670*3. 670*3 is 2010, so 1 + 2010 = 2011. Yep, that checks out. So 2011 is indeed the 671st term. Therefore, the correct answer is C.But hold on, the options are A:669, B:670, C:671, D:672. Hmm. So is there a chance I miscalculated 2010 divided by 3? Let me confirm. 3 times 600 is 1800, 2010 minus 1800 is 210. 3 times 70 is 210, so 600 +70 is 670. So 3*670=2010. So n-1=670, so n=671. That seems correct.Alternatively, maybe I made a mistake in setting up the formula. Let me check the formula again. The nth term is a1 + (n-1)d. So first term is when n=1: 1 + (1-1)*3 =1. Correct. Second term n=2:1 + (2-1)*3=4. Third term n=3:1 +2*3=7. Correct. So the formula is right. Therefore, 2011 is the 671st term. So option C. Therefore, the answer should be C: 671.Final Answerboxed{C}</think>To determine which term 2011 is in the arithmetic sequence 1, 4, 7, 10, ..., we start by noting the common difference (d = 3) and the first term (a_1 = 1). The general formula for the (n)-th term of an arithmetic sequence is:[a_n = a_1 + (n - 1)d]We need to find (n) such that (a_n = 2011). Substituting the known values into the formula:[2011 = 1 + (n - 1) cdot 3]Subtracting 1 from both sides:[2010 = (n - 1) cdot 3]Dividing both sides by 3:[670 = n - 1]Adding 1 to both sides:[n = 671]Thus, 2011 is the 671st term of the sequence. The correct answer is (boxed{C}).]

🔑:Given the arithmetic sequence: 1, 4, 7, 10… We have the first term a_1=1, and the common difference d=4-1=3, Therefore, the general formula for the nth term is a_n=a_1+(n-1)d=1+3(n-1)=3n-2. Let 3n-2=2011, solving this gives n=671. Therefore, 2011 is the 671st term of the arithmetic sequence: 1, 4, 7, 10… Hence, the correct choice is boxed{C}. From the arithmetic sequence: 1, 4, 7, 10…, we can determine the first term a_1=1 and the common difference d=4-1=3, which allows us to derive the general formula for the nth term a_n. Setting a_n=2011 and solving for n gives the answer. This question tests the understanding of the general formula for an arithmetic sequence, and it is considered a basic question.

🧠:[<think>Okay, let's try to figure out this physics problem. So, we have a small rubber ball moving between two vertical walls. One wall is stationary, and the other is moving away at a constant speed of 100 cm/s. The ball starts with an initial speed of 2017 cm/s, and we need to find its final speed after some collisions. The collisions are perfectly elastic, and the motion is horizontal. First, I need to visualize the setup. There's a stationary wall on one side, and a moving wall on the other. The moving wall is moving away at 100 cm/s. The ball is bouncing back and forth between these walls. Each time it hits a wall, it should reverse direction, right? And since the collisions are perfectly elastic, the speed relative to the wall should be preserved. But wait, one of the walls is moving. So when the ball collides with the moving wall, the speed of the wall might affect the ball's speed after the collision.Hmm, elastic collisions with moving walls. I remember that when a ball collides elastically with a moving wall, the relative speed between the ball and the wall is reversed. But since the wall is massive, its velocity doesn't change. So, if the wall is moving away, when the ball hits it, the ball's speed relative to the wall will reverse. Let me try to formalize this.Suppose the wall is moving with velocity ( u ). Let the ball's velocity before collision be ( v ). Then, in the reference frame of the wall, the ball is approaching with speed ( v - u ) (if the wall is moving away, then in the wall's frame, the ball is moving towards it at ( v - u )). After collision, the direction reverses, so the ball's velocity in the wall's frame becomes ( -(v - u) ). Then, converting back to the lab frame, the ball's velocity is ( -(v - u) + u = -v + 2u ).So, if the ball collides with the moving wall (moving away at speed ( u )), its velocity after collision is ( -v + 2u ). Similarly, when it collides with the stationary wall (which has speed 0), the velocity after collision is just ( -v ), since the same formula would give ( -v + 0 = -v ).So each collision with the stationary wall just reverses the direction, keeping the speed the same. Each collision with the moving wall reverses the direction and adds ( 2u ) to the speed. Wait, that seems like the ball gains speed each time it collides with the moving wall. But the moving wall is moving away, so maybe each collision happens when the wall is further away, and the time between collisions increases. However, the problem is asking for the final speed. Hmm. Maybe eventually, the ball can't catch up to the moving wall anymore, so the final speed would be when the ball's speed relative to the moving wall is less than the wall's speed, so it can't collide again. Then, the final speed would be whatever speed it has after the last collision.Alternatively, maybe the number of collisions is finite, and after some collisions, the ball's speed becomes such that it can't reach the moving wall again. So, we need to model each collision and track the speed until the point where the ball can no longer catch up with the moving wall.Let's break it down step by step.First, let's note that the moving wall is moving away at 100 cm/s. The ball starts with a speed of 2017 cm/s. Let's consider the first collision. The ball is moving towards the moving wall (which is initially at some position). Wait, but how far apart are the walls initially? The problem doesn't specify the initial distance between the walls. Hmm. Wait, maybe the initial positions don't matter because we are only concerned with the relative motion and the number of collisions. Wait, but the distance between the walls is increasing over time. So, when the ball bounces between the walls, each time it returns, the moving wall is further away. Therefore, the time between collisions increases each time. But without knowing the initial separation, how can we compute the number of collisions? Wait, maybe the problem is designed such that the number of collisions is finite, even without knowing the initial separation. Wait, but if the walls are moving apart at 100 cm/s, and the ball is bouncing between them, each time the ball collides with the moving wall, its speed increases by 2u. Wait, so after each collision with the moving wall, the ball's speed becomes v + 2u. Wait, no. Wait, let's go back.Earlier, we derived that after collision with the moving wall (speed u), the ball's velocity becomes -v + 2u. Wait, let's check that again.Suppose the wall is moving away at speed u. The ball is approaching the wall with speed v (in the lab frame). In the wall's frame, the ball is approaching at speed v - u (since the wall is moving away). After collision, the ball's speed in the wall's frame becomes -(v - u). Then, converting back to the lab frame: -(v - u) + u = -v + 2u. So yes, the ball's speed after collision is -v + 2u. The direction is reversed, and the speed becomes | -v + 2u |. Wait, if u is 100 cm/s and v is 2017 cm/s, then after the collision, the ball's speed would be | -2017 + 200 | = 1817 cm/s? Wait, no. Wait, direction matters here. Let me think.Wait, velocity is a vector, so if the ball is moving towards the moving wall (which is moving away), then the ball's velocity is positive (say, to the right), and the wall's velocity is also positive (to the right). Then, in the wall's frame, the ball's velocity is v - u (since the wall is moving to the right at u). If v > u, then the ball is catching up to the wall. After collision, in the wall's frame, the velocity becomes -(v - u). Then, converting back to the lab frame, it's -(v - u) + u = -v + 2u. So, the velocity becomes -v + 2u. Since the wall is moving away, if the ball's velocity after collision is less than u, the ball will be moving to the left, but at a speed less than the wall's speed to the right. Wait, so if -v + 2u is negative, that means the ball is moving to the left. But the speed is | -v + 2u |. But in any case, the speed after collision with the moving wall would be | -v + 2u |. Wait, but if 2u > v, then the velocity would be positive, meaning the ball is moving in the same direction as the wall, but with speed 2u - v. Hmm. Wait, maybe I need to think in terms of magnitudes.Wait, let's take numbers. Let's suppose the initial speed is 2017 cm/s to the right. The moving wall is moving to the right at 100 cm/s. So, first, the ball is moving towards the right, faster than the wall. So, the ball will catch up to the moving wall. At the moment of collision, the ball's speed is 2017 cm/s to the right, wall's speed is 100 cm/s to the right. In the wall's frame, the ball is approaching at 2017 - 100 = 1917 cm/s. After collision, it moves away at -1917 cm/s. Convert back to lab frame: -1917 + 100 = -1817 cm/s. So the ball's velocity after the first collision with the moving wall is -1817 cm/s (to the left). Then, the ball travels to the left towards the stationary wall. It collides with the stationary wall, reversing its velocity to +1817 cm/s (to the right). Then, the ball moves back towards the moving wall, which is now further away. The time between collisions with the moving wall will depend on how far the wall has moved during the ball's trip to the stationary wall and back.Wait, so after each collision with the moving wall, the ball's speed decreases by 2u? Wait, in the first collision, the speed went from 2017 to 1817, so a decrease of 200 cm/s, which is 2u (since u = 100). Then, in the next collision, if the ball is going 1817 cm/s towards the right, the wall is moving at 100 cm/s. Then, the relative speed is 1817 - 100 = 1717. After collision, in the wall's frame, it's -1717, so lab frame speed is -1717 + 100 = -1617 cm/s. So speed decreases by another 200 cm/s. So each collision with the moving wall reduces the speed by 200 cm/s. Then, this process continues until the ball's speed after a collision with the moving wall is such that it can't catch up to the wall again. Wait, but how does that happen?Wait, after each collision with the moving wall, the ball's speed is reduced by 200 cm/s. So starting from 2017 cm/s, after each collision with the moving wall, it becomes 2017 - 200n cm/s, where n is the number of collisions with the moving wall. But this can't go on forever, because at some point, the ball's speed after a collision would be less than the wall's speed (100 cm/s). Wait, but in the above example, even if the ball's speed is 117 cm/s, for example, then the relative speed would be 117 - 100 = 17 cm/s. Then, after collision, it would be -17 cm/s in the wall's frame, so lab frame speed would be -17 + 100 = 83 cm/s. Wait, so the speed would be 83 cm/s to the left. Then, when it bounces off the stationary wall, it becomes 83 cm/s to the right. Now, the wall is moving away at 100 cm/s, and the ball is moving towards it at 83 cm/s. But since the wall is moving away faster than the ball is approaching, the ball will never catch up. So, in that case, the ball's speed is 83 cm/s, and that's the final speed.Wait, but in that case, the final speed would be 2017 - 200n, where n is the number of collisions with the moving wall before the ball's speed becomes less than the wall's speed. Wait, but let's compute how many collisions with the moving wall occur before the ball's speed is less than the wall's speed.Wait, each collision reduces the speed by 200 cm/s. Starting from 2017 cm/s. So we need to find n such that 2017 - 200n < 100. Solving for n:2017 - 200n < 1002017 - 100 < 200n1917 < 200nn > 1917 / 200n > 9.585So n must be 10. So after 10 collisions with the moving wall, the speed would be 2017 - 200*10 = 2017 - 2000 = 17 cm/s. Then, after the 10th collision with the moving wall, the ball's speed would be 17 cm/s to the left. Then, it bounces off the stationary wall to 17 cm/s to the right. But the moving wall is moving away at 100 cm/s. Since the ball is moving at 17 cm/s to the right, which is slower than the wall's 100 cm/s, the distance between them is increasing, so the ball will never catch up. Hence, the final speed is 17 cm/s.Wait, but let's check this step by step. Let me verify with n=10.First collision: speed becomes 2017 - 200*1 = 1817Second collision: 1817 - 200 = 1617Third: 1417Fourth: 1217Fifth: 1017Sixth: 817Seventh: 617Eighth: 417Ninth: 217Tenth: 17Yes, after 10 collisions with the moving wall, the speed is 17 cm/s. Then, the ball is moving to the right at 17 cm/s, and the wall is moving away at 100 cm/s. Since 17 < 100, the ball can't catch up. Therefore, the final speed is 17 cm/s.But wait, the problem says "find its final speed". So is it 17 cm/s? Let me check.But let's think again. Each collision with the moving wall reduces the speed by 200 cm/s. So 2017 divided by 200 is 10.085. So 10 times 200 is 2000, so 2017 - 2000 = 17. Therefore, after 10 collisions, it's 17. Then, after that, the ball can't catch up to the moving wall, so it just bounces off the stationary wall once more? Wait, let's track the sequence of collisions.Start with the ball moving to the right at 2017 cm/s.1. Collides with moving wall (right wall), speed becomes 2017 - 2*100 = 1817 cm/s to the left.2. Collides with stationary wall (left wall), speed becomes 1817 cm/s to the right.3. Collides with moving wall again, speed