🧠:[<think>Okay, so I need to solve this problem about the sequence {a_n}. Let me first parse what's given here.Part (I) asks for the general formula of the sequence {a_n}. The conditions given are:1. The sum of the first n terms, S_n, satisfies the equation S_n = 2a_n - a_1.2. The terms a_1, a_2 + 1, a_3 form an arithmetic sequence.Alright, let's start with the first condition. The sum of the first n terms is S_n, which is equal to 2a_n - a_1. Hmm, that's an interesting relation. Normally, S_n is the sum up to the nth term, but here it's related to the nth term itself.Let me write down what I know about S_n. For any sequence, S_n = a_1 + a_2 + ... + a_n. Also, S_{n} = S_{n-1} + a_n. Maybe that recursive relation can help here.Given that S_n = 2a_n - a_1, let's try to write this in terms of S_{n-1}. Since S_n = S_{n-1} + a_n, substituting into the given equation:S_{n-1} + a_n = 2a_n - a_1Simplify this:S_{n-1} = 2a_n - a_1 - a_n = a_n - a_1So S_{n-1} = a_n - a_1. Hmm. Interesting. Let's see if we can relate S_{n-1} and S_n.Wait, from the original condition, S_n = 2a_n - a_1. Let's consider what S_{n-1} would be. If we replace n with n-1 in the equation, we get S_{n-1} = 2a_{n-1} - a_1.But earlier, we had S_{n-1} = a_n - a_1. Therefore:2a_{n-1} - a_1 = a_n - a_1Simplify this:2a_{n-1} - a_1 = a_n - a_1Subtract -a_1 from both sides:2a_{n-1} = a_nSo this gives us a recursive relation: a_n = 2a_{n-1}Wait, that's a simple geometric sequence! If a_n = 2a_{n-1}, then each term is double the previous term. So the sequence {a_n} is a geometric sequence with common ratio 2. Therefore, the general term would be a_n = a_1 * 2^{n-1}.But wait, hold on. Let me check if this holds with the given condition S_n = 2a_n - a_1.If {a_n} is a geometric sequence with ratio 2, then the sum S_n = a_1(2^n - 1). Let's see if that equals 2a_n - a_1.Compute 2a_n - a_1: 2*(a_1*2^{n-1}) - a_1 = a_1*2^n - a_1 = a_1(2^n - 1). Which is exactly the sum S_n. So that works out! Therefore, the general term is a geometric sequence with ratio 2. So a_n = a_1 * 2^{n-1}.But wait, there's another condition: a_1, a_2 + 1, a_3 form an arithmetic sequence. Let's verify this.Given that {a_n} is a geometric sequence with ratio 2, so:a_1 = a_1a_2 = 2a_1a_3 = 4a_1According to the problem, a_1, a_2 + 1, a_3 should be an arithmetic sequence. Let's write down these terms:First term: a_1Second term: a_2 + 1 = 2a_1 + 1Third term: a_3 = 4a_1For these three terms to form an arithmetic sequence, the difference between consecutive terms should be the same. So the difference between the second and first term should be equal to the difference between the third and second term.Compute the differences:Second term - first term: (2a_1 + 1) - a_1 = a_1 + 1Third term - second term: 4a_1 - (2a_1 + 1) = 2a_1 - 1Set them equal:a_1 + 1 = 2a_1 - 1Solving for a_1:a_1 + 1 = 2a_1 - 1Subtract a_1 from both sides:1 = a_1 - 1Add 1 to both sides:a_1 = 2So the first term a_1 is 2. Therefore, the general formula for {a_n} is a_n = 2 * 2^{n - 1} = 2^n.Wait, let me check if this satisfies both conditions.First, the sum S_n = 2a_n - a_1. Since a_n = 2^n, S_n should be sum from k=1 to n of 2^k = 2^{n+1} - 2. Let's compute 2a_n - a_1: 2*2^n - 2 = 2^{n+1} - 2, which matches. Good.Second, check the arithmetic sequence: a_1 = 2, a_2 + 1 = 2^2 + 1 = 4 + 1 = 5, a_3 = 2^3 = 8. So the three terms are 2, 5, 8. The differences are 5 - 2 = 3, and 8 - 5 = 3. So yes, arithmetic sequence with common difference 3. Perfect.Therefore, part (I) answer is a_n = 2^n.Now, moving on to part (II). We have to define b_n = log_2 a_n, and find the sum of the first n terms T_n of the sequence {a_n b_n}.First, let's write down what b_n is. Since a_n = 2^n, then log_2 a_n = log_2 2^n = n. Therefore, b_n = n.Therefore, the sequence {a_n b_n} is {2^n * n}. So we need to find the sum T_n = sum_{k=1}^n (2^k * k).Hmm, sum of k*2^k from k=1 to n. I remember that there's a formula for this kind of sum. Let me recall. The general formula for sum_{k=1}^n k r^k is r(1 - (n+1)r^n + n r^{n+1}) / (1 - r)^2. Let me verify this.Let S = sum_{k=1}^n k r^kMultiply both sides by r:rS = sum_{k=1}^n k r^{k+1} = sum_{k=2}^{n+1} (k-1) r^kSubtracting:S - rS = sum_{k=1}^n k r^k - sum_{k=2}^{n+1} (k-1) r^k= sum_{k=1}^n k r^k - sum_{k=2}^{n+1} (k-1) r^k= [1 r^1 + 2 r^2 + ... + n r^n] - [1 r^2 + 2 r^3 + ... + (n-1) r^n + n r^{n+1}]= r + (2 r^2 - r^2) + (3 r^3 - 2 r^3) + ... + (n r^n - (n-1) r^n) - n r^{n+1}= r + r^2 + r^3 + ... + r^n - n r^{n+1}This is a geometric series sum from k=1 to n of r^k minus n r^{n+1}So S(1 - r) = (r(1 - r^n)/(1 - r)) - n r^{n+1}Therefore, S = [r(1 - r^n)/(1 - r)^2 - n r^{n+1}/(1 - r)]Multiply numerator and denominator appropriately:S = r(1 - r^n) / (1 - r)^2 - n r^{n+1}/(1 - r)Combine terms:S = [r(1 - r^n) - n r^{n+1}(1 - r)] / (1 - r)^2Let me check if this simplifies to the formula I mentioned earlier.Alternatively, perhaps a better approach. Let me do the steps again.Let S = sum_{k=1}^n k r^kThen, S = r + 2r^2 + 3r^3 + ... + n r^nMultiply both sides by r:rS = r^2 + 2r^3 + 3r^4 + ... + n r^{n+1}Subtract the two equations:S - rS = (r + 2r^2 + 3r^3 + ... + n r^n) - (r^2 + 2r^3 + ... + n r^{n+1})= r + (2r^2 - r^2) + (3r^3 - 2r^3) + ... + (n r^n - (n-1) r^n) - n r^{n+1}= r + r^2 + r^3 + ... + r^n - n r^{n+1}Which is the sum of the geometric series from k=1 to n of r^k minus n r^{n+1}The sum of the geometric series is r(1 - r^n)/(1 - r). Therefore,S(1 - r) = r(1 - r^n)/(1 - r) - n r^{n+1}Multiply both sides by 1/(1 - r):S = [r(1 - r^n)] / (1 - r)^2 - [n r^{n+1}] / (1 - r)So S = [r(1 - r^n) - n r^{n+1}(1 - r)] / (1 - r)^2Simplify numerator:r - r^{n+1} - n r^{n+1} + n r^{n+2}= r - r^{n+1}(1 + n) + n r^{n+2}Alternatively, factor terms:= r - r^{n+1}(1 + n - n r)Hmm, not sure if that's helpful. Let's plug in r=2 for our case. Since here, r=2, so let's compute S = sum_{k=1}^n k*2^k.Plugging r=2 into the formula:S = [2(1 - 2^n)/(1 - 2)^2 - n*2^{n+1}/(1 - 2)]Compute denominator terms:(1 - 2)^2 = 1, so denominator is 1. Wait, no. Wait:Wait, the formula is:S = [2(1 - 2^n)/(1 - 2)^2 - n*2^{n+1}/(1 - 2)]Compute (1 - 2)^2 = (-1)^2 = 1So first term is 2(1 - 2^n)/1 = 2(1 - 2^n)Second term: -n*2^{n+1}/(-1) = n*2^{n+1}Therefore, S = 2(1 - 2^n) + n*2^{n+1}Simplify:= 2 - 2^{n+1} + n*2^{n+1}Factor 2^{n+1}:= 2 + (n - 1)2^{n+1}Wait, let's compute step by step:2(1 - 2^n) = 2 - 2^{n+1}Then add n*2^{n+1}:Total: 2 - 2^{n+1} + n*2^{n+1} = 2 + (n - 1)2^{n+1}Alternatively, factor 2^{n+1}:= 2 + 2^{n+1}(n - 1)Alternatively, write as (n - 1)2^{n+1} + 2Let me check with small n.Take n=1:sum_{k=1}^1 k*2^k = 1*2 = 2Using the formula: (1 - 1)2^{2} + 2 = 0 + 2 = 2. Correct.n=2:sum = 1*2 + 2*4 = 2 + 8 = 10Formula: (2 - 1)2^{3} + 2 = 1*8 + 2 = 10. Correct.n=3:sum = 2 + 8 + 3*8 = 2 + 8 + 24 = 34Formula: (3 - 1)2^{4} + 2 = 2*16 + 2 = 32 + 2 = 34. Correct.Good, so the formula works. Therefore, the sum T_n = (n - 1)2^{n+1} + 2.Alternatively, we can write it as 2 + (n - 1)2^{n+1} or factor 2^{n+1}:But maybe we can factor 2^{n+1}:T_n = 2^{n+1}(n - 1) + 2Alternatively, factor 2:T_n = 2[2^n(n - 1) + 1]But maybe it's better to leave it as is. Let me write it as:T_n = (n - 1)2^{n+1} + 2Alternatively, we can write it as (n - 1)2^{n+1} + 2 = n*2^{n+1} - 2^{n+1} + 2 = n*2^{n+1} - 2^{n+1} + 2 = n*2^{n+1} - 2(2^n - 1)But perhaps the first form is simplest. Let me check the original expression.Wait, the formula S = [2(1 - 2^n) + n*2^{n+1}]Wait, 2(1 - 2^n) + n*2^{n+1} = 2 - 2^{n+1} + n*2^{n+1} = 2 + 2^{n+1}(n - 1). Yes, same thing.So, the answer for part (II) is T_n = (n - 1)2^{n+1} + 2.Alternatively, if we factor 2 from the entire expression:= 2[ (n - 1)2^n + 1 ]But that might not be necessary. Let's verify with n=1:(n - 1)2^{n+1} + 2 = 0 + 2 = 2. Correct.n=2:(2 -1)2^3 + 2 = 8 + 2 = 10. Correct.n=3:(3 -1)2^4 + 2 = 2*16 +2 = 34. Correct.Therefore, the formula holds. So T_n = (n - 1)2^{n+1} + 2.Alternatively, maybe written as (n - 1)2^{n +1} + 2 is acceptable, but let me check if there's a more standard form.Wait, the general formula for sum_{k=1}^n k*2^k is indeed (n - 1)2^{n+1} + 2. Let me check another source.Wait, according to standard summation formulas, sum_{k=1}^n k*2^k = 2(1 - 2^n + n*2^n). Let me compute:2(1 - 2^n + n*2^n) = 2 - 2^{n+1} + n*2^{n+1} = same as before.So, 2 + (n - 1)2^{n+1} is another way. So both forms are equivalent. So either form is acceptable. Perhaps the problem expects the answer in a simplified form. Let me check the problem statement again.The problem says: "find the sum of the first n terms T_n of the sequence {a_n b_n}".Given that {a_n b_n} is {2^n * n}, so the sum is sum_{k=1}^n k*2^k = (n - 1)2^{n+1} + 2. So the answer is T_n = (n - 1)2^{n+1} + 2.Alternatively, factor 2:T_n = 2 + (n - 1)2^{n+1}But I don't think further simplification is necessary here. So the answer is as above.Therefore, for part (II), T_n is (n - 1)2^{n+1} + 2.Final Answer(I) The general term is boxed{a_n = 2^n}.(II) The sum of the first ( n ) terms ( T_n ) is boxed{(n - 1) cdot 2^{n+1} + 2}.</think>Given the sequence ({a_n}) with the sum of the first (n) terms (S_n) satisfying (S_n = 2a_n - a_1), and the terms (a_1), (a_2 + 1), (a_3) forming an arithmetic sequence:(I) To find the general formula for the sequence ({a_n}):1. The sum (S_n) of the first (n) terms is given by (S_n = 2a_n - a_1).2. Using the relationship (S_n = S_{n-1} + a_n), we substitute into the given equation to get (S_{n-1} + a_n = 2a_n - a_1), leading to (S_{n-1} = a_n - a_1).3. Replacing (n) with (n-1) in the original equation, we get (S_{n-1} = 2a_{n-1} - a_1). Equating this with the previous result, we find (2a_{n-1} - a_1 = a_n - a_1), simplifying to (a_n = 2a_{n-1}). This indicates a geometric sequence with ratio 2.4. Given (a_1), (a_2 + 1), (a_3) form an arithmetic sequence, we check the terms: (a_1), (2a_1 + 1), (4a_1). The differences must be equal, leading to (a_1 + 1 = 2a_1 - 1), solving to (a_1 = 2).5. Therefore, the sequence is (a_n = 2 cdot 2^{n-1} = 2^n).(II) To find the sum of the first (n) terms (T_n) of the sequence ({a_n b_n}) where (b_n = log_2 a_n):1. Since (a_n = 2^n), we have (b_n = n). Thus, (a_n b_n = 2^n cdot n).2. The sum (T_n = sum_{k=1}^n k cdot 2^k) is calculated using the formula for the sum of (k cdot r^k): [ T_n = (n - 1) cdot 2^{n+1} + 2 ] Final Answer(I) The general term is (boxed{a_n = 2^n}).(II) The sum of the first (n) terms (T_n) is (boxed{(n - 1) cdot 2^{n+1} + 2}).,<think>Okay, so I have this problem about a sequence {a_n} where the sum of the first n terms, S_n, is given by S_n = 2a_n - a_1. Also, it says that a_1, a_2 + 1, a_3 form an arithmetic sequence. I need to find the general formula for {a_n} in part (I), and then in part (II), define b_n as the logarithm base 2 of a_n and find the sum T_n of the first n terms of the sequence {a_n b_n}. Let me start with part (I). Hmm. So first, the sum of the first n terms S_n is 2a_n - a_1. Let me recall that S_n is the sum of the first n terms, so S_n = a_1 + a_2 + ... + a_n. The formula given here is S_n = 2a_n - a_1. That seems a bit unusual. Maybe this can help us relate a_n to the previous terms?Also, we are told that a_1, a_2 + 1, a_3 form an arithmetic sequence. An arithmetic sequence means that the difference between consecutive terms is constant. So the difference between a_2 + 1 and a_1 should be equal to the difference between a_3 and a_2 + 1. So, (a_2 + 1) - a_1 = a_3 - (a_2 + 1). Let me write that down:(a_2 + 1 - a_1) = (a_3 - a_2 - 1)Simplify that equation:Left side: a_2 + 1 - a_1Right side: a_3 - a_2 - 1So setting them equal:a_2 + 1 - a_1 = a_3 - a_2 - 1Bring all terms to one side:a_2 + 1 - a_1 - a_3 + a_2 + 1 = 0Combine like terms:2a_2 + 2 - a_1 - a_3 = 0Which gives:2a_2 + 2 = a_1 + a_3So that's one equation.Now, we need to use the other condition given, which is S_n = 2a_n - a_1. Let's see if we can find a recursive formula or something.First, note that S_n = a_1 + a_2 + ... + a_n = 2a_n - a_1Similarly, S_{n-1} = a_1 + a_2 + ... + a_{n-1} = 2a_{n-1} - a_1Therefore, the nth term a_n = S_n - S_{n-1} = (2a_n - a_1) - (2a_{n-1} - a_1) = 2a_n - a_1 - 2a_{n-1} + a_1 = 2a_n - 2a_{n-1}So, a_n = 2a_n - 2a_{n-1}Wait, that seems strange. Let's rearrange the equation:a_n = 2a_n - 2a_{n-1}Subtract 2a_n from both sides:-a_n = -2a_{n-1}Multiply both sides by -1:a_n = 2a_{n-1}So that gives a recurrence relation: a_n = 2a_{n-1}Which is a geometric sequence with common ratio 2. So, the sequence is geometric, each term is double the previous one.But wait, let's check this. If a_n = 2a_{n-1}, then the sequence is a geometric progression with ratio 2. Therefore, the general term should be a_n = a_1 * 2^{n-1}But we need to confirm this with the given conditions.But we also have that S_n = 2a_n - a_1. Let's check if this is consistent with the geometric sequence.If {a_n} is geometric with first term a_1 and ratio 2, then S_n = a_1(2^n - 1)/(2 - 1) = a_1(2^n - 1)But according to the problem, S_n = 2a_n - a_1. Let's compute 2a_n - a_1 for the geometric sequence:2a_n - a_1 = 2(a_1 * 2^{n-1}) - a_1 = a_1*2^n - a_1 = a_1(2^n - 1)Which matches the formula for S_n. So that works out. So that suggests that the sequence is indeed a geometric sequence with ratio 2.But wait, we also have the condition that a_1, a_2 + 1, a_3 form an arithmetic sequence. Let's check if this holds with the geometric sequence.If the sequence is geometric with ratio 2, then:a_1 = a_1a_2 = 2a_1a_3 = 4a_1So, the three terms given are a_1, a_2 + 1 = 2a_1 + 1, a_3 = 4a_1For these to form an arithmetic sequence, the difference between consecutive terms should be the same. So:(2a_1 + 1) - a_1 = 4a_1 - (2a_1 + 1)Simplify:Left side: 2a_1 + 1 - a_1 = a_1 + 1Right side: 4a_1 - 2a_1 - 1 = 2a_1 - 1Set equal:a_1 + 1 = 2a_1 - 1Solving for a_1:a_1 + 1 = 2a_1 - 1Subtract a_1 from both sides:1 = a_1 - 1Add 1 to both sides:a_1 = 2Therefore, the first term a_1 is 2. Then, since the sequence is geometric with ratio 2, the general term is a_n = 2 * 2^{n-1} = 2^n.But wait, let's check if this works with the arithmetic sequence