becomes 1817 - 200 = 1617 cm/s to the left.4. Collides with stationary wall, 1617 cm/s to the right.5. Collides with moving wall: 1617 - 200 = 1417 cm/s left.6. Collides with stationary: 1417 right.7. Moving wall collision: 1417 - 200 = 1217 left.8. Stationary: 1217 right.9. Moving: 1217 -200=1017 left.10. Stationary:1017 right.11. Moving:1017 -200=817 left.12. Stationary:817 right.13. Moving:817-200=617 left.14. Stationary:617 right.15. Moving:617-200=417 left.16. Stationary:417 right.17. Moving:417-200=217 left.18. Stationary:217 right.19. Moving:217-200=17 left.20. Stationary:17 right.Now, the ball is moving to the right at 17 cm/s. The moving wall is moving to the right at 100 cm/s. So, the distance between them is increasing. The ball can't catch up, so there are no more collisions. Therefore, the final speed is 17 cm/s.Therefore, the answer is 17 cm/s. But let me check if this makes sense. Each collision with the moving wall reduces the speed by 200 cm/s. So, after 10 collisions with the moving wall, the speed is 2017 - 10*200 = 17. But in the process, each collision with the moving wall is followed by a collision with the stationary wall. So the total number of collisions is 20, with 10 collisions on each wall. However, the final collision is with the stationary wall, after which the ball is moving towards the moving wall but can't reach it. Therefore, the final speed is 17 cm/s.But let's confirm with the initial equations. Each time the ball hits the moving wall, its speed is reduced by 200 cm/s. So the number of times it can hit the moving wall before its speed is less than 100 cm/s is floor((2017 - 100)/200) = floor(1917/200) = floor(9.585) = 9. But wait, that would mean 9 collisions? But in our previous count, it was 10 collisions. Hmm, there's a discrepancy here.Wait, let's calculate n such that 2017 - 200n < 100. Then, 2017 - 100 = 1917 < 200n => n > 1917/200 = 9.585. So n must be 10. Therefore, after 10 collisions with the moving wall, the speed is 17 cm/s. Therefore, the answer is 17 cm/s.But why does the formula suggest 10 collisions, but when we counted step by step, it took 10 collisions with the moving wall (each followed by a collision with the stationary wall), leading to 20 total collisions, with the final speed being 17 cm/s. So the answer is 17 cm/s.Alternatively, maybe we need to model each collision and the time between collisions. Wait, but since the problem doesn't give an initial distance between the walls, perhaps we can assume that the number of collisions is determined solely by the speed reduction. However, without knowing the initial separation, how can we know the time between collisions? Wait, but if we think about it, each time the ball goes from the stationary wall to the moving wall, the distance between the walls has increased, so the time it takes for the ball to travel is longer each time. However, since the speed of the ball is decreasing by 200 cm/s each time it collides with the moving wall, perhaps the number of collisions is finite because the speed of the ball will eventually be less than the speed of the moving wall, making it impossible to catch up.But if the problem states that the motion is always horizontal and collisions are perfectly elastic, then the ball will keep bouncing until it can't catch up. Therefore, the final speed would be the speed after the last collision with the moving wall, which is 17 cm/s. Therefore, the answer is boxed{17} cm/s.Wait, but the problem says "find its final speed". So, after the last collision, which is with the stationary wall, the ball is moving towards the moving wall but can't reach it. So, the final speed is 17 cm/s. Therefore, yes, the answer is 17 cm/s.But let me check if this approach is correct. In other words, does each collision with the moving wall reduce the ball's speed by 2u? Yes, because in the frame of the moving wall, the ball's speed is reversed, which translates to a reduction of 2u in the lab frame. Therefore, each collision with the moving wall reduces the speed by 200 cm/s. Therefore, the number of collisions with the moving wall is floor((2017 - 100)/200) + 1? Wait, floor(1917/200) = 9, so 9 + 1 = 10 collisions. Therefore, the final speed is 2017 - 10*200 = 17 cm/s.Yes, this seems consistent. Therefore, the final answer is 17 cm/s. So, boxed{17}.Final Answerboxed{17}</think>Given a small rubber ball moving between two vertical walls, where one wall is stationary and the other is moving away at a constant speed ( u = 100 , text{cm/s} ), we need to find the final speed of the ball after collisions. The initial speed of the ball is ( v_0 = 2017 , text{cm/s} ), and all collisions are perfectly elastic.1. Collision Analysis: - When the ball collides with the stationary wall, its speed remains the same, but its direction reverses. - When the ball collides with the moving wall, the relative speed between the ball and the wall is reversed. Converting back to the lab frame, the ball's speed after collision with the moving wall is reduced by ( 2u ).2. Speed Reduction: - Each collision with the moving wall reduces the ball's speed by ( 200 , text{cm/s} ) (since ( 2u = 200 , text{cm/s} )). - The number of collisions with the moving wall before the ball's speed becomes less than the wall's speed (100 cm/s) is calculated as follows: [ 2017 - 200n < 100 implies 2017 - 100 < 200n implies 1917 < 200n implies n > 9.585 ] Thus, ( n = 10 ) collisions are required.3. Final Speed Calculation: - After 10 collisions with the moving wall, the ball's speed is: [ 2017 - 200 times 10 = 2017 - 2000 = 17 , text{cm/s} ] - After this speed reduction, the ball's speed is 17 cm/s, which is less than the wall's speed of 100 cm/s. Therefore, the ball cannot catch up to the moving wall, and this is the final speed.The final speed of the ball is (boxed{17}) cm/s.,<think>Okay, so I have this physics problem here where a small rubber ball is moving between two vertical walls. One wall is stationary, and the other is moving away at a constant speed of 100 cm/s. The ball's motion is horizontal, and all collisions are perfectly elastic. The initial speed of the ball is 2017 cm/s, and I need to find its final speed. Hmm, let me try to figure this out step by step.First, let me visualize the setup. There are two walls: one is fixed, and the other is moving away to the right, say, at 100 cm/s. The ball is bouncing between them. Since the collisions are perfectly elastic, the kinetic energy should be conserved in each collision. But wait, one of the walls is moving. So when the ball collides with the moving wall, the frame of reference might be different, right? Because the wall itself is moving, so the relative speed of the ball with respect to the wall would matter for the collision.Let me recall how elastic collisions work when one of the objects is moving. In an elastic collision, both momentum and kinetic energy are conserved. But here, the wall is massive, so its mass can be considered infinite. In that case, when the ball collides with a stationary wall, it just reverses direction with the same speed. But if the wall is moving, even though it's massive, its motion affects the collision.Wait, but how does that work exactly? If the wall is moving, say, away from the ball, then when the ball hits the wall, the wall is receding. So the collision might not just reverse the direction but also transfer some energy. But since the wall is massive, its velocity doesn't change. Hmm, so maybe we need to use the relative velocity approach here.In the case of a perfectly elastic collision with a massive moving wall, the formula for the velocity of the ball after collision is given by:If the wall is moving with velocity ( u ), then the ball's velocity ( v' ) after collision is ( v' = 2u - v ). Wait, is that right?Wait, let me think. In the rest frame of the wall, the wall is stationary. The ball approaches with velocity ( v - u ) (since the wall is moving away at u, the relative speed is v - u). After collision, in the wall's frame, the velocity reverses, so it becomes ( -(v - u) ). Then, transforming back to the lab frame, we add the wall's velocity u, so the ball's velocity becomes ( - (v - u) + u = -v + 2u ). So yes, ( v' = 2u - v ).But wait, that formula is when the wall is moving. Let me check again. Suppose the wall is moving away from the ball. So if the ball is moving towards the wall with velocity v, and the wall is moving away with u. The relative velocity before collision is v - u. After collision, the relative velocity reverses, so it's -(v - u). Therefore, in the lab frame, the ball's velocity is -(v - u) + u = -v + 2u. That seems correct.But in this problem, one wall is stationary and the other is moving away. So the ball will collide with both walls multiple times. Each time it collides with the moving wall, its velocity changes according to ( v' = 2u - v ), and each time it collides with the stationary wall, it just reverses direction with the same speed.But wait, the moving wall is moving away, so the distance between the walls is increasing over time. Therefore, the time between collisions might be changing. The ball's speed will change after each collision with the moving wall, so each subsequent collision will happen after a longer time? Hmm, this seems complex. Maybe I need to model the motion step by step.Alternatively, maybe there's a trick here because of energy conservation or something else. Let me see.First, note that each collision with the moving wall imparts some speed to the ball. But since the wall is moving away, maybe the ball loses speed each time it collides with the moving wall? Wait, no. If the wall is moving away, when the ball collides with it, the relative speed is ( v - u ). After collision, the relative speed becomes ( -(v - u) ), so the ball's speed in the lab frame becomes ( - (v - u) + u = -v + 2u ). So if the ball was approaching the moving wall at speed v, after collision, it moves away with speed ( 2u - v ). If ( 2u - v ) is positive, the ball is moving towards the stationary wall. If it's negative, it's moving in the same direction as the moving wall, but since the wall is moving away, maybe the ball can't catch up again.Wait, so perhaps there's a point where after some collisions, the ball's speed becomes less than u, so it can no longer catch up to the moving wall, and thus the process stops. Then the final speed would be when it can't collide anymore. Hmm.So the problem is similar to a ball bouncing between two walls where one wall is moving away. Each time the ball collides with the moving wall, it loses some speed. The question is, after how many collisions will the ball's speed drop below the speed of the moving wall, making it impossible to catch up again. Then, the final speed would be the speed after the last collision.Alternatively, maybe energy considerations can help here. But since the wall is massive, kinetic energy isn't conserved in the usual sense. The wall's kinetic energy doesn't change, but the ball's does. Wait, but each collision with the moving wall transfers energy from the ball to the wall or vice versa?Wait, in the frame of the moving wall, when the ball collides, its velocity changes direction but not magnitude. So in that frame, the ball's kinetic energy is conserved. But when we transform back to the lab frame, the ball's kinetic energy changes because the reference frame is moving.Hmm, maybe not the best approach.Alternatively, let's think about the ball's speed after each collision with the moving wall. Let's denote the ball's speed after the nth collision with the moving wall as ( v_n ). Each time it collides with the moving wall, its speed becomes ( v_{n+1} = 2u - v_n ). But also, each time it collides with the stationary wall, it just reverses direction. So the sequence of speeds would be:Start with speed ( v_0 = 2017 ) cm/s towards the moving wall. Wait, initially, is the ball moving towards the moving wall or the stationary wall? The problem says the walls are vertical, and the ball is moving horizontally. The moving wall is moving away from the stationary one. So depending on the initial direction, the ball could start by moving towards either wall. But the problem doesn't specify the initial direction. Hmm. Wait, maybe it's implied that the ball is starting between the walls, so it's moving towards one of them. But since the walls are massive, and the collisions are elastic, maybe the initial direction doesn't matter? Wait, but the moving wall is moving away. Suppose the ball starts near the stationary wall, moving towards the moving wall. Then it would collide with the moving wall, which is moving away. Alternatively, if it starts moving towards the stationary wall, it would collide with the stationary wall first, reverse direction, then head towards the moving wall. Hmm. But the problem doesn't specify the initial position, just the initial speed. Maybe we can assume that the initial direction is towards the moving wall? Or maybe the problem is symmetric. Wait, maybe the answer doesn't depend on the initial direction. Hmm, perhaps not. Let me think.Suppose the ball starts moving towards the moving wall. Then, first collision is with the moving wall. The speed after collision would be ( v_1 = 2u - v_0 = 200 - 2017 = -1817 ) cm/s. Negative sign