condition.a_1 = 2a_2 = 2^2 = 4So a_2 + 1 = 4 + 1 = 5a_3 = 2^3 = 8So the three terms are 2, 5, 8. Is that an arithmetic sequence?Yes, because 5 - 2 = 3, and 8 - 5 = 3. The common difference is 3. So that works.Therefore, the general formula for the sequence is a_n = 2^n.Wait, but let me make sure that this is consistent with all n. Since we derived that a_n = 2a_{n-1} with a_1 = 2, so indeed the general term is a_n = 2^n.Therefore, part (I) answer is a_n = 2^n.But let me check the initial problem again. The problem states that S_n = 2a_n - a_1. Let's check for n = 1, 2, 3.For n=1: S_1 = a_1 = 2. The formula gives 2a_1 - a_1 = 2*2 -2 = 2. Correct.For n=2: S_2 = a_1 + a_2 = 2 + 4 = 6. The formula gives 2a_2 - a_1 = 2*4 -2 = 8 - 2 = 6. Correct.For n=3: S_3 = 2 + 4 + 8 = 14. Formula gives 2a_3 - a_1 = 2*8 -2 = 16 - 2 =14. Correct.So yes, that works. Therefore, part (I) is solved as a_n = 2^n.Now, part (II). Let b_n = log_2 a_n. Then, find the sum of the first n terms T_n of the sequence {a_n b_n}.First, let's express b_n. Since a_n = 2^n, then log_2 a_n = log_2 2^n = n. Therefore, b_n = n.Therefore, the sequence {a_n b_n} is {2^n * n}. So, T_n is the sum from k=1 to k=n of k * 2^k.So, we need to compute T_n = sum_{k=1}^n k*2^k.I remember that there's a formula for the sum of k*r^k from k=1 to n. Let me recall it.The general formula for sum_{k=1}^n k r^k is r(1 - (n+1) r^n + n r^{n+1}) / (1 - r)^2, when r ≠ 1.Since here r = 2, which is not 1, so we can use this formula.So plugging r=2:sum_{k=1}^n k*2^k = 2(1 - (n+1)2^n + n*2^{n+1}) / (1 - 2)^2Simplify denominator: (1 - 2)^2 = 1, so denominator is 1.Wait, let's check the formula again. Maybe I remember it wrong.Wait, perhaps I should derive it.Let me consider S = sum_{k=1}^n k*2^k.Let me write S = 2 + 2*2^2 + 3*2^3 + ... + n*2^nMultiply both sides by 2:2S = 2^2 + 2*2^3 + 3*2^4 + ... + n*2^{n+1}Subtract the original S from this equation:2S - S = S = (2^2 + 2*2^3 + 3*2^4 + ... + n*2^{n+1}) - (2 + 2*2^2 + 3*2^3 + ... + n*2^n)Let's compute each term:For 2^2: 2^2 - 0 = 4For 2*2^3: 2*8 - 2*4 = 16 - 8 = 8For 3*2^4: 3*16 - 3*8 = 48 - 24 = 24Wait, this might not be the best way. Let me write it out term by term.S = 2 + 2*4 + 3*8 + ... + n*2^n2S = 4 + 2*8 + 3*16 + ... + n*2^{n+1}Subtracting S from 2S:2S - S = S = (4 - 2) + (8 - 8) + (16 - 24) + ... + [n*2^{n+1} - (n-1)*2^n] - n*2^{n+1}Wait, maybe another approach. Let me use the standard method for such sums.Let S = sum_{k=1}^n k r^kMultiply by r:rS = sum_{k=1}^n k r^{k+1} = sum_{k=2}^{n+1} (k-1) r^kSubtract:rS - S = sum_{k=2}^{n+1} (k-1) r^k - sum_{k=1}^n k r^k= sum_{k=2}^{n} (k-1) r^k + n r^{n+1} - [sum_{k=1}^n k r^k]= sum_{k=2}^{n} (k-1) r^k + n r^{n+1} - sum_{k=1}^n k r^k= - sum_{k=1}^n k r^k + sum_{k=2}^n (k-1) r^k + n r^{n+1}Break down the first sum:sum_{k=1}^n k r^k = 1*r^1 + sum_{k=2}^n k r^kTherefore:- sum_{k=1}^n k r^k = -r - sum_{k=2}^n k r^kSo combining terms:-r - sum_{k=2}^n k r^k + sum_{k=2}^n (k -1) r^k + n r^{n+1}= -r + sum_{k=2}^n [ (k -1) - k ] r^k + n r^{n+1}= -r + sum_{k=2}^n (-1) r^k + n r^{n+1}= -r - sum_{k=2}^n r^k + n r^{n+1}The sum from k=2 to n of r^k is sum_{k=1}^n r^k - rWhich is (r^{n+1} - r)/(r - 1) - r = (r^{n+1} - r - r(r -1))/(r -1) ) ??? Wait, maybe better to recall that sum_{k=2}^n r^k = sum_{k=1}^n r^k - r = (r^{n+1} - r)/(r - 1) - r = [r^{n+1} - r - r(r -1)]/(r -1) = [r^{n+1} - r - r^2 + r]/(r -1) = [r^{n+1} - r^2]/(r -1)Hmm, maybe too complicated. Let's just proceed.So, S(r -1) = -r - [sum_{k=2}^n r^k] + n r^{n+1}But sum_{k=2}^n r^k = sum_{k=0}^n r^k - r^0 - r^1 = (r^{n+1} -1)/(r -1) -1 -r= (r^{n+1} -1 - (r -1) - r(r -1))/(r -1)Wait, maybe easier to note that sum_{k=2}^n r^k = (r^2 + r^3 + ... + r^n) = r^2( r^{n -1} -1 )/(r -1 )But regardless, let's substitute back into S(r -1):S(r -1) = -r - [ (r^{n+1} - r)/(r -1) - r ] + n r^{n+1}Wait, let me try again. sum_{k=2}^n r^k = sum_{k=1}^n r^k - r = (r^{n+1} - r)/(r -1) - rSo,S(r -1) = -r - [ (r^{n+1} - r)/(r -1) - r ] + n r^{n+1}= -r - (r^{n+1} - r)/(r -1) + r + n r^{n+1}Simplify: -r + r cancels, so:= - (r^{n+1} - r)/(r -1) + n r^{n+1}Multiply numerator and denominator:= [ - (r^{n+1} - r) + n r^{n+1}(r -1) ] / (r -1)Expand the numerator:= -r^{n+1} + r + n r^{n+2} - n r^{n+1}Factor terms:= r + (-r^{n+1} - n r^{n+1}) + n r^{n+2}= r - r^{n+1}(1 + n) + n r^{n+2}Factor out r^{n+1}:= r + r^{n+1}( - (1 + n) + n r )= r + r^{n+1}( n r - n -1 )Therefore,S(r -1) = [ r + r^{n+1}(n r -n -1) ] / (r -1 )Hence,S = [ r + r^{n+1}(n r -n -1) ] / (r -1)^2Therefore, substituting r =2:S = [2 + 2^{n+1}(2n -n -1) ] / (2 -1)^2Simplify:Denominator is 1, so:S = 2 + 2^{n+1}(n -1)But wait, 2n -n -1 = n -1. So,S = 2 + 2^{n+1}(n -1)Factor 2:= 2 + 2^{n+1}(n -1) = 2 + (n -1)2^{n+1}Alternatively, factor 2^{n+1}:But perhaps write it as (n -1)2^{n+1} + 2Alternatively, factor 2:= 2[1 + (n -1)2^n]But maybe leave it as 2 + (n -1)2^{n+1}Wait, let me check with a small n. Let's take n=1:Sum_{k=1}^1 k*2^k = 1*2 =2. The formula gives 2 + (1 -1)2^{2} = 2 +0=2. Correct.n=2:Sum is 2 + 4*2= 2 +8=10. The formula gives 2 + (2 -1)2^{3}=2 +8=10. Correct.n=3:Sum is 2 + 8 + 24=34. Formula:2 + (3 -1)2^4=2 + 2*16=2+32=34. Correct.n=4: sum is 2 + 8 +24 +64= 98. Formula:2 +3*32=2+96=98. Correct. So the formula works.Therefore, T_n = 2 + (n -1)2^{n+1}Alternatively, factor out 2:= 2 + 2^{n+1}(n -1)We can write this as 2^{n+1}(n -1) +2But maybe simplify further:= 2^{n+1}(n -1) +2 = 2^{n+1}n -2^{n+1} +2 = n2^{n+1} -2^{n+1} +2Factor 2^{n+1} terms:= n2^{n+1} -2^{n+1} +2 = 2^{n+1}(n -1) +2Alternatively, factor 2 from the first two terms and the last term:=2( n2^n -2^n ) +2 =2*2^n(n -1) +2=2^{n+1}(n -1) +2Alternatively, we can write this as:=2^{n+1}(n -1) +2=2^{n+1}(n -1) +2*1=2[2^n(n -1) +1]But maybe the original expression is acceptable.Alternatively, combine terms:=2^{n+1}(n -1) +2=2^{n+1}(n -1) +2=2^{n+1}(n -1) +2*1=2(2^n(n -1) +1 )But perhaps the simplest form is (n -1)2^{n+1} +2, which is the same as 2 + (n -1)2^{n+1}So, that's the formula.Alternatively, let's write it as T_n = (n -1)2^{n+1} +2.Therefore, the sum T_n is (n -1)2^{n+1} +2.Alternatively, factor 2:=2[ (n -1)2^n +1 ]But perhaps the original form is better.Alternatively, note that 2 + (n -1)2^{n+1} can be rewritten as (n -1)2^{n+1} +2.Alternatively, perhaps the problem expects a different form.Wait, let's check again.We had S = sum_{k=1}^n k2^k = 2 + (n -1)2^{n+1}Wait, for n=1: 2 + (1 -1)2^{2}=2+0=2, correct.n=2: 2 + (2 -1)2^{3}=2+8=10, correct.n=3:2 + (3 -1)2^4=2 +2*16=34, correct.So yes, the formula is T_n = (n -1)2^{n+1} +2.Alternatively, some sources present this formula as (n -1)2^{n+1} +2, but I think it's better to write it as:T_n = 2^{n+1}(n -1) + 2.Alternatively, factor 2^{n+1}:But not sure if it's needed. So maybe write the answer as (n -1)2^{n+1} +2.Alternatively, maybe we can write it in another way.Let me see:(n -1)2^{n+1} +2 = n2^{n+1} -2^{n+1} +2 = n2^{n+1} -2^{n+1} +2 = n2^{n+1} -2(2^{n} -1)But I don't know if that helps. Probably the original expression is the simplest.Therefore, the sum T_n is (n -1)2^{n+1} +2.Thus, part (II)'s answer is T_n = (n -1)2^{n+1} +2.But let me check the problem statement again. The problem says "the sum of the first n terms T_n of the sequence {a_n b_n}".Since a_n =2^n, b_n =n, so a_n b_n =n2^n, and T_n is sum_{k=1}^n k2^k, which we found to be (n -1)2^{n+1} +2. So that's correct.But maybe we can write it as (n -1)2^{n+1} +2 = n2^{n+1} -2^{n+1} +2 =n2^{n+1} -2^{n+1} +2= n2^{n+1} -2(2^n -1). Hmm, but perhaps not necessary. So the answer is T_n = (n -1)2^{n+1} +2.Alternatively, if we factor 2, we get 2[(n -1)2^n +1]. But both forms are acceptable. However, the standard formula is often written as (n -1)2^{n+1} +2.Therefore, part (II) answer is T_n = (n -1)2^{n+1} +2.Final Answer(I) boxed{a_n = 2^n}(II) boxed{T_n = (n - 1) cdot 2^{n+1} + 2}</think>Given the sequence ({a_n}) where the sum of the first (n) terms (S_n) satisfies (S_n = 2a_n - a_1), and (a_1), (a_2 + 1), (a_3) form an arithmetic sequence, we need to find the general formula for ({a_n}) and the sum of the first (n) terms (T_n) of the sequence ({a_n b_n}) where (b_n = log_2 a_n). Part (I)1. Sum Condition: The sum (S_n = 2a_n - a_1). For (n geq 1), we have: [ S_n = a_1 + a_2 + cdots + a_n ] Using the given condition, we also know: [ S_{n-1} = 2a_{n-1} - a_1 ] Subtracting these, we get: [ a_n = S_n - S_{n-1} = (2a_n - a_1) - (2a_{n-1} - a_1) implies a_n = 2a_n - 2a_{n-1} implies a_n = 2a_{n-1} ] This indicates a geometric sequence with ratio 2.2. Arithmetic Sequence Condition: Given (a_1), (a_2 + 1), (a_3) form an arithmetic sequence: [ (a_2 + 1) - a_1 = a_3 - (a_2 + 1) ] Substituting (a_2 = 2a_1) and (a_3 = 4a_1) (from the geometric sequence): [ (2a_1 + 1) - a_1 = 4a_1 - (2a_1 + 1) implies a_1 + 1 = 2a_1 - 1 implies a_1 = 2 ] Thus, the sequence is (a_n = 2^n). Part (II)1. Define (b_n): Since (a_n = 2^n), we have (b_n = log_2 a_n = n).2. Sum (T_n): The sequence ({a_n b_n}) is (n cdot 2^n). The sum (T_n) is: [ T_n = sum_{k=1}^n k cdot 2^k ] Using the formula for the sum of (k cdot r^k): [ T_n = (n - 1) cdot 2^{n+1} + 2 ] Final Answers(I) (boxed{a_n = 2^n})(II) (boxed{T_n = (n - 1) cdot 2^{n+1} + 2})]

🔑:Solution: (I) Given S_n=2a_n-a_1, when ngeqslant 2, a_n=S_n-S_{n-1}=2a_n-2a_{n-1}, thus a_n=2a_{n-1}, then a_2=2a_1, a_3=2a_2=4a_1, since a_1, a_2+1, a_3 form an arithmetic sequence, i.e., a_1+a_3=2(a_2+1), therefore a_1+4a_1=2(a_2+1), solving this gives: a_1=2. Thus, the sequence {a_n} is a geometric sequence with the first term as 2 and common ratio as 2. Therefore, a_n=2^n; (II) According to the problem, b_n=log_2a_n=n, a_nb_n=ncdot2^n, the sum of the first n terms T_n of the sequence {a_nb_n} is T_n=1cdot2+2cdot2^2+ldots+ncdot2^n, thus 2T_n=1cdot2^2+2cdot2^3+ldots+(n-1)cdot2^n+ncdot2^{n+1}, therefore -T_n=2+2^2+2^3+ldots+2^n-ncdot2^{n+1}, = frac{2(1-2^n)}{1-2}-ncdot2^{n+1}, =(1-n)cdot2^{n+1}-2, thus T_n=(n-1)cdot2^{n+1}+2, The sum of the first n terms T_n of the sequence {a_nb_n} is T_n=(n-1)cdot2^{n+1}+2. Therefore, the final answers are: (I) a_n=2^n, so the general formula for the sequence {a_n} is boxed{2^n}. (II) The sum of the first n terms T_n of the sequence {a_nb_n} is boxed{(n-1)cdot2^{n+1}+2}.

🧠:[<think>Okay, so I have this problem here about a right triangle with sides a, b, and c. It says that for a natural number n greater than 2, the equation (aⁿ + bⁿ + cⁿ)² = 2(a²ⁿ + b²ⁿ + c²ⁿ) holds. I need to find what n is. Hmm, let's start breaking this down.First, since it's a right triangle, by the Pythagorean theorem, we know that a² + b² = c², assuming c is the hypotenuse. So maybe that relationship will come in handy here.The given equation is (aⁿ + bⁿ + cⁿ)² = 2(a²ⁿ + b²ⁿ + c²ⁿ). Let me expand the left-hand side and see if I can simplify it. Expanding (aⁿ + bⁿ + cⁿ)² gives a²ⁿ + b²ⁿ + c²ⁿ + 2aⁿbⁿ + 2aⁿcⁿ + 2bⁿcⁿ. So setting that equal to the right-hand side, which is 2a²ⁿ + 2b²ⁿ + 2c²ⁿ, we can subtract the right-hand side from both sides to get:(a²ⁿ + b²ⁿ + c²ⁿ + 2aⁿbⁿ + 2aⁿcⁿ + 2bⁿcⁿ) - 2(a²ⁿ + b²ⁿ + c²ⁿ) = 0Simplifying that:- a²ⁿ - b²ⁿ - c²ⁿ + 2aⁿbⁿ + 2aⁿcⁿ + 2bⁿcⁿ = 0Which can be rewritten as:2aⁿbⁿ + 2aⁿcⁿ + 2bⁿcⁿ = a²ⁿ + b²ⁿ + c²ⁿHmm, maybe I can factor out a 2 on the left side:2(aⁿbⁿ + aⁿcⁿ + bⁿcⁿ) = a²ⁿ + b²ⁿ + c²ⁿNow, since we know that in a right triangle, a² + b² = c², perhaps we can use substitution or some relation here. Let me think. Maybe if I let c² = a² + b², then cⁿ can be expressed in terms of a and b? But cⁿ is (sqrt(a² + b²))ⁿ, which might complicate things. Unless there's a specific n that simplifies this.Alternatively, maybe we can try specific values of n. Since n is a natural number greater than 2, possible candidates could be 3, 4, etc. Let me try n=2 first, even though n has to be greater than 2, just to see what happens.If n=2, then the equation becomes (a² + b² + c²)² = 2(a⁴ + b⁴ + c⁴). But in a right triangle, c² = a² + b², so substituting that in:( (a² + b²) + (a² + b²) )² = 2(a⁴ + b⁴ + (a² + b²)² )Left side: (2a² + 2b²)² = 4(a² + b²)²Right side: 2(a⁴ + b⁴ + a⁴ + 2a²b² + b⁴) = 2(2a⁴ + 2b⁴ + 2a²b²) = 4a⁴ + 4b⁴ + 4a²b²Left side is 4(a⁴ + 2a²b² + b⁴) = 4a⁴ + 8a²b² + 4b⁴Comparing both sides: 4a⁴ + 8a²b² + 4b⁴ vs 4a⁴ + 4b⁴ + 4a²b². They are not equal unless a or b is zero, which they can't be in a triangle. So n=2 doesn't work, which is expected since n>2. But this might suggest that for higher n, similar cancellation needs to occur.Let me try n=3. Let's see.Compute left side: (a³ + b³ + c³)²Right side: 2(a⁶ + b⁶ + c⁶)Again, since c² = a² + b², so c³ = c * c² = c(a² + b²) = a²c + b²c. But not sure if that helps. Maybe instead, let's pick a specific right triangle, like a 3-4-5 triangle, and compute both sides for n=3. Maybe that will give some insight.Take a=3, b=4, c=5.Left side: (3³ + 4³ + 5³)² = (27 + 64 + 125)² = (216)² = 46656Right side: 2(3⁶ + 4⁶ + 5⁶) = 2(729 + 4096 + 15625) = 2(20450) = 40900But 46656 ≠ 40900, so n=3 doesn't work here. Hmm, so maybe n=4?Testing n=4 with 3-4-5 triangle.Left side: (3⁴ + 4⁴ + 5⁴)² = (81 + 256 + 625)² = (962)² = 925444Right side: 2(3⁸ + 4⁸ + 5⁸) = 2(6561 + 65536 + 390625) = 2(462722) = 925444Wait, they are equal! For n=4, in the 3-4-5 triangle, the equation holds. Interesting. So n=4 might be a candidate. But we need to confirm if this holds for all right triangles, not just 3-4-5.Let me check another right triangle, say 5-12-13.Left side for n=4: (5⁴ + 12⁴ +13⁴)²Calculating each term:5⁴ = 625, 12⁴ = 20736, 13⁴ = 28561Sum: 625 + 20736 + 28561 = 625 + 20736 = 21361; 21361 + 28561 = 49922So left side: (49922)²Right side: 2(5⁸ + 12⁸ +13⁸)Calculating each term:5⁸ = (5⁴)² = 625² = 39062512⁸ = (12⁴)² = 20736². Let me compute that: 20736 * 20736. Hmm, 20000² = 400,000,000, 736² = 541,696, and cross terms: 2*20000*736 = 29,440,000. So total is 400,000,000 + 29,440,000 + 541,696 = 429,981,69613⁸ = (13⁴)² = 28561². Let's calculate that: 28561*28561. Hmm, 28560² = (28560)^2. Let me compute 2856^2 * 10^2. 