indicates direction opposite to the initial direction, so now the ball is moving towards the stationary wall. Then, it collides with the stationary wall, reversing direction again, so speed becomes 1817 cm/s towards the moving wall. Then, before the next collision with the moving wall, the distance between the walls has increased because the moving wall is moving away at 100 cm/s. So the time between collisions increases each time.Alternatively, maybe there's a pattern here. Each time the ball collides with the moving wall, its speed decreases by ( 2(v_n - u) )? Wait, no. Wait, the formula is ( v_{n+1} = 2u - v_n ). So each collision with the moving wall inverts the speed relative to the moving wall.Wait, let's write down the sequence. Starting with ( v_0 = 2017 ).First collision with moving wall: ( v_1 = 2u - v_0 = 200 - 2017 = -1817 ). Then collision with stationary wall reverses direction: ( v_2 = 1817 ).Second collision with moving wall: ( v_3 = 2u - v_2 = 200 - 1817 = -1617 ). Then collision with stationary wall: ( v_4 = 1617 ).Third collision: ( v_5 = 2u - v_4 = 200 - 1617 = -1417 ). Then ( v_6 = 1417 ).Continuing this pattern:v7 = 200 - 1417 = -1217v8 = 1217v9 = 200 - 1217 = -1017v10 = 1017v11 = 200 - 1017 = -817v12 = 817v13 = 200 - 817 = -617v14 = 617v15 = 200 - 617 = -417v16 = 417v17 = 200 - 417 = -217v18 = 217v19 = 200 - 217 = -17v20 = 17v21 = 200 - 17 = 183Wait, here after v20 =17 cm/s towards the moving wall. Then, after collision with moving wall, speed becomes 2u -17 = 200 -17=183 cm/s away from the moving wall, which is towards the stationary wall. Then, collision with stationary wall reverses direction: 183 cm/s towards the moving wall. Then, the next collision with the moving wall would be 200 -183=17 cm/s. Wait, so now the speed alternates between 17 and 183? Wait, that can't be. Because after 17 cm/s towards moving wall, collision gives 200 -17=183 cm/s away. Then collision with stationary wall reverses to 183 towards moving wall. Then collision with moving wall gives 200 -183=17. So it's entering a cycle between 17 and 183? Hmm, but that would be perpetual. But in reality, the distance between the walls is increasing because the moving wall is moving away at 100 cm/s. So even if the ball's speed is 17 cm/s towards the moving wall, the wall is moving away at 100 cm/s, so the ball can't catch up. Wait, so once the ball's speed towards the moving wall is less than the wall's speed, it will never collide again. Therefore, the last collision occurs when the ball's speed after collision is still greater than the wall's speed. Wait, but how does that work?Wait, let's think about the process. Each time the ball collides with the moving wall, it's moving towards it with some speed. The moving wall is moving away at 100 cm/s. For the ball to collide again with the moving wall, after bouncing off the stationary wall, it needs to cover the distance between the walls, which is increasing. So even if the ball's speed is higher than the wall's speed, the time it takes to go from the stationary wall to the moving wall depends on both the ball's speed and the wall's speed.Wait, this is getting complicated. Maybe instead of tracking each collision, we can model the ball's speed and the distance between the walls over time.Alternatively, think in terms of relative velocity. When the ball is going towards the moving wall, its speed relative to the wall is ( v - u ). After collision, the relative speed becomes ( -(v - u) ), so in the lab frame, the ball's speed is ( - (v - u) + u = -v + 2u ). Then, the ball travels towards the stationary wall, which is now at a distance that has increased by ( u times Delta t ), where ( Delta t ) is the time taken for the ball to go from the moving wall to the stationary wall and back. Wait, but the time to go from moving wall to stationary wall after the first collision is ( frac{d_0 + u Delta t_1}{v_1} ), but this is getting recursive.Alternatively, perhaps there's a way to model the number of collisions until the ball's speed is less than the wall's speed. Let's see.Looking at the sequence we had earlier:v0 = 2017After first collision with moving wall: v1 = -1817 (moving towards stationary wall)Then collision with stationary wall: v2 = 1817 (moving towards moving wall)Second collision with moving wall: v3 = -1617v4=1617Third collision: v5=-1417v6=1417Continuing this, each time the speed decreases by 200 cm/s: 2017, 1817, 1617, 1417,... until we reach a point where the speed is less than u=100 cm/s.Wait, but in the sequence above, when we got to v19 = -17, then v20=17. Then next collision would be v21=200 -17=183. Wait, that's higher than 100 cm/s. Then 183, collision with moving wall gives 200 -183=17. So it alternates between 17 and 183. But 17 is less than 100, so the next time the ball is moving towards the moving wall with 17 cm/s, while the wall is moving away at 100 cm/s. Therefore, the ball can't catch up. Hence, after speed becomes 17 cm/s towards the moving wall, there are no more collisions. So the final speed is 17 cm/s. But wait, in the problem statement, it's asking for the final speed. If the last collision is when the ball's speed is 17 cm/s away from the moving wall (towards the stationary wall), then after that, it moves towards the stationary wall, bounces back with 17 cm/s towards the moving wall, but since the wall is moving away faster (100 cm/s) than the ball is approaching (17 cm/s), the ball can't catch up. Therefore, the final speed is 17 cm/s. But wait, in the sequence, after v20=17, it collides with the moving wall and becomes 183 cm/s. Wait, no. Wait, after v19=-17, the ball is moving towards the stationary wall at 17 cm/s, collides with the stationary wall, reversing to 17 cm/s towards the moving wall (v20=17). Then, it's moving towards the moving wall at 17 cm/s. But the moving wall is moving away at 100 cm/s. So the relative speed is 17 - 100 = -83 cm/s. Wait, that means the ball is moving towards the wall at 17 cm/s, but the wall is moving away at 100 cm/s, so the distance between them is increasing. So the ball will never reach the wall again. Therefore, after the ball is moving towards the moving wall at 17 cm/s, it can't catch up, so there are no more collisions. Hence, the final speed is 17 cm/s. Therefore, the answer is 17 cm/s.Wait, but according to the sequence above, after v19=-17 (moving towards stationary wall at 17 cm/s), collides with stationary wall, becomes v20=17 (towards moving wall). Then, since the moving wall is moving away at 100 cm/s, the ball can't catch up, so there's no collision. Therefore, the final speed is 17 cm/s.Therefore, the answer is boxed{17}.Wait, but let me check the steps again to make sure.Starting at 2017 cm/s towards the moving wall.First collision: speed becomes 2*100 -2017 = 200 -2017 = -1817 cm/s (moving towards stationary wall).Collision with stationary wall: reverses to 1817 cm/s towards moving wall.Second collision: 2*100 -1817 = -1617, then reverses to 1617.Continuing this, each time subtracting 200.2017, 1817, 1617, 1417,..., decreasing by 200 each time.Wait, wait. Wait, the difference between each collision is 200 cm/s decrease. So starting from 2017, subtract 200 each time.Wait, 2017 - 200 = 1817, then 1817 -200=1617, etc. So how many times can we subtract 200 from 2017 until we get below 100?Wait, but according to our earlier sequence, after 10 subtractions:2017 -10*200=2017-2000=17. So after 10 collisions with the moving wall, the speed is 17 cm/s. But then, after that, the next collision would result in 2*100 -17=183. Wait, but that contradicts the previous thought. Wait, perhaps the step-by-step approach was wrong?Wait, no. Let me re-examine.Wait, each collision with the moving wall changes the speed as v' = 2u - v. So starting from 2017:1st collision: 200 -2017= -1817Then collision with stationary wall: 18172nd collision: 200 -1817= -1617Collision with stationary wall: 16173rd collision: 200 -1617= -1417...Continuing this, each time, after collision with moving wall, the speed is 2u - v. So starting from v0=2017, the speeds after each collision with moving wall are:v1=2u -v0=200 -2017=-1817v2=2u -v1=200 -(-1817)=200 +1817=2017? Wait, wait, no.Wait, hold on. Wait, maybe I made a mistake here. Because after the first collision, the ball is moving towards the stationary wall with speed 1817 cm/s (after collision with stationary wall). Then, when it collides again with the moving wall, which is now further away. But in the formula, is it 2u -v each time?Wait, perhaps not. Wait, the formula v' = 2u -v is when the ball is moving towards the moving wall with speed v. But after the first collision, the ball is moving away from the moving wall with speed 1817 cm/s towards the stationary wall. Then, when it comes back, it's moving towards the moving wall again with speed 1817 cm/s. Then, the next collision with the moving wall would result in speed 2u -v = 200 -1817= -1617. So that's correct.But then, after collision with the stationary wall, it's 1617 towards the moving wall. Then, collision with moving wall: 200 -1617= -1417. So each time, the speed reduces by 200 cm/s. So 2017, 1817, 1617,... until we get to a point where the speed is less than u=100 cm/s.So 2017 -200*n <100.Solve for n:2017 -200n <1002017 -100 <200n1917 <200nn >1917/200=9.585.So n=10. So after 10 collisions with the moving wall, the speed would be 2017 -200*10=2017-2000=17 cm/s. Then, moving towards the moving wall at 17 cm/s. Since the wall is moving away at 100 cm/s, the ball cannot catch up. Hence, the final speed is 17 cm/s.Therefore, the answer is 17. But in the problem statement, the initial speed is 2017 cm/s, which is a prime number, maybe a red herring. Anyway, the calculation seems to show that after 10 collisions with the moving wall, the speed is 17 cm/s, which is less than the wall's speed, so the ball can't collide again. Hence, the final speed is 17 cm/s.But let me confirm this with a different approach. Let's model the time between collisions.Suppose the distance between the walls at time t is D(t) = D0 + u*t, where D0 is the initial distance. But wait, the problem doesn't mention the initial distance. Hmm, maybe it cancels out. Let's assume the initial distance is D0.First collision: ball starts at D0, moving towards the moving wall with speed v0=2017 cm/s. The time until collision is t1 = D0/(v0 + u). Because during time t1, the ball travels v0*t1, and the wall moves u*t1, so the distance covered by the ball plus the wall's movement equals D0. Hence, t1 = D0/(v0 + u).After collision, the ball's speed is v1 = 2u -v0 = 200 -2017= -1817 cm/s (moving towards the stationary wall). The distance between the walls at this time is D1 = D0 + u*t1.Then, the ball travels back to the stationary wall, which is at distance D1. The time taken is t2 = D1 / |v1| = (D0 + u*t1)/1817.But D1 = D0 + u*t1 = D0 + u*(D0/(2017 + 100)) = D0*(1 + 100/(2117)) = D0*(2117 +100)/2117 = D0*2217/2117.Then t2 = D1 /1817 = (D0*2217/2117)/1817 = D0*2217/(2117*1817).After that, the ball collides with the stationary wall and reverses direction to v2=1817 cm/s towards the moving wall. The distance between walls is now D2 = D1 + u*t2.Then, the time until next collision with the moving wall is t3 = (D2)/(v2 + u) = (D1 + u*t2)/(1817 + 100) = (D0*2217/2117 + u*(D0*2217/(2117*1817)))/1917.This is getting really complicated. The time between collisions increases each time, and the distance increases. However, the key point is that each time the ball collides with the moving wall, its speed decreases by 200 cm/s. Therefore, after n collisions with the moving wall, the speed is v_n = 2017 - 200n.We need to find n such that v_n < u =100. So 2017 -200n <100 ⇒200n>2017 -100=1917⇒n>1917/200=9.585. So n=10. Hence, the final speed is 2017 -200*10=17 cm/s.Therefore, regardless of the initial distance or the time between collisions, the speed after each collision with the moving wall decreases by 200 cm/s, leading to the final speed of 17 cm/s.Yes, this seems to confirm the earlier result. Therefore, the final speed of the ball is 17 cm/s.Final Answerboxed{17}</think>The problem involves a small rubber ball moving between two vertical walls, one stationary and the other moving away at a constant speed ( u = 100 , text{cm/s} ). The ball's initial speed is ( v_0 = 2017 , text{cm/s} ), and we need to find its final speed after all collisions. Key steps:1. Elastic Collisions: When the ball collides with the moving wall, its velocity changes according to the formula ( v' = 2u - v ). For the stationary wall, the ball's velocity simply reverses direction.2. Sequence of Speeds: Each collision with the moving wall reduces the ball's speed by 200 cm/s. The sequence of speeds after each collision with the moving wall is: ( 2017, 1817, 1617, 1417, ldots ).3. Termination Condition: The ball will continue to collide with the walls until its speed is less than the wall's speed ( u = 100 , text{cm/s} ).4. Final Speed Calculation: The speed after ( n ) collisions with the moving wall is ( v_n = 2017 - 200n ). Solving for ( n ) when ( v_n < 100 ): [ 2017 - 200n < 100 implies 200n > 1917 implies n > 9.585 ] The smallest integer ( n ) is 10. Thus, the final speed is: [ 2017 - 200 times 10 = 17 , text{cm/s} ]Therefore, the final speed of the ball is (boxed{17}).]

🔑:1. With each elastic collision with the moving wall, the direction of the ball's velocity changes to the opposite, and its speed relative to the Earth decreases by 2 u, i.e., by 200 cm/s. Upon collision with the stationary wall, only the direction changes. Thus, after n=10 collisions with the moving wall and the same number with the stationary wall, the ball's speed will become v=v_{0}-2 u n=17 mathrm{~cm} / mathrm{s}, and the ball will no longer be able to catch up with the receding wall. (15 points)

🧠:[<think>Okay, so I need to figure out the range of values for the real number ( a ) such that the function ( f(x) = x^2 + 2x + a ) is always greater than 0 for any ( x ) in the interval ([1, +infty)). Hmm, let's start by understanding what the problem is asking. First, the function is a quadratic in terms of ( x ). The general form of a quadratic function is ( ax^2 + bx + c ), and in this case, the coefficients are: leading coefficient (for ( x^2 )) is 1, the coefficient for ( x ) is 2, and the constant term is ( a ). The problem states that this quadratic must be positive for all ( x ) starting from 1 and going to infinity. Quadratic functions can open upwards or downwards depending on the sign of the leading coefficient. Here, since the leading coefficient is 1, which is positive, the parabola opens upwards. That means the function has a minimum point (vertex) and tends to infinity as ( x ) approaches both positive and negative infinity. However, our domain here is restricted to ( x geq 1 ), so we only need to ensure that the function is positive from ( x = 1 ) onwards. Since the parabola opens upwards, the minimum value of the function on the interval ([1, +infty)) will occur at the vertex if the vertex is within the interval, or at ( x = 1 ) if the vertex is to the left of ( x = 1 ). Therefore, we need to find the vertex of the quadratic and check where it is located relative to ( x = 1 ). The vertex of a quadratic ( ax^2 + bx + c ) occurs at ( x = -frac{b}{2a} ). For our function ( f(x) = x^2 + 2x + a ), the vertex is at ( x = -frac{2}{2*1} = -1 ). So the vertex is at ( x = -1 ), which is definitely to the left of ( x = 1 ). That means that on the interval ([1, +infty)), the function is increasing because the vertex is the minimum point, and once we pass the vertex, the function increases as ( x ) increases. Since the vertex is at ( x = -1 ), from ( x = 1 ) onwards, the function will be increasing. Therefore, the minimum value of the function on the interval ([1, +infty)) occurs at ( x = 1 ). So, if we can ensure that ( f(1) > 0 ), then the function will be positive for all ( x geq 1 ). Let's check that. Calculating ( f(1) ):( f(1) = (1)^2 + 2*(1) + a = 1 + 2 + a = 3 + a ).So, to have ( f(1) > 0 ), we need:( 3 + a > 0 )Which implies:( a > -3 ).Wait, but is that all? Let me verify. Since the function is increasing on ([1, +infty)), if the function is positive at ( x = 1 ), then it will stay positive for all larger ( x ). However, maybe I should check if there's a possibility of the function dipping below zero somewhere else in the interval. But since the vertex is at ( x = -1 ), which is outside the interval ([1, +infty)), and the function is increasing from ( x = -1 ) onwards, it's indeed increasing throughout ([1, +infty)). Therefore, the minimal value on that interval is at ( x = 1 ), so ensuring ( f(1) > 0 ) should be sufficient. But let's think again. Suppose someone challenges this reasoning. Let me consider if there's a case where even though the vertex is outside the interval, the function might still dip below zero. For example, if the function had a maximum instead of a minimum. But in our case, since it's a parabola opening upwards, the vertex is a minimum. If the vertex were in the interval, then we would need to check the value at the vertex. But here, the vertex is at ( x = -1 ), so to the left of our interval. Therefore, the function is increasing on ([1, +infty)), so the minimal value is at ( x = 1 ). So yes, if ( f(1) > 0 ), then the function is positive for all ( x geq 1 ). But wait, another thought: even if the vertex is outside the interval, could there be a point in the interval where the function is negative? Since the parabola opens upwards, once we pass the vertex, the function increases. Therefore, from ( x = -1 ) onwards, it's increasing. Since our interval starts at ( x = 1 ), which is to the right of the vertex, the function is already increasing there. Therefore, the minimal value in the interval is indeed at ( x = 1 ). Therefore, the answer should be ( a > -3 ). But let me check with another approach. Let's solve the inequality ( x^2 + 2x + a > 0 ) for ( x geq 1 ). Rewriting the inequality: ( a > -x^2 - 2x ). Therefore, for each ( x geq 1 ), ( a ) must be greater than ( -x^2 - 2x ). So to satisfy this inequality for all ( x geq 1 ), ( a ) must be greater than the maximum value of ( -x^2 - 2x ) on the interval ([1, +infty)). Let’s find the maximum of ( g(x) = -x^2 - 2x ) on ([1, +infty)). Since ( g(x) ) is a quadratic function opening downwards (because the coefficient of ( x^2 ) is -1), it has a maximum at its vertex. The vertex of ( g(x) ) is at ( x = -b/(2a) = -(-2)/(2*(-1)) = 2/(-2) = -1 ). So the vertex is at ( x = -1 ), which is outside the interval ([1, +infty)). Therefore, on the interval ([1, +infty)), ( g(x) ) is decreasing because the parabola opens downward, and the vertex is at ( x = -1 ). Thus, the maximum value of ( g(x) ) on ([1, +infty)) occurs at ( x = 1 ). Calculating ( g(1) = -(1)^2 - 2*(1) = -1 - 2 = -3 ). So the maximum value of ( g(x) ) on ([1, +infty)) is -3. Therefore, to satisfy ( a > -x^2 - 2x ) for all ( x geq 1 ), ( a ) must be greater than -3. This confirms the earlier result. So both approaches lead to the conclusion that ( a > -3 ). But let me test with some specific values to make sure. If ( a = -3 ), then the function becomes ( f(x) = x^2 + 2x - 3 ). Let's check ( f(1) = 1 + 2 - 3 = 0 ). So at ( x = 1 ), the function is zero. Since the function is increasing for ( x geq 1 ), when ( a = -3 ), the function is zero at ( x = 1 ) and positive for ( x > 1 ). But the problem states that ( f(x) > 0 ) must hold true for any ( x in [1, +infty) ). Therefore, ( a = -3 ) is not allowed because at ( x = 1 ), the function is zero, not positive. Thus, ( a ) must be greater than -3. If we take ( a = -4 ), then ( f(1) = 1 + 2 - 4 = -1 ), which is negative. Then, since the function is increasing, it will eventually become positive as ( x ) increases, but since the problem requires ( f(x) > 0 ) for all ( x geq 1 ), ( a = -4 ) is invalid. If we take ( a = -2 ), then ( f(1) = 1 + 2 - 2 = 1 > 0 ), and since the function is increasing, it remains positive for all ( x geq 1 ). So this works. Therefore, the range of values for ( a ) is all real numbers greater than -3. Another way to approach this could be by completing the square. Let's try that. Starting with ( f(x) = x^2 + 2x + a ). Completing the square for the quadratic part:( x^2 + 2x = (x + 1)^2 - 1 )Therefore, ( f(x) = (x + 1)^2 - 1 + a = (x + 1)^2 + (a - 1) )Since ( (x + 1)^2 ) is always non-negative, the minimum value of ( f(x) ) occurs when ( (x + 1)^2 = 0 ), which is at ( x = -1 ). The minimum value of ( f(x) ) is ( 0 + (a - 1) = a - 1 ). However, this minimum is at ( x = -1 ), which is outside our interval of interest ([1, +infty)). But in our interval, the minimal value is at ( x = 1 ), as we previously determined. Plugging ( x = 1 ) into the completed square form:( f(1) = (1 + 1)^2 + (a - 1) = 4 + a - 1 = 3 + a ), which matches our earlier calculation. So indeed, ensuring ( 3 + a > 0 ) gives ( a > -3 ). Therefore, all the methods lead to the same conclusion. Just to recap:1. The quadratic opens upwards, so it has a minimum at ( x = -1 ).2. The interval of concern is ([1, +infty)), which is to the right of the vertex.3. The function is increasing on this interval, so the minimal value is at ( x = 1 ).4. Ensuring the function is positive at ( x = 1 ) gives the condition ( a > -3 ).5. Verified with multiple methods: direct evaluation, rewriting the inequality, completing the square.Therefore, the range of values for ( a ) is all real numbers greater than -3.Final AnswerThe range of values for ( a ) is boxed{(-3, +infty)}.