2856^2: (2800 + 56)^2 = 2800² + 2*2800*56 + 56² = 7,840,000 + 313,600 + 3,136 = 8,156,736. Then multiply by 100: 815,673,600. Then add (28561² - 28560²) = (28561 - 28560)(28561 + 28560) = 1*57,121 = 57,121. So 815,673,600 + 57,121 = 815,730,721Therefore, 5⁸ +12⁸ +13⁸ = 390,625 + 429,981,696 + 815,730,721. Let's add them:390,625 + 429,981,696 = 430,372,321; 430,372,321 + 815,730,721 = 1,246,103,042Multiply by 2: 2,492,206,084Left side was (49,922)². Let's compute that: 50,000² = 2,500,000,000. Subtract (50,000 - 49,922)² and adjust. Wait, alternatively, 49,922 * 49,922. Let's do (50,000 - 78)^2 = 50,000² - 2*50,000*78 + 78² = 2,500,000,000 - 7,800,000 + 6,084 = 2,500,000,000 - 7,800,000 = 2,492,200,000; then +6,084 = 2,492,206,084. Which matches the right side! So in the 5-12-13 triangle, n=4 also works.Hmm, so n=4 seems to satisfy the equation in two different Pythagorean triples. Maybe n=4 is the answer. But to confirm, perhaps we can try a non-integer right triangle? Although a, b, c are lengths, they don't have to be integers. Let's take an isosceles right triangle, where a = b = 1, so c = √2.Check for n=4:Left side: (1⁴ + 1⁴ + (√2)^4)^2 = (1 + 1 + 4)^2 = (6)^2 = 36Right side: 2(1⁸ + 1⁸ + (√2)^8) = 2(1 + 1 + (4)^2) = 2(1 + 1 + 16) = 2*18 = 36Equal again. So n=4 works here too. Let's try n=4 in a general right triangle. Let me assume that c is the hypotenuse, so c² = a² + b².We need to verify whether (a⁴ + b⁴ + c⁴)² = 2(a⁸ + b⁸ + c⁸)First, let's compute left side:(a⁴ + b⁴ + c⁴)²But c⁴ = (a² + b²)² = a⁴ + 2a²b² + b⁴So c⁴ = a⁴ + b⁴ + 2a²b²Therefore, a⁴ + b⁴ + c⁴ = a⁴ + b⁴ + (a⁴ + b⁴ + 2a²b²) = 2a⁴ + 2b⁴ + 2a²b² = 2(a⁴ + b⁴ + a²b²)So left side becomes [2(a⁴ + b⁴ + a²b²)]² = 4(a⁴ + b⁴ + a²b²)^2Right side: 2(a⁸ + b⁸ + c⁸)Compute c⁸: since c² = a² + b², c⁸ = (a² + b²)^4Expand (a² + b²)^4:= a^8 + 4a^6b² + 6a^4b^4 + 4a²b^6 + b^8So c⁸ = a^8 + 4a^6b² + 6a^4b^4 + 4a²b^6 + b^8Therefore, a⁸ + b⁸ + c⁸ = a⁸ + b⁸ + a⁸ + 4a^6b² + 6a^4b^4 + 4a²b^6 + b⁸= 2a⁸ + 2b⁸ + 4a^6b² + 6a^4b^4 + 4a²b^6Multiply by 2: 4a⁸ + 4b⁸ + 8a^6b² + 12a^4b^4 + 8a²b^6Now, let's compute the left side: 4(a⁴ + b⁴ + a²b²)^2First compute (a⁴ + b⁴ + a²b²)^2:= (a⁴ + b⁴)^2 + 2(a⁴ + b⁴)(a²b²) + (a²b²)^2= a^8 + 2a^4b^4 + b^8 + 2a^6b² + 2a²b^6 + a^4b^4= a^8 + b^8 + 3a^4b^4 + 2a^6b² + 2a²b^6Multiply by 4:4a^8 + 4b^8 + 12a^4b^4 + 8a^6b² + 8a²b^6Comparing left and right sides:Left: 4a^8 + 4b^8 + 12a^4b^4 + 8a^6b² + 8a²b^6Right: 4a⁸ + 4b⁸ + 8a^6b² + 12a^4b^4 + 8a²b^6They are the same! So yes, for any right triangle, when n=4, the equation holds. Therefore, n must be 4.But wait, the problem says "for a natural number n greater than 2", so n=4 is the answer. Let me check n=5 just to be thorough. Let's test n=5 in the 3-4-5 triangle.Left side: (3⁵ + 4⁵ +5⁵)^2Compute each term:3⁵ = 243, 4⁵ = 1024, 5⁵ = 3125Sum: 243 + 1024 = 1267; 1267 + 3125 = 4392Left side: 4392² = ?Right side: 2(3¹⁰ +4¹⁰ +5¹⁰)Compute each term:3¹⁰ = 59049, 4¹⁰ = 1048576, 5¹⁰ = 9765625Sum: 59049 + 1048576 = 1107625; 1107625 + 9765625 = 10873250Multiply by 2: 21746500Left side: 4392². Let's compute 4000² = 16,000,000, 392² = 153,664, and cross term 2*4000*392 = 3,136,000. So total is 16,000,000 + 3,136,000 + 153,664 = 19,289,664. Which is not equal to 21,746,500. So n=5 doesn't work.Therefore, n=4 is the only natural number greater than 2 that satisfies the equation for right triangles. Thus, the answer is 4.Final AnswerThe value of ( n ) is boxed{4}.</think>Given that (a), (b), and (c) are the lengths of the sides of a right triangle, we need to find the natural number (n) greater than 2 such that the equation ((a^n + b^n + c^n)^2 = 2(a^{2n} + b^{2n} + c^{2n})) holds.First, we use the Pythagorean theorem, which states (a^2 + b^2 = c^2). Expanding the left-hand side of the given equation, we get:[(a^n + b^n + c^n)^2 = a^{2n} + b^{2n} + c^{2n} + 2a^nb^n + 2a^nc^n + 2b^nc^n]Setting this equal to the right-hand side (2(a^{2n} + b^{2n} + c^{2n})), we subtract (2(a^{2n} + b^{2n} + c^{2n})) from both sides:[a^{2n} + b^{2n} + c^{2n} + 2a^nb^n + 2a^nc^n + 2b^nc^n - 2(a^{2n} + b^{2n} + c^{2n}) = 0]Simplifying, we get:[-a^{2n} - b^{2n} - c^{2n} + 2a^nb^n + 2a^nc^n + 2b^nc^n = 0]Rewriting, we have:[2(a^nb^n + a^nc^n + b^nc^n) = a^{2n} + b^{2n} + c^{2n}]Testing specific values of (n) with different right triangles (e.g., 3-4-5, 5-12-13, isosceles right triangle) shows that (n = 4) satisfies the equation. For (n = 4), we verify the equation holds for general right triangles by expanding both sides and confirming they are equal.For example, in a general right triangle with (c^2 = a^2 + b^2), we compute (c^4 = (a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4). Substituting (c^4) into the left-hand side and right-hand side of the equation confirms both sides are equal, verifying (n = 4) works for all right triangles.Thus, the value of (n) is (boxed{4}).,<think>Okay, let's see. So the problem is about a right triangle with sides a, b, c, and there's this equation involving exponents n: (aⁿ + bⁿ + cⁿ)² = 2(a²ⁿ + b²ⁿ + c²ⁿ). We need to find the natural number n > 2 that satisfies this. Alright, let's start breaking this down.First, since it's a right triangle, by Pythagoras' theorem, we know that a² + b² = c² if c is the hypotenuse. Wait, but maybe the hypotenuse could be a or b as well? Hmm, but the problem says a, b, c are the lengths of the three sides, so depending on which is the hypotenuse. But maybe we can assume, without loss of generality, that c is the hypotenuse. So let's set c as the hypotenuse. So then, a² + b² = c². That's a given.Now, the equation we have is (aⁿ + bⁿ + cⁿ)² = 2(a²ⁿ + b²ⁿ + c²ⁿ). Let's expand the left side and see what we get. Expanding (aⁿ + bⁿ + cⁿ)² would give a²ⁿ + b²ⁿ + c²ⁿ + 2aⁿbⁿ + 2aⁿcⁿ + 2bⁿcⁿ. So the left-hand side (LHS) is a²ⁿ + b²ⁿ + c²ⁿ + 2aⁿbⁿ + 2aⁿcⁿ + 2bⁿcⁿ. The right-hand side (RHS) is 2a²ⁿ + 2b²ⁿ + 2c²ⁿ. If we subtract RHS from LHS, we get:(a²ⁿ + b²ⁿ + c²ⁿ + 2aⁿbⁿ + 2aⁿcⁿ + 2bⁿcⁿ) - 2(a²ⁿ + b²ⁿ + c²ⁿ) = 0Simplifying that:- a²ⁿ - b²ⁿ - c²ⁿ + 2aⁿbⁿ + 2aⁿcⁿ + 2bⁿcⁿ = 0So:2aⁿbⁿ + 2aⁿcⁿ + 2bⁿcⁿ = a²ⁿ + b²ⁿ + c²ⁿDivide both sides by 2:aⁿbⁿ + aⁿcⁿ + bⁿcⁿ = (a²ⁿ + b²ⁿ + c²ⁿ)/2Hmm, not sure if that helps. Maybe there's a different approach.Since we know it's a right triangle, maybe we can express c in terms of a and b. Let's substitute c² = a² + b² into the equation. Wait, but we have cⁿ in the equation. So maybe if we can express cⁿ in terms of a and b? But unless n is 2, that's not straightforward. Maybe we need to pick specific values of n to test. Since n is a natural number greater than 2, let's try n=3, n=4, etc., and see if the equation holds.Alternatively, maybe there's a general approach. Let's see. Let's consider the given equation:(aⁿ + bⁿ + cⁿ)² = 2(a²ⁿ + b²ⁿ + c²ⁿ)Expanding the left side, as before:a²ⁿ + b²ⁿ + c²ⁿ + 2aⁿbⁿ + 2aⁿcⁿ + 2bⁿcⁿ = 2a²ⁿ + 2b²ⁿ + 2c²ⁿSubtracting a²ⁿ + b²ⁿ + c²ⁿ from both sides:2aⁿbⁿ + 2aⁿcⁿ + 2bⁿcⁿ = a²ⁿ + b²ⁿ + c²ⁿSo:a²ⁿ + b²ⁿ + c²ⁿ - 2aⁿbⁿ - 2aⁿcⁿ - 2bⁿcⁿ = 0Hmm, maybe factor this somehow. Let me rearrange terms:a²ⁿ - 2aⁿbⁿ + b²ⁿ + c²ⁿ - 2aⁿcⁿ - 2bⁿcⁿ = 0Wait, the first three terms look like (aⁿ - bⁿ)², but let's check:(aⁿ - bⁿ)² = a²ⁿ - 2aⁿbⁿ + b²ⁿ. So that's the first three terms. Then the remaining terms are c²ⁿ - 2aⁿcⁿ - 2bⁿcⁿ. So:(aⁿ - bⁿ)² + c²ⁿ - 2aⁿcⁿ - 2bⁿcⁿ = 0But c²ⁿ - 2aⁿcⁿ - 2bⁿcⁿ can be written as c²ⁿ - 2cⁿ(aⁿ + bⁿ). So:(aⁿ - bⁿ)² + c²ⁿ - 2cⁿ(aⁿ + bⁿ) = 0Hmm, maybe factor this expression. Let's see. Let's note that c² = a² + b², so c is sqrt(a² + b²). But when we have cⁿ, that's (a² + b²)^{n/2}. Not sure if that helps. Maybe substitute c² = a² + b² into the equation.Alternatively, let's consider specific cases. Let's try n=3 first.Case n=3:We need to check if (a³ + b³ + c³)² = 2(a⁶ + b⁶ + c⁶)But let's compute both sides. Let me choose a specific right triangle to test. For simplicity, take a=1, b=1, then c=√2. Then compute both sides.Left side: (1³ + 1³ + (√2)³)² = (1 + 1 + (2√2))² = (2 + 2√2)² = 4 + 8√2 + 8 = 12 + 8√2 ≈ 12 + 11.31 = 23.31Right side: 2(1⁶ + 1⁶ + (√2)^6) = 2(1 + 1 + (8)) = 2(10) = 20But 23.31 ≈ 20? No, not equal. So n=3 does not satisfy.Wait, maybe my choice of a=1, b=1 is not good. Maybe pick a Pythagorean triplet where a, b, c are integers, like 3,4,5.Let’s try with a=3, b=4, c=5.For n=3:Left side: (27 + 64 + 125)² = (216)² = 46656Right side: 2(729 + 4096 + 15625) = 2(20450) = 40900But 46656 ≠ 40900. So n=3 doesn't work.Hmm. Let's try n=4.Compute for n=4 with a=3, b=4, c=5.Left side: (81 + 256 + 625)² = (962)². Let's compute that: 962*962. 900²=810000, 62²=3844, and 2*900*62=111600. So total is 810000 + 111600 + 3844 = 925444.Right side: 2(3^8 + 4^8 + 5^8). Compute each term:3^8 = 65614^8 = 655365^8 = 390625Sum: 6561 + 65536 = 72097; 72097 + 390625 = 462722. Multiply by 2: 925444. Hey, that's equal! So for n=4, the equation holds for the 3-4-5 triangle. Interesting. So maybe n=4 is the answer? Let's check another triangle to confirm.Take a=5, b=12, c=13. Let's compute for n=4.Left side: (5⁴ + 12⁴ +13⁴)². Compute each term:5⁴=625, 12⁴=20736, 13⁴=28561. Sum: 625 + 20736 = 21361; 21361 +28561=49922. Then square it: 49922².Right side: 2*(5^8 +12^8 +13^8). Compute each term:5^8=390625, 12^8=429981696, 13^8=815730721. Sum: 390625 +429981696=430372321; 430372321 +815730721=1,246,103,042. Multiply by 2: 2,492,206,084.Left side: 49922². Let's compute 49922*49922. Let's note that (50000 - 78)^2 = 50000² - 2*50000*78 + 78² = 2,500,000,000 - 7,800,000 + 6,084 = 2,500,000,000 -7,800,000 = 2,492,200,000; then +6,084 = 2,492,206,084. Which matches the right side! So for n=4, the equation holds for 5-12-13 triangle as well. So that's a good sign. Let's check another case, like a=1, b=1, c=√2. Maybe the previous n=3 didn't work, but n=4.Left side: (1⁴ + 1⁴ + (√2)^4)^2. Compute each term: 1 +1 + (4) =6. Square it: 36.Right side: 2*(1^8 +1^8 + (√2)^8). Compute each term: 1 +1 + (16) =18. Multiply by 2: 36. So equal. So even in this case, it works. So n=4 seems to work.Let’s check n=5 just to be thorough. Take the 3-4-5 triangle.Left side: (3⁵ +4⁵ +5⁵)². 3⁵=243, 4⁵=1024, 5⁵=3125. Sum:243+1024=1267; 1267+3125=4392. Square:4392²= (4400 -8)^2=4400² -2*4400*8 +8²=19,360,000 -70,400 +64=19,289,664.Right side: 2*(3^10 +4^10 +5^10). 3^10=59049, 4^10=1,048,576, 5^10=9,765,625. Sum:59049+1,048,576=1,107,625; 1,107,625 +9,765,625=10,873,250. Multiply by 2:21,746,500. Compare to left side 19,289,664. Not equal. So n=5 doesn't work.Therefore, n=4 seems to be the answer. Let me check if there's a general way to see why n=4 works.We can try to prove that for any right triangle, (a⁴ + b⁴ + c⁴)² = 2(a⁸ + b⁸ +c⁸). Wait, but how?Wait, but in the cases we checked, it worked. Let's see if there's an identity here. Let's recall that in a right triangle, c² =a² +b². Let's see if c⁴ can be expressed in terms of a and b.c⁴ = (a² +b²)^2 = a⁴ + 2a²b² + b⁴.Therefore, a⁴ + b⁴ + c⁴ = a⁴ + b⁴ + a⁴ + 2a²b² + b⁴ = 2a⁴ + 2b⁴ + 2a²b² = 2(a⁴ + b⁴ + a²b²).So (a⁴ + b⁴ + c⁴)² = [2(a⁴ + b⁴ + a²b²)]² = 4(a⁴ + b⁴ +a²b²)^2.Now let's compute 2(a⁸ + b⁸ +c⁸). Let's compute c⁸.c⁸ = (a² + b²)^4 = a^8 + 4a^6b² +6a^4b^4 +4a²b^6 +b^8.Therefore, 2(a⁸ +b⁸ +c⁸) = 2(a⁸ +b⁸ + a^8 +4a^6b² +6a^4b^4 +4a^2b^6 +b^8) = 2(2a^8 + 2b^8 +4a^6b² +6a^4b^4 +4a^2b^6).Simplify: 4a^8 +4b^8 +8a^6b² +12a^4b^4 +8a^2b^6.Now, let's compute [2(a⁴ +b⁴ +a²b²)]²:First, expand inside the square: 2(a⁴ +b⁴ +a²b²) = 2a⁴ +2b⁴ +2a²b².Square it: (2a⁴ +2b⁴ +2a²b²)^2.Expand term by term:= (2a⁴)^2 + (2b⁴)^2 + (2a²b²)^2 + 2*(2a⁴)*(2b⁴) + 2*(2a⁴)*(2a²b²) + 2*(2b⁴)*(2a²b²)= 4a^8 +4b^8 +4a^4b^4 + 8a^4b^4 + 8a^6b² + 8a^2b^6Combine like terms:4a^8 +4b^8 + (4a^4b^4 +8a^4b^4) +8a^6b² +8a^2b^6=4a^8 +4b^8 +12a^4b^4 +8a^6b² +8a^2b^6Which is exactly the same as 2(a⁸ +b⁸ +c⁸). So, indeed, (a⁴ +b⁴ +c⁴)^2 = 2(a⁸ +b⁸ +c⁸). Therefore, n=4 satisfies the equation for any right triangle. So the answer is n=4.Final AnswerThe natural number ( n ) is boxed{4}.</think>Given that (a), (b), and (c) are the lengths of the sides of a right triangle, we need to find the natural number (n > 2) such that ((a^n + b^n + c^n)^2 = 2(a^{2n} + b^{2n} + c^{2n})).First, we use the Pythagorean theorem, which states (a^2 + b^2 = c^2). We then expand the left-hand side (LHS) and right-hand side (RHS) of the given equation:[(a^n + b^n + c^n)^2 = a^{2n} + b^{2n} + c^{2n} + 2a^n b^n + 2a^n c^n + 2b^n c^n][2(a^{2n} + b^{2n} + c^{2n})]Subtracting the RHS from the LHS, we get:[a^{2n} + b^{2n} + c^{2n} + 2a^n b^n + 2a^n c^n + 2b^n c^n - 2(a^{2n} + b^{2n} + c^{2n}) = 0]Simplifying, we find:[-a^{2n} - b^{2n} - c^{2n} + 2a^n b^n + 2a^n c^n + 2b^n c^n = 0][2a^n b^n + 2a^n c^n + 2b^n c^n = a^{2n} + b^{2n} + c^{2n}]We test specific values of (n). For (n = 4), we check if the equation holds for various right triangles. For example, using the 3-4-5 triangle:- LHS: ((3^4 + 4^4 + 5^4)^2 = (81 + 256 + 625)^2 = 962^2 = 925444)- RHS: (2(3^8 + 4^8 + 5^8) = 2(6561 + 65536 + 390625) = 2(462722) = 925444)Both sides are equal. Further verification with other triangles (e.g., 5-12-13 and 1-1-(sqrt{2})) confirms the equation holds for (n = 4).Additionally, we use algebraic manipulation to show that for (n = 4), the identity holds due to the structure of the Pythagorean theorem. Expanding (c^4 = (a^2 + b^2)^2) and substituting into the equation verifies the equality.Thus, the natural number (n) is (boxed{4}).]

🔑:6. 4 .Let x=a^{frac{n}{2}}, y=b^{frac{n}{2}}, z=c^{frac{pi}{2}}, thenbegin{aligned}0= & 2left(a^{2 n}+b^{2 n}+c^{2 n}right)-left(a^{n}+b^{n}+c^{n}right)^{2} = & 2left(x^{4}+y^{4}+z^{4}right)-left(x^{2}+y^{2}+z^{2}right)^{2} = & x^{4}+y^{4}+z^{4}-2 x^{2} y^{2}-2 x^{2} z^{2}-2 y^{2} z^{2} = & -(x+y+z)(x+y-z)(y+z-x)(z+x & -y) .end{aligned}Assume c is the hypotenuse, then z>x, z>y. It follows thatx+y+z>0, y+z-x>0, z+x-y>0 text {. }therefore(*) is equivalent to z=x+y,which means left(frac{a}{c}right)^{frac{n}{2}}+left(frac{b}{c}right)^{frac{n}{2}}=1.On the other hand, a^{2}+b^{2}=c^{2} holds, or left(frac{a}{c}right)^{2}+left(frac{b}{c}right)^{2}= 1.Since 0<frac{a}{c}<1,0<frac{b}{c}<1, y=left(frac{a}{c}right)^{x}+left(frac{b}{c}right)^{x} is a monotonically decreasing function, and it takes y=1 at only one point x, therefore, frac{n}{2}=2, n=4.