</think>Given the function ( f(x) = x^2 + 2x + a ), we need to determine the range of values for ( a ) such that ( f(x) > 0 ) for all ( x in [1, +infty) ).1. Quadratic Function Analysis: - The function ( f(x) = x^2 + 2x + a ) is a quadratic function opening upwards (since the coefficient of ( x^2 ) is positive). - The vertex of the quadratic function is at ( x = -frac{b}{2a} = -1 ), which is outside the interval ([1, +infty)).2. Behavior on the Interval: - Since the vertex is at ( x = -1 ), the function is increasing for ( x geq -1 ). Therefore, on the interval ([1, +infty)), the function is increasing. - The minimum value of ( f(x) ) on ([1, +infty)) occurs at ( x = 1 ).3. Evaluating at ( x = 1 ): - Calculate ( f(1) ): [ f(1) = 1^2 + 2 cdot 1 + a = 3 + a ] - To ensure ( f(x) > 0 ) for all ( x geq 1 ), we need ( f(1) > 0 ): [ 3 + a > 0 implies a > -3 ]4. Verification: - Rewriting the inequality ( x^2 + 2x + a > 0 ) as ( a > -x^2 - 2x ), we find the maximum value of ( -x^2 - 2x ) on ([1, +infty)). The maximum value occurs at ( x = 1 ) and is (-3). - Completing the square confirms the minimum value at ( x = 1 ) and the condition ( a > -3 ).Thus, the range of values for ( a ) is (boxed{(-3, +infty)}).,<think>Okay, so I have this problem: Given the function f(x) = x² + 2x + a, I need to find the range of values for the real number a such that f(x) > 0 for any x in the interval [1, +∞). Hmm, let me think about how to approach this.First, I remember that quadratic functions can be analyzed by looking at their graphs. Since the coefficient of x² here is 1, which is positive, the parabola opens upwards. That means the function has a minimum point (vertex) and tends to infinity as x increases. So, if the minimum value of f(x) on the interval [1, +∞) is greater than 0, then the entire function will be positive on that interval. Therefore, my goal should be to find the conditions on a such that the minimum value of f(x) in [1, +∞) is positive.To find the vertex of the quadratic function f(x) = x² + 2x + a, I can use the formula for the vertex of a parabola. The x-coordinate of the vertex is given by -b/(2a), where in the standard form ax² + bx + c. Wait, in this case, the coefficient of x² is 1, so a (in the standard form) is 1, and the coefficient of x is 2. So, the x-coordinate of the vertex is -2/(2*1) = -1. So the vertex is at x = -1.But the interval we're concerned with is [1, +∞). The vertex at x = -1 is actually to the left of this interval. So that means on the interval [1, +∞), the function is increasing because the parabola opens upwards and the vertex is at x = -1, which is the minimum point. Therefore, the function is increasing for all x > -1. Since our interval starts at x = 1, which is greater than -1, the function will be increasing on [1, +∞). So, the minimum value of f(x) on [1, +∞) will occur at x = 1.Therefore, if I ensure that f(1) > 0, then since the function is increasing for all x ≥ 1, the function will be positive on the entire interval. So, substituting x = 1 into f(x):f(1) = (1)² + 2*(1) + a = 1 + 2 + a = 3 + aSo, we need 3 + a > 0, which implies a > -3.Wait, is that all? Let me double-check. So, if the vertex is at x = -1, which is outside the interval [1, +∞), then the function is indeed increasing on [1, +∞), so the minimal value is at x = 1. Thus, the condition is simply that f(1) > 0, which leads to a > -3.But wait, let me verify this with another approach to make sure I didn't make a mistake.Alternatively, we can consider the inequality x² + 2x + a > 0 for all x ≥ 1. Rearranging this inequality, we get a > -x² - 2x for all x ≥ 1.So, the right-hand side, -x² - 2x, is a function of x. For a to satisfy the inequality for all x ≥ 1, a must be greater than the maximum value of -x² - 2x on [1, +∞).Therefore, let's consider the function g(x) = -x² - 2x. We need to find the maximum value of g(x) on [1, +∞). Then, a must be greater than this maximum value.To find the maximum of g(x), since it's a quadratic function opening downward (coefficient of x² is -1), its vertex will give the maximum point. The x-coordinate of the vertex is at -b/(2a) where here, in the function g(x) = -x² -2x, the coefficients are a = -1, b = -2. So, the vertex is at x = -(-2)/(2*(-1)) = 2 / (-2) = -1. So the maximum of g(x) is at x = -1, which again is outside our interval [1, +∞). Therefore, on [1, +∞), the function g(x) is decreasing because the parabola opens downward. Therefore, the maximum value on [1, +∞) is at x = 1.Therefore, the maximum value of g(x) on [1, +∞) is g(1) = -(1)^2 -2*(1) = -1 -2 = -3. Hence, a must be greater than -3.So both approaches lead me to the same conclusion: a > -3.But let me check with another example. Suppose a = -3. Then, the function becomes f(x) = x² + 2x -3. Let's check f(1): 1 + 2 - 3 = 0. So, at x = 1, f(1) = 0. But the problem states that f(x) > 0 for any x ∈ [1, +∞). Therefore, a = -3 is not acceptable because f(1) = 0, which violates the strict inequality. Therefore, a must be greater than -3.What if a is slightly larger than -3, say a = -2.9? Then f(1) = 3 + (-2.9) = 0.1 > 0. Since the function is increasing beyond x = 1, it will stay positive. So, that's good.But let me check another point. Let's take a = -3. Then, f(x) = x² + 2x -3. Let's check x = 2: f(2) = 4 + 4 -3 = 5 > 0. So even though at x =1 it's zero, but for x >1, it's positive. But the problem states that f(x) > 0 for any x ∈ [1, +∞). Therefore, since x=1 is included in the interval, and f(1) =0, which is not greater than 0, so a = -3 is invalid. Therefore, a must be greater than -3.Is there a possibility that even if a is greater than -3, the function might dip below zero somewhere else in the interval [1, +∞)? For instance, maybe there's another critical point?But we already considered that the vertex is at x = -1, which is outside the interval [1, +∞). Since the parabola opens upward, the function is increasing on [1, +∞). Therefore, once it's positive at x =1, it will stay positive for all x >1.Therefore, the only critical point is at x=1. So, as long as f(1) >0, the entire function is positive on [1, +∞).Wait, but just to be thorough, let's consider the derivative of f(x). f'(x) = 2x + 2. Setting this equal to zero gives 2x + 2 = 0 => x = -1. So, the critical point is at x = -1, which again is outside the interval [1, +∞). Therefore, on [1, +∞), the derivative is 2x + 2, which is always positive because x ≥1. Therefore, the function is strictly increasing on [1, +∞). Therefore, the minimal value on that interval is at x=1, and the rest is increasing. Hence, the condition is indeed f(1) >0 => a > -3.Therefore, the range of values for a is all real numbers greater than -3. In interval notation, that's (-3, +∞). So, the answer is a > -3.But let me check if there's an alternative method. Maybe completing the square?Let's try that. The function is f(x) = x² + 2x + a.Completing the square: x² + 2x = (x + 1)^2 -1. Therefore, f(x) = (x + 1)^2 -1 + a = (x + 1)^2 + (a -1).So, since (x + 1)^2 is always non-negative, the minimum value of f(x) is (a -1) when x = -1. But again, since x is in [1, +∞), the minimal value is at x =1. So plugging x=1 into the completed square: (1 +1)^2 + (a -1) = 4 + a -1 = 3 + a. Which again leads us to 3 + a >0 => a > -3.Same result. So, seems solid.Alternatively, suppose I use discriminant analysis. For a quadratic equation to be always positive, the discriminant must be negative. Wait, but in this case, the quadratic is not always positive for all x, but only for x in [1, ∞). So discriminant analysis might not directly apply here, because the quadratic can have roots, but as long as those roots are not in the interval [1, ∞), then f(x) >0 on that interval.Wait, maybe that's another approach. Let's see.Suppose the quadratic equation x² + 2x + a =0. The discriminant is D = 4 -4*1*a = 4 -4a.If D <0, then the quadratic is always positive. But since the leading coefficient is positive, D <0 implies the quadratic never crosses the x-axis. So, if 4 -4a <0 => 4 <4a => a >1. So, if a >1, then f(x) >0 for all real x, which is a stronger condition. But the problem only requires f(x) >0 for x ∈ [1, +∞). Therefore, even if the quadratic has roots, as long as those roots are less than 1, then on [1, +∞), f(x) would be positive.Therefore, another way to look at this is that the quadratic x² + 2x + a has its roots (if any) at x = [-2 ± sqrt(4 -4a)]/2 = [-2 ± 2*sqrt(1 -a)]/2 = -1 ± sqrt(1 -a). So, the roots are at x = -1 + sqrt(1 - a) and x = -1 - sqrt(1 -a). Since sqrt(1 -a) is real only if 1 -a ≥0 => a ≤1.Therefore, when a ≤1, the quadratic has real roots. The rightmost root is x = -1 + sqrt(1 -a). To ensure that f(x) >0 for all x ≥1, we need that the largest root (if any) is less than 1. Therefore, set the largest root to be less than 1:-1 + sqrt(1 -a) <1Solve for a:sqrt(1 -a) <2Square both sides (since sqrt(1 -a) is non-negative):1 -a <4Therefore, -a <3 => a >-3.But we also need that sqrt(1 -a) is real, so 1 -a ≥0 => a ≤1. Therefore, combining these two conditions:a >-3 and a ≤1.But wait, but if a >1, then there are no real roots, and since the quadratic opens upwards, f(x) >0 for all real x, which certainly includes [1, ∞). Therefore, when a >1, the condition is satisfied.But when a ≤1, we need the largest root to be less than 1. Which gives a >-3. So combining all cases:If a >1: f(x) >0 for all x.If -3 <a ≤1: f(x) has roots, but they are left of x=1, so f(x) >0 on [1, ∞).If a ≤-3: Then the largest root is ≥1, so f(x) would be ≤0 at some point in [1, ∞).Therefore, the full range of a is a >-3. Because when a >1, it's included in a >-3.Wait, but when a >1, it's a subset of a >-3. So the total range is a >-3. Whether there are real roots or not, the critical condition is a >-3. So even though the discriminant approach seems to split into cases, in the end, combining all cases where f(x) >0 on [1, ∞), we get a >-3.Therefore, this method also leads to the same conclusion: a must be greater than -3.Hence, I think the answer is a > -3, which in boxed form is boxed{(-3, +infty)}.Final AnswerThe range of values for the real number ( a ) is boxed{(-3, +infty)}.</think>Given the function ( f(x) = x^2 + 2x + a ), we need to determine the range of values for ( a ) such that ( f(x) > 0 ) for any ( x in [1, +infty) ).1. Vertex Analysis: - The vertex of the quadratic function ( f(x) = x^2 + 2x + a ) is at ( x = -1 ). Since the parabola opens upwards, the function is increasing for ( x > -1 ). - On the interval ([1, +infty)), the function is increasing because the vertex is at ( x = -1 ), which is outside the interval. Therefore, the minimum value on ([1, +infty)) occurs at ( x = 1 ).2. Evaluating at ( x = 1 ): - Substituting ( x = 1 ) into the function: ( f(1) = 1^2 + 2 cdot 1 + a = 3 + a ). - To ensure ( f(x) > 0 ) for all ( x in [1, +infty) ), we need ( f(1) > 0 ), which gives ( 3 + a > 0 ). Solving this inequality, we get ( a > -3 ).3. Alternative Approach (Completing the Square): - Rewriting ( f(x) ) by completing the square: ( f(x) = (x + 1)^2 + (a - 1) ). - The minimum value of ( f(x) ) at ( x = -1 ) is ( a - 1 ). However, since we are concerned with the interval ([1, +infty)), the minimum value at ( x = 1 ) is ( 3 + a ), leading to the same condition ( a > -3 ).4. Discriminant Analysis: - The quadratic equation ( x^2 + 2x + a = 0 ) has roots ( x = -1 pm sqrt{1 - a} ). - For ( f(x) ) to be positive on ([1, +infty)), the largest root must be less than 1. Solving ( -1 + sqrt{1 - a} < 1 ) gives ( sqrt{1 - a} < 2 ), leading to ( a > -3 ).Combining all methods, the range of values for ( a ) is (boxed{(-3, +infty)}).]

🔑:AnalysisThis problem involves the monotonicity and minimum value of a function. First, we find the axis of symmetry of the function to be x=-1. This implies that f(x) is an increasing function on [1,+infty). We can then transform the always-true condition into a problem about the function's minimum value.SolutionSince the axis of symmetry of f(x) = (x+1)^{2} + a - 1 is x=-1, f(x) is monotonically increasing on [1,+infty) and its minimum value is f(1).In order for f(x) > 0 to hold true,we only need f(1) > 0,1 + 2 + a > 0,a > -3.Thus, the range of values for a is boxed{(-3,+infty)}.

🧠:[<think>嗯，我现在需要解决的问题是在集合{1000, 1001, 1002, ..., 2000}中找出有多少对连续的整数，当这两个数相加时不需要进位。首先，我要理解清楚题目的意思。题目是说，当两个连续的整数相加的时候，每一位上的数字相加都不超过9，也就是不需要进位。比如，比如说1234和1235相加的话，每一位分别是1+1=2，2+2=4，3+3=6，4+5=9，都不超过9，所以这对数符合条件。反之，比如129和130相加的话，个位9+0=9没问题，十位2+3=5也没问题，百位1+1=2，没问题，所以不用进位。不过这里题目里的数都是四位数，对吧？因为集合是从1000到2000，所以每个数都是四位数，除了2000可能算四位数还是五位数？不过2000在这里应该被当作四位数处理，所以每个数都是四位数的，也就是从1000到2000，包括2000吗？题目里写的是“{1000,1001,1002, …, 2000}”所以2000是最后一个数，所以每个数都是四位数的，对吧？但是2000其实是四位数，所以没问题。接下来，我要找出所有这样的连续整数对，也就是形如(n, n+1)的对，其中n和n+1都属于这个集合，并且当n和n+1相加的时候，每一位上的数字相加都不超过9，这样就不会有进位。首先，我需要明确，这里的连续整数对是像（1000,1001），（1001,1002），一直到（1999,2000）这样的对吗？对的，所以总共有2000 - 1000 = 1000个数，所以这样的连续对应该是1000对，对吗？不过实际上，从1000到2000共有1001个数，所以连续对应该是1000对吗？是的，比如从a到b共有n个数的话，连续对就是n-1个，所以这里2000 - 1000 + 1 = 1001个数，所以连续对是1000对。但题目是要在这1000对中找出符合条件的对，也就是当n和n+1相加时，每一位相加都不进位。这里需要注意的是，题目中的“加”是直接相加，而不是考虑二进制或者其他进制，是十进制加法，对吗？接下来，我需要分析当两个连续的整数n和n+1相加时，每一位的情况。因为n和n+1是连续的，所以它们的各位数字有什么规律呢？例如，假设n是一个四位数abcd（每一位分别是a,b,c,d），那么n+1可能是abce，其中如果d不是9的话，那么个位就是d+1，其他位不变。但如果d是9的话，那么个位变为0，然后进位到十位，也就是c+1，如果c不是9的话，否则继续进位，以此类推。比如，n=1239，那么n+1=1240；n=1299，n+1=1300；n=1999，n+1=2000。现在的问题是，当我们将n和n+1相加的时候，每一位的数字相加是否都不超过9。也就是说，每一位上，n的该位数字加上n+1的该位数字是否<=9，这样就不会产生进位。需要注意的是，这里可能会有进位的影响，但由于n和n+1是连续的，因此它们之间本身的进位可能只在某些位上有变化。不过，这个时候，我需要仔细分析每一位的情况。例如，假设n的各位数字是a, b, c, d，那么n+1的各位数字可能是：- 如果d < 9，则n+1的各位是a, b, c, d+1；- 如果d=9且c < 9，则n+1的各位是a, b, c+1, 0；- 如果d=9，c=9，且b <9，则n+1的各位是a, b+1, 0, 0；- 如果d=9，c=9，b=9，且a <9，则n+1的各位是a+1, 0, 0, 0；- 在本题中，由于n的范围是1000到2000，当n=1999时，n+1=2000，所以此时a=1，变成2，而其他各位都是0。现在，当我们将n和n+1相加时，各个位的数字相加情况如何？例如，当n的个位d不是9的时候，n+1的个位是d+1，那么个位的相加就是d + (d+1) = 2d +1。如果这个结果<=9，那么个位不需要进位。类似地，其他位的情况要看是否有进位，但由于n和n+1是连续的，所以如果n的某一位之后的所有位都是9，那么n+1会在该位加1，后面所有位变为0。比如，如果n是1239，那么n+1=1240，此时十位是3+1=4，个位变为0。这时候，当n和n+1相加时，各个位的数字相加情况为：n: 1 2 3 9n+1: 1 2 4 0相加结果各位分别是：千位：1 + 1 = 2百位：2 + 2 = 4十位：3 + 4 = 7个位：9 + 0 = 9这里每个位相加都是<=9，所以不需要进位，对吗？不过这时候，这个例子中的n=1239和n+1=1240相加是否真的不需要进位呢？但根据这个例子，每个位相加的结果都是<=9，所以确实不需要进位。这说明当n的个位是9的时候，n+1的个位是0，同时十位如果是c，在n+1中十位是c+1吗？不过可能要看是否有进位。例如，如果n是1239，那么n+1=1240，十位本来是3，变为4，对吗？所以当n=1239时，n+1的十位是4，个位是0，对吗？所以当n和n+1相加时，各个位的数字相加是：千位：1+1=2百位：2+2=4十位：3+4=7个位：9+0=9都不超过9，所以不需要进位，所以这对数1239和1240是符合要求的对吗？看来这样的例子是存在的，所以需要考虑这种情况。那么现在的问题转化为，如何系统地找出所有这样的n，使得n和n+1的各位数字相加都不超过9。首先，n是一个四位数，从1000到1999，以及2000吗？不过2000的下一个数2001不在集合中，所以2000和2001这对不存在于集合中，所以实际需要考虑的n是从1000到1999，因为n+1必须是2000的话，n=1999，所以n的范围是1000到1999，而n+1的范围是1001到2000，所以总共有1000对。现在，我需要分析每一个可能的n的各位数字，使得当n和n+1相加时，各位数字的和都不超过9。首先，我们需要明确n和n+1的各位数字之间的关系。假设n的各位数字是a（千位）、b（百位）、c（十位）、d（个位）。那么n+1的各位数字取决于d是否为9，c是否为9，等等。例如：- 如果d < 9，那么n+1的各位数字是a, b, c, d+1；- 如果d = 9，但c < 9，那么n+1的各位数字是a, b, c+1, 0；- 如果d = 9，c = 9，但b < 9，那么n+1的各位数字是a, b+1, 0, 0；- 如果d = 9，c = 9, b = 9，但a < 9，那么n+1的各位数字是a+1, 0, 0, 0；- 在本题中，a的取值范围是1或2，因为n在1000到2000之间，当a=1时，n在1000到1999之间；当a=2时，n=2000，但2000的下一个数是2001，不在集合中，所以a=2的情况只有n=2000，但没有对应的n+1在集合里，所以实际a只能是1或者当n=1999时，n+1=2000，这时候a=1变成2吗？不过这里需要注意，当n=1999时，n+1=2000，所以此时n的千位是1，而n+1的千位是2，百位、十位、个位都是0。现在，针对n和n+1的各位数字相加的情况：1. 当n的个位d <9时，n+1的个位是d+1，所以个位相加为d + (d+1) = 2d +1。要满足2d +1 <=9，即d <=4。因为当d=4时，2*4+1=9，满足；当d=5时，2*5+1=11>9，不满足。所以个位d必须是0到4之间，这样当d <9时，个位相加才不会进位。同时，在这种情况下，n+1的其他位（十位、百位、千位）都和n相同，所以其他位相加分别为a + a, b + b, c + c，对吗？例如，n的千位是a，n+1的千位也是a，所以千位相加是a + a <=9，所以a <=4。不过这里a是千位数字，而n的范围是1000到1999，所以a只能是1，对吗？当a=2时，n=2000，但2000的下一个数不在集合里，所以a只能是1。因此，千位上的数字相加是1 +1=2，满足<=9，没问题。同样地，百位上的数字相加是b + b <=9，所以b <=4，对吗？因为如果b >=5的话，b + b =2b >=10，就会进位。所以百位上的数字b必须<=4。同理，十位上的数字c + c <=9，所以c <=4。个位上的数字当d <9时，d <=4，所以个位相加不会进位；当d >=5时，如果d <9，那么d+1会导致个位相加超过9，对吗？比如d=5的话，n的个位是5，n+1的个位是6，相加5+6=11，这时候个位会有进位，所以不符合条件。所以在d <9的情况下，只有当d <=4时，个位相加才不会进位。但是如果n的个位d=9，那么n+1的个位变为0，同时十位c会加1，如果十位原本是c，现在变为c+1，除非c=9，这时候会继续进位到百位，以此类推。这个时候，当n的个位是9时，n+1的个位是0，十位是c+1（如果c<9），那么十位相加的情况是c + (c+1)吗？例如，假设n的十位c=3，个位d=9，那么n+1的十位变为4，个位变为0，所以十位相加是3 +4=7 <=9，对吗？这样没问题。但是如果十位c=5，d=9，那么n+1的十位是6，所以十位相加是5 +6=11，这时候会产生进位，所以不符合条件。所以，当n的个位是9时，十位需要满足c + (c+1) <=9，即2c +1 <=9，解得c <=4。同样，如果十位c=9，那么n+1的十位会变成0，而百位b会增加1，这时候需要考虑百位的情况。例如，n=1b99，其中b是百位，那么n+1=1(b+1)00，所以百位相加是b + (b+1) =2b +1 <=9，所以b <=4。同样，如果百位b=9，十位c=9，个位d=9，那么n+1=1(a+1)000，但原来的千位a=1，所以n+1=2000，这时候千位相加是1 +2=3 <=9，对吗？所以千位没问题，但百位、十位、个位都是0，所以相加的时候，百位是9 +0=9，对吗？但这里需要注意，当n=1999时，n+1=2000，相加的话各位分别是：n: 1 9 9 9n+1: 2 0 0 0相加各位的结果是：千位：1 +2=3百位：9 +0=9十位：9 +0=9个位：9 +0=9都不超过9，所以1999和2000这对符合条件吗？是的，所以这对是符合条件的。因此，即使百位、十位、个位都是9的情况下，当n=1999，n+1=2000，相加各位都是9或者3，都不超过9，对吗？所以，这个时候，我需要系统地考虑所有可能的进位情况，并确定每一位上的数字相加是否都不超过9。总结一下，当n和n+1相加时，各位相加的结果需要满足：- 千位：1 +1=2（当n的千位是1，n+1的千位也是1，除了当n=1999时，n+1=2000，此时千位相加是1+2=3）- 其他位的情况取决于是否有进位发生。但是实际上，当n=1999时，n+1=2000，这时候：n的千位是1，百位是9，十位是9，个位是9；n+1的千位是2，百位是0，十位是0，个位是0；相加时各位分别是：千位：1 +2=3百位：9 +0=9十位：9 +0=9个位：9 +0=9都是<=9，所以不需要进位，对吗？所以这对是符合条件的。因此，我需要分情况讨论：1. 当n的个位d <9时，n+1的个位是d+1，所以个位相加为d + (d+1)=2d+1 <=9，解得d <=4。同时，十位、百位、千位在n和n+1中是相同的，所以十位相加是c + c <=9，即c <=4；同理，百位b <=4；千位a=1，所以相加是1+1=2，没问题。因此，在这种情况下，各位数字必须满足：- 千位a=1（总是如此）- 百位b <=4- 十位c <=4- 个位d <=4这样的数有多少个呢？因为a=1，b可以是0到4，c可以是0到4，d可以是0到4，所以这样的数的数量是：b有5种选择（0,1,2,3,4），c有5种，d有5种，所以总数是5×5×5=125个。每个这样的数对应的n和n+1都满足各位相加不需要进位，对吗？比如，当n=1000，那么n+1=1001，各位相加：千位1+1=2，百位0+0=0，十位0+0=0，个位0+1=1，都不超过9，所以符合条件。当n=1444，n+1=1445，各位相加：1+1=2，4+4=8，4+4=8，4+5=9，都不超过9，符合条件。而当n=1455，这时候个位d=5，不满足d <=4，所以这类数不在这个情况里，所以不会被算入。好的，所以这种情况有125个符合条件的数，对吗？但是这里需要注意的是，这种情况下每个符合条件的数n对应一个符合条件的对(n, n+1)，所以共有125个这样的对吗？不过，当n的各位都是<=4的时候，n+1的各位除了个位d+1之外，其他位都不变，所以确实满足各位相加都不超过9的条件。好的，这是第一种情况。接下来，第二种情况是当n的个位d=9，这时候n+1的个位变为0，十位c变为c+1，但前提是c <9，否则会继续进位到百位，依此类推。在这种情况下，我们需要检查各位相加的情况：假设n的个位d=9，十位c <9，所以n+1的十位是c+1，个位是0，其他位不变。这时，各位相加的情况是：- 千位：1 +1=2- 百位：b + b- 十位：c + (c +1)- 个位：9 +0=9要满足十位相加c + (c +1) <=9，即2c +1 <=9，解得c <=4。同时，百位相加b + b <=9，即b <=4。同样，十位c必须满足c <=4，并且d=9，所以此时的条件是：- 千位a=1- 百位b <=4- 十位c <=4- 个位d=9这样的数有多少个呢？b有5种选择（0-4），c有5种（0-4），个位固定为9，所以数量是5×5×1=25个。每个这样的数n对应一个对(n, n+1)，其中n+1的十位是c+1，个位是0，其他位不变。此时各位相加的结果是：千位2，百位2b，十位2c +1，个位9，都不超过9，对吗？比如，n=1239，n+1=1240，相加各位：1+1=2，2+2=4，3+4=7，9+0=9，都<=9，符合条件。好的，所以这种情况有25个对吗？第三种情况是当n的个位d=9，十位c=9，此时n+1的十位变为0，百位b+1，前提是b <9。此时，各位相加的情况：- 千位：1 +1=2- 百位：b + (b +1)- 十位：9 +0=9- 个位：9 +0=9要满足百位相加b + (b +1) <=9，即2b +1 <=9，解得b <=4。同时，由于十位c=9，所以十位在n中是9，在n+1中是0，所以十位相加为9 +0=9，没问题。此时的条件是：- 千位a=1- 百位b <=4- 十位c=9- 个位d=9这样的数有多少个呢？百位b有5种选择（0-4），十位和个位固定为9，所以数量是5×1×1=5个。例如，n=1 0 9 9，n+1=1 1 0 0，相加各位：1+1=2，0+1=1，9+0=9，9+0=9，对吗？不，这里n=1099，n+1=1100，所以相加的话：千位1+1=2，百位0+1=1，十位9+0=9，个位9+0=9，确实符合条件。但原题中的n是在1000到2000之间，所以百位b=0是可以的，比如1099。所以这种情况有5个对吗？第四种情况是当n的个位d=9，十位c=9，百位b=9，此时n+1的百位变为0，千位a+1，即从1变为2，得到2000。此时，n=1999，n+1=2000，各位相加的情况：千位1 +2=3百位9 +0=9十位9 +0=9个位9 +0=9都不超过9，所以符合条件。这种情况只有一种，即n=1999，对吗？所以第四种情况的数量是1个。现在，我需要将所有这些情况加起来：第一种情况（d <=4，其他位<=4）：125个第二种情况（d=9，c <=4，b <=4）：25个第三种情况（d=9，c=9，b <=4）：5个第四种情况（d=9，c=9，b=9，a=1）：1个所以总共有125 +25 +5 +1=156对吗？但这里需要验证一下，是否所有情况都被考虑到，有没有重复或者遗漏。例如，当n=1999时，属于第四种情况，对吗？是的，这个例子是正确的，因为所有低位都是9，进位到千位，导致n+1=2000，而这对数相加时各位都不超过9。再比如，当n=1499时，十位和个位都是9，百位b=4，所以属于第三种情况，对吗？n=1499，n+1=1500，相加各位：千位1+1=2，百位4+5=9，十位9+0=9，个位9+0=9，对吗？是的，符合条件，所以百位相加是4 +5=9，符合条件。所以第三种情况的5个数是对的。那是否还有其他情况需要考虑呢？比如，当n的百位b=9，十位c=9，个位d=9，但千位a=1，这时n+1=2000，对吗？这种情况下，这个数n=1999，对吗？是的，已经算在第四种情况里了。那有没有其他可能的情况呢？比如，当n的十位c=5，个位d=9，那么n+1的十位变为6，此时十位相加是5+6=11，超过了9，所以不符合条件，对吗？因此这种情况下这个数不会被计入第二种情况，因为第二种情况要求c <=4，所以十位c=5的情况不符合条件。所以，所有的条件都被正确分类了，对吗？现在，将四种情况相加：125 +25=150150+5=155155+1=156所以总共有156对符合条件？不过，我之前算出来的是156，但需要再仔细检查一下是否有错误。首先，第一种情况是当个位d <=4，并且百位b <=4，十位c <=4，这时候每个位置都有5种选择（0-4），所以总数是5×5×5=125，对吗？是的，这个是对的。第二种情况是个位d=9，十位c <=4，百位b <=4，这时候百位有5种，十位有5种，个位固定为9，所以5×5=25，对吗？是的。第三种情况是十位c=9，个位d=9，百位b <=4，这时候百位有5种，十位和个位固定，所以5种对吗？是的，比如b=0到4，所以5个数。第四种情况是百位b=9，十位c=9，个位d=9，千位a=1，这时候只有n=1999，对吗？是的，所以1种。所以加起来是125+25+5+1=156。不过，我突然想到，当百位b=5的情况下，如果十位c=9，个位d=9，这时n的百位b=5，十位c=9，个位d=9，那么n+1会变成什么样子？比如n=1599，那么n+1=1600，对吗？这时，n和n+1相加的话，各位相加是：千位1+1=2，百位5+6=11，超过9，所以不符合条件，对吗？所以这种情况下，即使十位和个位都是9，但百位b=5的话，百位相加5+6=11，不符合条件，所以这种情况不会被计入第三种情况中，对吗？第三种情况的条件是百位b <=4，所以只有当b=0到4时，十位c=9，个位d=9的情况下，才会被计入，此时百位相加是b + (b+1) <=9，对吗？是的，所以第三种情况只包括百位b<=4的情况，所以当b=5时，十位c=9，个位d=9的情况下，虽然十位和个位都是9，但百位相加会超过9，所以不被计入。因此，我的计算是正确的，总共有156对。不过，我需要再检查一下是否还有其他情况需要考虑，或者是否有重复计算的情况。比如，当n的个位d=9，十位c=4，百位b=4时，属于第二种情况，对吗？n=1449，n+1=1450，此时各位相加：千位1+1=2，百位4+4=8，十位4+5=9，个位9+0=9，对吗？都满足条件，所以计入第二种情况，对吗？是的，这时候十位相加是4+5=9，刚好满足条件，没有问题。再比如，n=1499，n+1=1500，各位相加：千位1+1=2，百位4+5=9，十位9+0=9，个位9+0=9，符合条件，对吗？是的，计入第三种情况，对吗？是的，因为十位c=9，个位d=9，百位b=4 <=4，所以计入第三种情况。再比如，n=1999，n+1=2000，各位相加：千位1+2=3，百位9+0=9，十位9+0=9，个位9+0=9，符合条件，对吗？是的，计入第四种情况。现在，我需要确认是否有其他可能的情况，比如当n的十位c=9，但个位d不是9的情况，或者类似的情况，是否需要考虑？比如，当n的十位c=9，个位d=8，那么n+1的个位是9，十位仍然是9，对吗？这时候，n和n+1相加的话，十位相加是9+9=18，超过9，所以不符合条件，所以不会被计入任何情况中，对吗？是的，所以这种情况不会被计入，因为十位相加会超过9，所以只有当十位在n和n+1中的数字相加不超过9时，才会被计入。而这种情况只有当n+1的十位数字是c+1的情况下，即当n的个位是9时，n+1的十位才会增加，所以这种情况已经被第二种情况覆盖了。因此，我认为我的分类是正确的，总共有156对符合条件的连续整数。不过，为了确保正确，我可以尝试手动计算一些例子，然后看看是否符合预期。例如，当n=1000，n+1=1001，相加各位：1+1=2，0+0=0，0+0=0，0+1=1，都<=9，符合条件，计入第一种情况。当n=1004，n+1=1005，个位相加4+5=9，符合条件，计入第一种情况。当n=1005，n+1=1006，个位相加5+6=11，超过9，不符合条件，所以不计入。当n=1009，n+1=1010，各位相加：1+1=2，0+0=0，0+1=1，9+0=9，都<=9，对吗？这里十位是0 vs 1，所以十位相加0+1=1，对吗？是的，符合条件，计入第二种情况，对吗？是的，因为n=1009，个位d=9，十位c=0 <=4，百位b=0 <=4，所以计入第二种情况，对吗？是的，所以这个例子是对的。当n=1049，n+1=1050，各位相加：1+1=2，0+0=0，4+5=9，9+0=9，符合条件，对吗？是的，计入第二种情况。当n=1099，n+1=1100，各位相加：1+1=2，0+1=1，9+0=9，9+0=9，符合条件，计入第三种情况，对吗？是的，因为十位c=9，个位d=9，百位b=0 <=4。当n=1499，n+1=1500，各位相加：1+1=2，4+5=9，9+0=9，9+0=9，符合条件，计入第三种情况。当n=1999，n+1=2000，符合条件，计入第四种情况。另外，当n=1444，n+1=1445，符合条件，计入第一种情况。当n=1449，n+1=1450，十位相加4+5=9，符合条件，计入第二种情况。当n=1454，n+1=1455，个位相加4+5=9，符合条件，计入第一种情况。当n=1455，n+1=1456，个位相加5+6=11，不符合条件，所以不计入。这样看来，我的分类是正确的，例子都符合预期。因此，总共有156对符合条件的连续整数对。不过，我需要再检查一下计算是否正确，特别是各个情况的数量：第一种情况：5×5×5=125第二种情况：5×5×1=25第三种情况：5×1×1=5第四种情况：1总计：125+25=150，150+5=155，155+1=156。是的，没错。因此，答案应该是156，对吗？不过，为了确保万无一失，我需要考虑是否存在其他可能的情况，比如当进位影响到更高位的情况，但根据我的分析，所有可能的情况已经被覆盖。例如，当n=1999，n+1=2000，属于第四种情况，对吗？是的，而其他情况则是在不同位置有进位，但所有可能的进位都被考虑到了。因此，我认为正确答案是156，即boxed{156}。答案boxed{156}</think>我们需要找到集合 {1000, 1001, 1002, ldots, 2000} 中连续整数对 (n, n+1) 的数量，使得这两个数相加时不需要进位。 分析过程1. 情况一：个位 d < 9 - 条件：个位 d leq 4，十位 c leq 4，百位 b leq 4，千位 a = 1。 - 数量：5 times 5 times 5 = 125 对。2. 情况二：个位 d = 9，十位 c leq 4 - 条件：十位 c leq 4，百位 b leq 4，千位 a = 1。 - 数量：5 times 5 = 25 对。3. 情况三：个位 d = 9，十位 c = 9，百位 b leq 4 - 条件：百位 b leq 4，千位 a = 1。 - 数量：5 对。4. 情况四：个位 d = 9，十位 c = 9，百位 b = 9 - 条件：千位 a = 1。 - 数量：1 对（即 n = 1999 和 n+1 = 2000）。 总计将上述所有情况相加：[ 125 + 25 + 5 + 1 = 156 ] 最终答案[boxed{156}],<think>嗯，好的，我现在要解决的问题是：在集合{1000, 1001, 1002, ..., 2000}中，有多少对连续整数相加时不进位。题目说的是连续整数，也就是像（1000,1001）、（1001,1002）这样的相邻数对，当这两个数相加的时候不需要进位。那首先我需要明确几个问题点：首先，什么是“进位”？在加法中，当某一位上的数字相加大于等于10时，就要向高位进1，这就是进位。比如，比如15+6=21，个位5+6=11，需要进位，所以十位本来是1+0=1，但进了一位后变成2，个位变成1。这里如果两个连续整数相加的时候，每一位相加都不超过9，也就是说每一位相加都不需要进位，这样的数对有多少个。然后，题目中的集合是从1000到2000的所有整数，包括2000。所以总共有2000 - 1000 + 1 = 1001个数，因此可能的连续数对是1000对，也就是从（1000,1001）到（1999,2000），共1000对。但实际有多少对满足条件呢？接下来我需要找出这些数对中的每一对，当两个连续整数相加时，每一位上的数字相加都不超过9。因为这两个数是连续的，所以它们的差是1，也就是说第二个数比第一个数大1。比如n和n+1，所以我们要计算n + (n+1)的时候，是否每一位相加都不进位。不过，这里需要注意一点，当两个数相加的时候，每一位的相加是否会产生进位。不过，这里的情况是，n和n+1相加。因为n和n+1是连续的，所以它们的各个数位可能存在进位的情况，尤其是在末尾有几个9的情况下。例如，比如n=1999，n+1=2000，这时候n + (n+1) = 1999 + 2000 = 3999。这个时候每一位相加的情况是怎样的呢？不过可能我的思路有问题，因为题目说的是两个连续整数相加的时候不需要进位，也就是说当把n和n+1这两个数相加时，每一位相加都不需要进位。对吗？不过这里可能存在一个误解，是不是题目中的“两个整数相加时不需要进位”指的是它们的每一位相加都不产生进位？例如，比如1234和1235相加，每一位的各位分别是4+5=9，十位3+3=6，百位2+2=4，千位1+1=2，所以每一位相加都不超过9，不需要进位，这样的数对就是满足条件的。那现在问题转化为：在1000到2000之间的所有整数中，有多少个n使得n和n+1相加时每一位相加都不超过9？接下来我需要分析，当n和n+1相加时，每一位的情况是怎样的。因为n和n+1是连续的，所以当n的最后k位都是9的时候，n+1会导致最后k位变成0，并向前进1。例如，n=1299，n+1=1300，那么n + (n+1) = 1299 + 1300 = 2599。这时候各个位相加的情况是：个位：9 + 0 = 9，不进位；十位：9 + 0 = 9，不进位；百位：2 + 3 = 5，不进位；千位：1 + 1 = 2，不进位；所以这个例子中，每一位相加都是不超过9的，不需要进位，对吗？那这样的话，这样的数对是符合条件的。那问题来了，当n的最后有k个9时，n+1会导致最后k位变为0，并且第k+1位加1。所以当计算n + (n+1)时，每一位的具体情况是什么样的？例如，假设n的最后一位是9，那么n+1的最后一位就是0，进位到十位。如果十位原来也是9，那么进位到百位，依此类推。所以n + (n+1)的各位相加其实可以拆分为各个数位的数字相加，考虑是否有进位的情况。不过可能更简单的方法是从每一位的角度来看，当n和n+1相加时，各位是否满足相加的和不超过9。这时候，我需要分析每个数位上的数字在n和n+1中的情况。不过因为n和n+1是连续的，所以它们的各个数位可能有以下情况：假设n的各位数字为a（千位）、b（百位）、c（十位）、d（个位），那么n可以表示为1000a + 100b + 10c + d。那么n+1的各位数字则取决于是否有进位：- 如果d < 9，那么n+1的各位是a, b, c, d+1；- 如果d=9，但c <9，则n+1是a, b, c+1, 0；- 如果d=9，c=9，但b <9，则n+1是a, b+1, 0, 0；- 如果d=9，c=9，b=9，a <9，则n+1是a+1, 0, 0, 0；- 但题目中的n是从1000到2000，所以a的范围是1或2（因为2000也是一个数，不过当a=2时，n=2000，n+1=2001，但2001可能超过题目中的集合范围吗？题目中的集合是到2000，所以最大的n是2000，那么对应的n+1=2001不在集合里，所以可能最后一对是1999和2000，对吗？不过题目中的集合是到2000，所以当n=1999时，n+1=2000，这一对是在集合内的。而当n=2000时，n+1=2001，不在集合内，所以可能数对的范围是从1000到1999，每个n对应n+1，直到1999和2000，共有1000对，对吗？不过不管怎样，现在我需要考虑的是，当n和n+1相加时，每一位相加的结果是否都不超过9。当n的各位数字是a, b, c, d，那么n+1的各位数字可能是不同的，这取决于进位的情况。例如：情况1：d <9，则n+1的个位是d+1，其他位不变。此时n和n+1相加时，各位相加的情况是：- 个位：d + (d+1) → 这个和是2d +1；不过这时候，实际上n和n+1相加的和应该是n + (n+1) = 2n +1。