🧠:[<think>Okay, so I need to figure out which numbers in the sequence 101, 10101, 1010101, ... are prime. Let me start by understanding the structure of these numbers. Each subsequent number in the sequence adds another "10" before the final "1". So, the first term is 101, which is three digits. The next one is 10101, which is five digits, then 1010101 with seven digits, and so on. So, each term has an odd number of digits, increasing by two each time. First, let me write down the general form of these numbers. The first term is 101, which can be written as 1*100 + 0*10 + 1 = 100 + 1 = 101. The second term is 10101, which is 1*10,000 + 0*1,000 + 1*100 + 0*10 + 1 = 10,000 + 100 + 1 = 10,101. Wait, but the second term in the problem is listed as 10101, which is 10,101. So, each term alternates 1s and 0s, starting and ending with 1. Therefore, each term can be represented as a series of 1s separated by 0s, with the number of 1s increasing each time. Alternatively, since each term is built by appending "10" each time before the final "1", we can think of the nth term as having n "10" pairs followed by a "1". Wait, let's check:First term: 101. That's "10" followed by "1", so n=1. Second term: "10" "10" "1", so n=2. Third term: "10" "10" "10" "1", n=3. So, the nth term has n "10"s and then a "1". Therefore, the number of digits is 2n + 1. For example, n=1: 3 digits, n=2: 5 digits, n=3:7 digits, etc. So the nth term is a number composed of n "10"s followed by a "1".Now, to express this number mathematically. Let's denote the nth term as A_n. Then, A_n can be written as:A_n = 10^{2n} + 10^{2n - 2} + 10^{2n - 4} + ... + 10^2 + 1.This is a geometric series where each term is 10^2 times the previous term, starting from 1 and having n+1 terms. Wait, let's check:Wait, when n=1: 10^{2*1} + 1 = 100 + 1 = 101. Correct.When n=2: 10^{4} + 10^{2} + 1 = 10,000 + 100 + 1 = 10,101. Correct.Similarly, n=3: 10^6 + 10^4 + 10^2 + 1 = 1,000,000 + 10,000 + 100 + 1 = 1,010,101. Which is 1010101. Correct.So, in general, A_n is a sum of a geometric series with first term 1, ratio 100, and number of terms n+1. Therefore, the formula for A_n is:A_n = (100^{n+1} - 1) / (100 - 1) = (100^{n+1} - 1)/99.Let me verify this formula for n=1: (100^2 - 1)/99 = (10,000 - 1)/99 = 9,999/99 = 101. Correct.For n=2: (100^3 -1)/99 = (1,000,000 - 1)/99 = 999,999/99 = 10,101. Correct.So yes, A_n = (100^{n+1} - 1)/99. Therefore, each term in the sequence is equal to (100^{n+1} - 1)/99.Now, the question is which of these numbers are prime. Let's check for the first few terms.n=1: 101. Is 101 prime? Yes, 101 is a prime number. It's not divisible by 2, 3, 5, 7, 11. Let me check: 101 ÷ 2 = 50.5, not integer. 101 ÷ 3 ≈ 33.666... not integer. 101 ÷ 5 = 20.2, nope. 7? 101 ÷7 ≈14.428... nope. 11? 101 ÷11≈9.18... So, 101 is prime. So n=1 term is prime.n=2: 10,101. Let's check if 10101 is prime. First, check divisibility by small primes. Let's see:Divisible by 3? Sum of digits: 1+0+1+0+1=3. 3 is divisible by 3, so 10101 ÷3= 3367. So, 10101=3*3367. Therefore, not prime. So n=2 term is composite.n=3: 1,010,101. Let's check if 1010101 is prime. Again, check divisibility. Sum of digits: 1+0+1+0+1+0+1=4. Not divisible by 3. Ends with 1, so not divisible by 2 or 5. Let's check 7: 1010101 ÷7. Let me compute 1010101 ÷7. 7*144300=1,010,100. So 1010101 - 1,010,100 =1. So remainder 1. Not divisible by 7.Check 11: Alternating sum: (1 + 1 + 1) - (0 + 0 + 0) =3 -0=3. Not divisible by 11. How about 13: 13*77623=1,009,099. 1,010,101 -1,009,099=1,002. 1,002 ÷13≈77.07. Not integer. So 13*77=1001. 1,002-1001=1. So remainder 1. Not divisible by 13.Check 17: Let me compute 17*59,412=1,010,004. Then 1,010,101 -1,010,004=97. 97 ÷17=5.7. Not divisible. So remainder 97. Not divisible by 17.19: 19*53163=1,010,097. Then 1,010,101 -1,010,097=4. Not divisible.23: 23*43,913=1,010, 0... Hmm, maybe this is time-consuming. Alternatively, perhaps we can factorize 1010101.Alternatively, let's check if A_3 = (100^4 -1)/99. So 100^4=100,000,000. 100,000,000 -1=99,999,999. 99,999,999 /99=1,010,101. So, A_3=1,010,101.But 100^4 -1 can be factored as (100^2 -1)(100^2 +1). So, 100^4 -1= (100-1)(100+1)(100^2 +1)=99*101*10,001. Therefore, A_3=(99*101*10,001)/99=101*10,001. Therefore, 1,010,101=101*10,001. So, composite. Therefore, n=3 term is composite.Wait, that's a useful factorization. Let me check:(100^4 -1)/99 = (100^2)^2 -1)/99 = (100^2 -1)(100^2 +1)/99 = (99)(101)(10,001)/99 =101*10,001. Therefore, yes, A_3=101*10,001. So composite.Similarly, let's check if A_n can be factored in general. Since A_n = (100^{n+1} -1)/99. If n+1 is composite, say n+1 = k*m, then 100^{k*m} -1 factors as (100^k -1)(100^{k(m-1)} + 100^{k(m-2)} + ... +1). Therefore, if n+1 is composite, then A_n is composite.But even if n+1 is prime, A_n might still be composite. So, for A_n to be prime, n+1 must be prime, and also (100^{n+1} -1)/99 must be prime.So, first, n+1 must be prime. So, n must be one less than a prime number.But even then, (100^{prime} -1)/99 might not be prime. Let's see.For example, take n+1=2 (prime). Then n=1. A_1=(100^2 -1)/99= (10,000 -1)/99=9,999/99=101, which is prime. So that works.n+1=3 (prime). Then n=2. A_2=(100^3 -1)/99= (1,000,000 -1)/99=999,999/99=10,101, which we saw factors into 3*3367, so composite.n+1=5 (prime). Then n=4. Let's compute A_4=(100^5 -1)/99=100^5=10,000,000,000. 10,000,000,000 -1=9,999,999,999. Divide by 99: 9,999,999,999 ÷99=101,010,101.01? Wait, no, let's compute.Wait 99*101,010,101=99*100,000,000 +99*1,010,101=9,900,000,000 +99,999,999=9,900,000,000 +99,999,999=9,999,999,999. So yes, 9,999,999,999/99=101,010,101. So A_4=101010101.Is 101010101 prime? Let's check. Sum of digits: 1+0+1+0+1+0+1+0+1=5. Not divisible by 3. Ends with 1, so not divisible by 2 or 5. Check divisibility by small primes.Check 7: 101010101 ÷7. Let's see, 7*14,430,000=101,010,000. Then 101,010,101 -101,010,000=101. 101 ÷7≈14.428. So remainder 101-7*14=101-98=3. Not divisible by 7.11: Alternating sum: (1+1+1+1) - (0+0+0+0)=4-0=4. Not divisible by 11.13: Let's try 13. 13*7,769,238=101,000,094. 101,010,101 -101,000,094=10,007. 10,007 ÷13=769.769... Not integer. So remainder exists.17: 17*5,941,182=101,000, 094. Hmm, similar to before. This is tedious. Maybe try another approach.Alternatively, since A_n=(100^{n+1}-1)/99. If n+1 is prime, does that mean A_n could be prime? Not necessarily. For example, when n+1=3, which is prime, A_n=10,101=3*3367. So composite.Similarly, n+1=5, which is prime, A_4=101010101. Let's check if it's divisible by 101. 101*1,000,000=101,000,000. 101,010,101 -101,000,000=10,101. 10,101 ÷101=100.0099... Wait, 101*100=10,100. So 10,101 -10,100=1. So remainder 1. Therefore, not divisible by 101.How about 21? Wait, 21 is not prime. Maybe 7, 11, 13, 17, 19, 23, 29, etc.Alternatively, let's try to factor 101010101. Let me check with a known prime factorization. Alternatively, perhaps using the fact that 101010101 can be written as (10^8 + 10^6 + 10^4 + 10^2 +1). Maybe this can be factored further.Alternatively, perhaps there's a pattern. For example, when n+1 is prime, but 100 is a base, so 100 ≡ 1 mod some primes. Let's consider modulo primes.Alternatively, perhaps using the fact that 100 ≡ 1 mod 11. So, 100 ≡ 1 mod 11, so 100^{k} ≡1^k=1 mod 11. Therefore, (100^{n+1} -1)/99 ≡ (1 -1)/99 ≡0/99≡0 mod11 when n+1 is any integer. Wait, (100^{n+1} -1)/99 = A_n. If 100≡1 mod11, then 100^{n+1}≡1^{n+1}≡1 mod11. Therefore, 100^{n+1}-1≡0 mod11. Therefore, 11 divides (100^{n+1}-1). Then, since 99=9*11, (100^{n+1}-1)/99 = [(100^{n+1}-1)/11]/9. Therefore, if 11 divides (100^{n+1}-1)/11, then 11 divides A_n. Wait, perhaps this is getting complicated.Wait, (100^{n+1} -1)/99 = ((10^2)^{n+1} -1)/99 = (10^{2(n+1)} -1)/99. If 2(n+1) is even, which it is, so 10^{even} -1 is divisible by 11, as 10^even ≡1 mod11. So, 10^{2(n+1)} -1 ≡0 mod11, so 11 divides (10^{2(n+1)} -1), hence 11 divides 99*(A_n). Since 99 is 9*11, so 11 divides 11*(9*A_n), so 11 divides 9*A_n. Since 11 is prime and doesn't divide 9, 11 must divide A_n. Therefore, A_n is divisible by 11 for all n >=1. Wait, but hold on, A_1=101, which is not divisible by 11. So this contradicts? Wait, maybe my reasoning is wrong.Wait, let me check A_1=101. 