而我们需要的是在每一位相加时不进位，也就是说，每一位上的数字相加后小于10。不过这里的每一位上的数字相加可能需要考虑进位的问题，对吗？或者说，这里的“进位”是指在计算n + (n+1)时，每一位相加是否需要进位？即，当我们将n和n+1这两个数相加的时候，像普通加法一样，每一位相加是否会产生进位？例如，n = 1234，n+1 = 1235，那么n + n+1 = 2469，这里每一位的相加是1+1=2，2+2=4，3+3=6，4+5=9，都不进位。而如果n=1239，n+1=1240，那么n + n+1 = 2479，各位相加：千位：1+1=2，没进位；百位：2+2=4，没进位；十位：3+4=7，没进位；个位：9+0=9，没进位；所以这也是符合条件的。但是如果n=1299，n+1=1300，那么n + n+1=2599：千位：1+1=2；百位：2+3=5；十位：9+0=9；个位：9+0=9；都不进位，对吗？但是如果n=1999，n+1=2000，相加得3999：千位：1+2=3；百位：9+0=9；十位：9+0=9；个位：9+0=9；都不进位，对吗？那这样的情况都是满足条件的。那么接下来问题转化为，寻找所有n在1000到1999之间的数，使得n和n+1相加时每一位相加都不产生进位。不过这里需要注意的是，当n=1999时，n+1=2000，这时候相加是3999，各位都不进位。所以这样的数对是存在的。那如何系统地计算这样的数对的数量呢？可以考虑每一位的可能情况，然后组合起来。首先，我们需要分析，当n和n+1相加时，每一位上的数字相加是否都不超过9。需要注意的是，这里的相加并不是指单独每一位相加，而是指在正常的加法过程中，每一位的相加是否产生进位。但实际上，因为n和n+1是连续的，所以它们的各位数字可能在某些位置上有进位关系，所以在相加的时候可能会影响高位。不过，或许更简单的方式是直接分析n和n+1的每一位数字在相加时是否会导致进位。例如，对于每一位来说，n的这一位的数字和n+1的这一位的数字相加是否小于10。不过，这里需要注意的是，当n的某一位在加1时可能会产生进位，所以n+1的某一位可能比n的这一位小（如果后面有进位的话）。例如，如果n的个位是9，那么n+1的个位是0，而十位如果原来是9的话也会进位，依此类推。所以，对于每一位来说，n和n+1的对应位数字可能不同，需要具体分析。例如，考虑四位数的n，每一位分别是a（千位）、b（百位）、c（十位）、d（个位）。则n+1的各位可能是：如果d <9，那么n+1的各位是a, b, c, d+1；如果d=9，c<9，则n+1是a, b, c+1, 0；如果d=9, c=9, b<9，则n+1是a, b+1, 0, 0；如果d=9, c=9, b=9, a<9，则n+1是a+1, 0, 0, 0；在题目中，n的范围是1000到2000，所以a的可能值是1或2（当n=2000时，a=2，但此时n+1=2001不在集合中，所以n的最大值是1999，对应的n+1=2000）。不过题目中的集合是到2000，所以当n=1999时，n+1=2000是有效的数对。现在，我们需要计算当n + (n+1)时，每一位相加都不需要进位。也就是说，对于每一位i，n的第i位数字加上n+1的第i位数字的和必须小于10。那如何确定n的每一位和n+1的对应位数字的和呢？例如，考虑n的个位d，当d <9时，n+1的个位是d+1，所以个位相加为d + (d+1) = 2d +1。这个值必须小于10，即2d +1 ≤9 → 2d ≤8 → d ≤4。因此，当d ≤4时，个位相加的和为2d+1 ≤9，不会进位。而当d ≥5时，2d+1 ≥11，这时候个位相加会有进位，因此这样的数对不满足条件。但是如果d=9，那么n+1的个位是0，并且十位c会被进位加1。此时，n的个位是9，n+1的个位是0，因此个位相加为9+0=9，不会进位。不过此时十位的情况需要看原来的十位c是否被进位影响。例如，如果n的十位c原本是9，那么n+1的十位会变成0，并且百位b被进位加1，依此类推。这时候，当n的个位是9时，n+1的个位是0，所以个位相加是9+0=9，没问题，不进位。但十位的n的c和n+1的c'可能会有不同的情况，具体取决于是否发生了进位。例如，如果n是a b c 9，那么n+1是a b (c+1) 0，如果c <9的话。此时，十位的数字相加是c + (c+1) = 2c +1，必须满足2c +1 ≤9 → c ≤4。但如果c=5，则2c+1=11，十位相加会有进位，此时十位的和是11，意味着需要进位到百位，因此整个加法中十位会有进位，导致不符合条件。但是如果n的十位是9，那么n+1的十位会因为个位的进位变成0，并且百位进位。例如，n是a b 9 9，那么n+1是a (b+1) 0 0。此时，十位的相加是9（来自n的十位）和0（来自n+1的十位），所以9+0=9，不进位。而百位则是b + (b+1) = 2b +1，需要满足2b +1 ≤9 → b ≤4。同样，如果百位是5，则2b+1=11，进位，不符合条件。同理，如果百位是9的话，比如n是a 9 9 9，那么n+1是(a+1) 0 0 0。此时，百位的相加是9 +0=9（因为n的百位是9，n+1的百位是0，进位到千位），所以百位的和是9，不进位。千位则是a + (a+1) = 2a +1，需要满足2a +1 ≤9 → a ≤4。不过题目中的n的千位a是1或者2（当n=2000时，a=2）。如果a=1，那么n+1的千位还是1，除非n=1999，这时候n+1=2000，千位变成2。所以对于a=1的情况，当n的千位是1时，n+1的千位仍然是1，除非后面三位都是9，这时候千位变为2。现在，我需要系统地分析每一位的可能情况，并找出满足条件的n的个数。首先，我们可以将问题分解为四位数的每一位，然后确定每一位的条件。假设n的各位为a b c d，其中a=1或2（但n<=1999，所以a=1或者当n=2000时a=2，但n=2000时n+1=2001不在集合中，所以a=1的情况是主要的）。接下来，我们需要考虑n和n+1的各位相加时，每一位的和都不超过9。首先，考虑个位的情况：当d <9时，n+1的个位是d+1，所以个位相加为d + (d+1) = 2d +1 ≤9 → d ≤4。当d=9时，n+1的个位是0，所以个位相加为9+0=9 ≤9，满足条件。但此时会发生进位到十位，导致十位的c变为c+1，如果c+1后十位数字改变的话。接下来考虑十位的情况：当d <9时，十位相加为c + c = 2c（因为n+1的十位还是c，因为个位没有进位）。但如果d <9，那么十位在n和n+1中都是c，所以十位相加是c + c = 2c ≤9 → c ≤4。但是如果d=9，那么十位在n中是c，在n+1中是c+1（如果c <9），或者在n+1中是0（如果c=9）。此时，十位相加为c + (c+1) = 2c +1 ≤9 → c ≤4（当c <9时），或者当c=9时，十位相加为9 +0=9，满足条件。同样，百位的情况：当十位没有进位时（即d <9或者d=9但c <9且处理完进位后十位相加满足条件），百位相加的情况是b + b = 2b ≤9 → b ≤4，或者如果十位有进位（当c=9且d=9时），百位在n中是b，在n+1中是b+1，所以百位相加为b + (b+1) = 2b +1 ≤9 → b ≤4。或者当十位有进位导致百位进位，比如如果b=9，那么n+1的百位会是0，并且千位进位，此时百位相加为9 +0=9，满足条件。同理，千位的情况：当百位没有进位时，千位相加是a + a = 2a ≤9 → a ≤4，但这里的a是1或2，所以当a=1时，2a=2 ≤9，符合条件；当a=2时，2a=4 ≤9，也符合条件。但如果百位有进位，比如百位原来是9，并且被进位后变为0，那么千位在n中是a，在n+1中是 a+1，此时千位相加为a + (a+1) = 2a +1。对于a=1来说，2a+1=3 ≤9；对于a=2来说，2a+1=5 ≤9，所以也符合条件。不过需要注意的是，当n=1999时，n+1=2000，此时千位相加是1 +2=3，符合条件；百位是9+0=9，十位9+0=9，个位9+0=9，都符合条件。所以这个数对是有效的。现在，我需要将这些情况系统地分解，考虑各个数位上的可能情况，并计算符合条件的数的个数。可能的方法是分情况讨论，根据末尾有多少个9：1. 末尾没有9的情况：即d <9。此时，n+1的个位是d+1，其他位不变。此时，个位相加为2d+1 ≤9 → d ≤4。同时，十位相加为2c ≤9 → c ≤4；百位相加为2b ≤9 → b ≤4；千位相加为2a ≤9 → a ≤4，但a=1或2，所以都满足。因此，这种情况下，各个位的数字都要满足d ≤4，c ≤4，b ≤4，a可以是1或2。不过当a=2时，n只能是2000，但2000的下一个数是2001，不在集合中，所以当a=2时，n只能是2000，此时没有下一个数，所以有效的a只能是1。因此，在这种情况下，当a=1，b ≤4，c ≤4，d ≤4时，这样的数有多少个？各个位的取值范围：a=1；b: 0到4；c: 0到4；d: 0到4；所以数量是1（a=1） ×5（b=0到4） ×5（c=0到4） ×5（d=0到4） = 5×5×5=125个。不过注意n的范围是从1000到1999，所以当a=1时，b可以是0到9，但这里b必须≤4，所以b从0到4，c从0到4，d从0到4，因此这样的数共有5×5×5=125个。这种情况下，每个这样的n对应的n+1也是有效的，都在集合中，并且相加时各位都不进位。接下来考虑末尾有一个9的情况：即d=9，但c <9。此时，n+1的个位是0，十位是c+1，百位和千位不变。此时，个位相加为9+0=9，符合条件；十位相加为c + (c+1) = 2c +1 ≤9 → c ≤4；百位相加为2b ≤9 → b ≤4；千位相加为2a ≤9 → a=1，所以这种情况下，条件为：a=1；b ≤4；c ≤4；d=9；数量是1（a=1） ×5（b=0到4） ×5（c=0到4） ×1（d=9） =5×5=25个。接下来考虑末尾有两个9的情况：即d=9，c=9，但b <9。此时，n+1的个位和十位都是0，百位是b+1，千位不变。此时，个位相加9+0=9；十位相加9+0=9；百位相加为b + (b+1) =2b +1 ≤9 → b ≤4；千位相加为2a ≤9 → a=1。因此，条件为：a=1；b ≤4；c=9；d=9；数量是1（a=1） ×5（b=0到4） ×1（c=9） ×1（d=9） =5个。接下来考虑末尾有三个9的情况：即d=9，c=9，b=9，但a <9。此时，n+1的个位、十位、百位都是0，千位是a+1。题目中的n的范围是1000到1999，所以a=1，所以n=1999，n+1=2000。此时，各位相加：个位9+0=9；十位9+0=9；百位9+0=9；千位1+2=3（因为n的千位是1，n+1的千位是2），所以3 ≤9，符合条件。所以这种情况只有一种可能，即n=1999，对应的数对（1999,2000）。因此，数量是1。最后，末尾有四个9的情况是不可能的，因为n最大是1999，所以不可能有四个9。综上所述，总共有四种情况：1. 末尾没有9：125个；2. 末尾有一个9：25个；3. 末尾有两个9：5个；4. 末尾有三个9：1个；所以总共有125 +25 +5 +1 = 156个数对满足条件。不过需要验证一下是否正确。例如，末尾没有9的情况，即d ≤4，同时c ≤4，b ≤4，a=1，这样的数有多少个？是的，每个位置有5种可能（0-4），所以5×5×5=125。末尾有一个9，即d=9，c ≤4，b ≤4，a=1，所以是5×5=25。末尾有两个9，即d=9，c=9，b ≤4，a=1，所以5个。末尾有三个9，即d=9，c=9，b=9，a=1，只有1999，所以1个。加起来125+25=150，+5=155，+1=156。所以总数是156对吗？不过等一下，可能这里有一个错误，因为在处理末尾有三个9的情况时，是否只有1999符合条件？是的，当a=1，b=9，c=9，d=9时，n=1999，对应的n+1=2000，确实在集合中，所以这个数对是有效的，因此确实应该算进去，所以总数是156。不过我需要再验证几个例子是否正确。例如，当n=1444时，n+1=1445，相加得1444+1445=2889，各位相加：千位1+1=2；百位4+4=8；十位4+4=8；个位4+5=9；都不进位，符合条件，所以这个数对应该被计算在内，属于末尾没有9的情况，即d=4 ≤4，所以属于第一种情况。再比如，n=1459，n+1=1460，相加得1459+1460=2919，各位相加：千位1+1=2；百位4+4=8；十位5+6=11，这里十位相加是11，所以需要进位，对吗？这样十位会产生进位，导致不符合条件，但是根据上面的分析，当d=9时，需要c ≤4，因为十位相加是c + (c+1) = 2c +1 ≤9 → c ≤4。但在这个例子中，n=1459，c=5，所以2c +1 =11，这时候十位相加会有进位，因此这个数对不符合条件，因此不能被计算在内。但是根据我的之前的计算，当末尾有一个9时，c必须≤4，所以当c=5时，这种情况不会被计算在内，所以上面的例子中的n=1459不在符合条件的数对中，因此我的分析是正确的。再比如，n=1499，n+1=1500，相加得1499+1500=2999，各位相加：千位1+1=2；百位4+5=9；十位9+0=9；个位9+0=9；都不进位，符合条件。这里n=1499，属于末尾有两个9的情况，即c=9，d=9，但b=4 ≤4，所以被计算在内，属于第三种情况，所以是正确的。另一个例子，n=1999，n+1=2000，相加得3999，各位都不进位，符合条件，属于第四种情况。再检查一种情况，比如n=1399，n+1=1400，相加得1399+1400=2799，各位相加：千位1+1=2；百位3+4=7；十位9+0=9；个位9+0=9；都不进位，符合条件。这里n=1399，属于末尾有两个9的情况，b=3 ≤4，所以符合条件，属于第三种情况，对吗？是的，所以计算正确。所以总数应该是125 +25 +5 +1=156对吗？不过，我需要再确认一下各个情况的计算是否正确。第一种情况：末尾没有9，即d ≤4，c ≤4，b ≤4，a=1：b的范围是0到4，共5种；c的范围是0到4，共5种；d的范围是0到4，共5种；所以数量是5×5×5=125，正确。第二种情况：末尾有一个9，即d=9，c ≤4，b ≤4，a=1：c的范围0到4，共5种；b的范围0到4，共5种；所以数量是5×5=25，正确。第三种情况：末尾有两个9，即d=9，c=9，b ≤4，a=1：b的范围0到4，共5种；数量是5，正确。第四种情况：末尾有三个9，即d=9，c=9，b=9，a=1：只有1种，即1999，正确。所以总和是125+25+5+1=156。不过题目中的集合是从1000到2000，包括2000，所以当n=1999时，n+1=2000是有效的，确实在集合中，所以这个数对应该被计算在内，所以总数是156。不过，我发现可能有一个问题，就是当a=2时，也就是n=2000，但是n+1=2001不在集合中，所以n=2000不是有效的数对，因此在所有情况中，a只能是1，对应的n从1000到1999，对吗？所以我的计算都是基于a=1的情况，因此没有问题，所以总数是156。不过让我再检查一下当a=1时，千位相加是1+1=2，或者当n=1999时，千位是1+2=3，对吗？在n=1999时，n+1=2000，所以千位相加是1+2=3，不超过9，对吗？是的，所以符合条件。所以最终的答案应该是156。不过，我需要再确认一下是否漏掉了什么情况，或者重复计算了某些情况。例如，当末尾有一个9的情况，即d=9，c ≤4，此时十位相加是c + (c+1) =2c +1 ≤9 → c ≤4，对吗？是的，所以当c=4时，2×4+1=9，符合条件，而当c=5时，2×5+1=11，不符合，所以c必须≤4，这样是正确的。同样，当末尾有两个9时，即c=9，d=9，此时百位相加是b + (b+1) =2b +1 ≤9 → b ≤4，对吗？是的。当末尾有三个9时，即b=9，c=9，d=9，此时千位相加是1 +2=3，符合条件，对吗？是的。所以所有情况都覆盖到了，并且计算正确，因此总数是156。不过，我突然想到，题目中的集合是否包括2000？是的，集合是{1000,1001,...,2000}，所以当n=1999时，n+1=2000是有效的，所以应该被计算在内，而n=2000没有下一个数，所以不影响总数。因此，我认为正确的答案是156，即boxed{156}。不过，我再次检查一下计算是否正确：125 +25 =150150 +5=155155 +1=156是的，没错。所以最终的答案应该是156。答案boxed{156}</think>在集合 {1000, 1001, 1002, ldots, 2000} 中，寻找连续整数对 (n, n+1) 使得相加时不需要进位的数量。我们需要确保每一位上的数字相加都不超过9。分析各位的情况：1. 末尾没有9的情况： - 个位 d leq 4，十位 c leq 4，百位 b leq 4，千位 a = 1。 - 数量：5 times 5 times 5 = 125。2. 末尾有一个9的情况： - 个位 d = 9，十位 c leq 4，百位 b leq 4，千位 a = 1。 - 数量：5 times 5 = 25。3. 末尾有两个9的情况： - 个位 d = 9，十位 c = 9，百位 b leq 4，千位 a = 1。 - 数量：5。4. 末尾有三个9的情况： - 个位 d = 9，十位 c = 9，百位 b = 9，千位 a = 1（即 n = 1999）。 - 数量：1。总数为以上情况之和：[ 125 + 25 + 5 + 1 = 156 ]最终答案：[boxed{156}]]

🔑:3. 1563. Suppose n and n+1 are to be added, and write n in the form 1 mathrm{ABC}. If any one of mathrm{A}, mathrm{B} and mathrm{C} is equal to 5,6,7 or 8 , then carrying is needed in the addition. If mathrm{B} is 9 and mathrm{C} is not 9 , or if mathrm{A} is 9 and mathrm{B} and mathrm{C} are not both 9 , carrying is also required. Therefore, if no carrying is required, n must be of one of the following forms:1 mathrm{ABC} quad 1 mathrm{AB} 9 quad 1 mathrm{~A} 99 quad 1999where each of mathrm{A}, mathrm{B}, mathrm{C} is from 0 to 4 . It follows that the answer is 5^{3}+5^{2}+5+1=156.