101 divided by 11 is approximately 9.18, which is not an integer. So 101 is not divisible by 11. But according to the previous reasoning, 10^{2(n+1)} -1 is divisible by 11. For n=1, 10^{4} -1=10,000 -1=9,999. 9,999 divided by 99=101. So 9,999=99*101. So 9,999 is divisible by 11? 9,999 ÷11=909, which is 909. So yes, 9,999=11*909. But then 9,999/99=101, which is not divisible by 11. So how does that work?Wait, (100^{n+1} -1)/99 = A_n. For n=1: (100^2 -1)/99=9999/99=101. Now, 9999 is divisible by 11, as 9999 ÷11=909. So 9999=11*909. But 909 ÷9=101. So 9999=11*9*101=99*101. Therefore, when we divide by 99, we get 101. So in this case, A_1=101 is not divisible by 11, even though 100^{n+1}-1 is divisible by 11. Because when we divide by 99 (which is 9*11), the 11 cancels out. Therefore, A_n = (100^{n+1}-1)/99 = (11*K)/99 = K/9. Therefore, if K is divisible by 9, then A_n is integer. So, perhaps the earlier reasoning was wrong in assuming that A_n is divisible by 11. It's actually (100^{n+1} -1)/99. Since 100 ≡1 mod11, then 100^{n+1} ≡1 mod11, so 100^{n+1}-1 ≡0 mod11. Therefore, 11 divides (100^{n+1}-1). Therefore, (100^{n+1}-1)/99 = [(100^{n+1}-1)/11]/9. Since 11 divides the numerator, [(100^{n+1}-1)/11] is an integer, and then we divide by 9. Therefore, A_n is an integer. However, [(100^{n+1}-1)/11] is not necessarily divisible by 9. For example, when n=1: (100^2 -1)/11=9999/11=909, which is 909. Then 909 divided by 9=101. So, in this case, it's divisible by 9, hence A_n=101 is integer. When n=2: (100^3 -1)/11=999999/11=90909. 90909 divided by9=10101. So, again, divisible by9. So, for all n, [(100^{n+1}-1)/11] is divisible by9? Let me check for n=3: (100^4 -1)/11=99,999,999/11=9,090,909. 9,090,909 divided by9=1,010,101. Yes. So it seems [(100^{n+1}-1)/11] is divisible by9 for all n>=1. Therefore, A_n is always an integer, as expected.But does that mean A_n is divisible by11? Wait, in the case of n=1, A_n=101, which is not divisible by11. So, even though 100^{n+1}-1 is divisible by11, when we divide by99, the 11 cancels out. So A_n=(100^{n+1}-1)/99= [(100^{n+1}-1)/11]/9. So, the numerator is divisible by11, but after dividing by11 and then by9, A_n is not necessarily divisible by11. For example, n=1: [(100^2 -1)/11]/9=909/9=101. 101 is not divisible by11. Similarly, n=2: [(100^3 -1)/11]/9=90909/9=10101. 10101 ÷11=918.272..., not integer. So 10101 is not divisible by11. Wait, but the original number (100^{n+1}-1) is divisible by11, but after dividing by99, it's not necessarily divisible by11.Therefore, my previous conclusion that A_n is divisible by11 for all n is wrong. So, scratch that. Therefore, maybe A_n is not necessarily divisible by11. Therefore, perhaps the earlier factorization approach is better.So, for A_n=(100^{n+1}-1)/99. If n+1 is composite, then 100^{n+1}-1 can be factored into smaller exponents, leading to A_n being composite. So, if n+1 is composite, then A_n is composite. Therefore, if A_n is prime, then n+1 must be prime. But the converse isn't necessarily true. For example, when n+1=3 (prime), A_n=10101=3*3367, which is composite.Therefore, the only candidates for prime A_n are those where n+1 is prime. So, the first few n+1 primes are 2,3,5,7,11, etc., so n=1,2,4,6,10,...But even in those cases, A_n might not be prime. For example, n=1: prime. n=2: composite. Let's check n=4 (n+1=5). A_4=(100^5 -1)/99=101010101. Is this prime? Let's check.Alternatively, perhaps we can find a factorization for A_4. Let me try dividing by small primes. 101010101 ÷7=14,430,014.428... No. ÷11=9,182,736.454... No. ÷13=7,769,238.538... No. ÷17=5,941,770.647... No. ÷19=5,316,321.105... No. ÷23=4,391,743.52... No. ÷29=3,483,106.931... No. ÷31=3,258,390.354... No. ÷37=2,730,002.702... No. ÷41=2,463,173.195... No. ÷43=2,349,072.581... No. ÷53=1,906,038.698... No.Alternatively, maybe there's a pattern here. Let me consider that 101010101 can be written as 10^8 +10^6 +10^4 +10^2 +1. Let's see if this can be expressed as a product of polynomials. For example, x^4 +x^3 +x^2 +x +1 factors into cyclotomic polynomials, but here it's x^8 +x^6 +x^4 +x^2 +1. Which is x^2*(x^6 +x^4 +x^2 +1) +1. Hmm, not sure. Alternatively, note that x^8 +x^6 +x^4 +x^2 +1 = (x^10 -1)/(x^2 -1). Because (x^2 -1)(x^8 +x^6 +x^4 +x^2 +1)=x^10 -1. So yes, x^8 +x^6 +x^4 +x^2 +1=(x^10 -1)/(x^2 -1). Therefore, substituting x=10, we get (10^10 -1)/(10^2 -1)= (10,000,000,000 -1)/99=999,999,999,999/99=10,101,010,101. Wait, but that's A_4=101,010,101. Wait, but (10^10 -1)/99=10101010101, which is a different number. Wait, perhaps I made a miscalculation. Wait, x=10: x^10 -1=10,000,000,000 -1=9,999,999,999. Then divided by (x^2 -1)=99. So 9,999,999,999 /99=101,010,101.01? Wait, no. Wait 9,999,999,999 ÷99=101,010,101.01? Wait, but 99*101,010,101=9,999,999,999. So, 10^10 -1=99*101,010,101 +0. So, (10^10 -1)/99=101,010,101. But A_4 is 101010101, which is 101,010,101. So, yes, A_4=(10^10 -1)/99. Therefore, (10^10 -1)/99=101010101. So, using the factorization of x^10 -1, which is (x^5 -1)(x^5 +1). Then, x^10 -1=(x-1)(x+1)(x^4 +x^3 +x^2 +x +1)(x^4 -x^3 +x^2 -x +1). Therefore, substituting x=10, we have 10^10 -1=9*11*11111*9091. Therefore, (10^10 -1)/99= (9*11*11111*9091)/99= (9/9)*(11/11)*11111*9091=1*1*11111*9091=11111*9091. Therefore, 101,010,101=11111*9091. Therefore, composite. Therefore, A_4 is composite.So, even when n+1=5 (prime), A_n is composite. Let's check n=6 (n+1=7). Then A_6=(100^7 -1)/99. That's a big number. Let's see if we can factor it. Using the same approach:100^7 -1=10^{14} -1=(10^7)^2 -1=(10^7 -1)(10^7 +1). Therefore, (10^{14} -1)/99= (10^7 -1)(10^7 +1)/99. Now, 10^7 -1=9,999,999=9999999. 9999999=9*111,111. Wait, 9*111,111=999,999. Wait, 10^7 -1=9,999,999=9*1,111,111. Because 1,111,111*9=9,999,999. Then 1,111,111 is 1111111. Is that prime? Let me check. 1,111,111. Let's check divisibility by 239: 239*4644=1,111, 116. Over. Maybe 1111111 ÷101=11,000.11... Not exact. Alternatively, 1111111 is known to be 239*4649. Let me check: 239*4000=956,000. 239*600=143,400. 239*49=11,711. Total:956,000+143,400=1,099,400 +11,711=1,111,111. Yes, so 1,111,111=239*4649. Therefore, 10^7 -1=9*239*4649. Then, 10^7 +1=10,000,001. Let's factor that. 10,000,001. Divided by 11: 10,000,001 ÷11=909,091. So, 10,000,001=11*909,091. Is 909,091 prime? Let me check. Divided by 7: 909,091 ÷7≈129,870.14. Not integer. 13: 909,091 ÷13≈69,930.076. Not integer. 17: 909,091 ÷17≈53,476. So 17*53,476=909, 092. Close, but not exact. Maybe 909,091 is prime? Not sure, but let's assume it is for now. Therefore, (10^7 -1)(10^7 +1)/99=(9*239*4649)(11*909,091)/99=9*11*239*4649*909,091/99= (99/99)*239*4649*909,091=239*4649*909,091. Therefore, A_6=239*4649*909,091. Therefore, composite. So, even when n+1=7 (prime), A_n is composite.Therefore, it seems that even when n+1 is prime, A_n is composite. The only time A_n was prime is when n=1 (n+1=2). Let's check n=10 (n+1=11). A_10=(100^{11} -1)/99. That's a huge number. It's unlikely to be prime, but let's see if we can find a pattern.Alternatively, maybe the only prime in the sequence is the first term, 101. Let me check n=5 (n+1=6, which is composite). Then A_5=(100^6 -1)/99. 100^6=1e12, so (1e12 -1)/99=10101010101.01? Wait, actually, (1e12 -1)/99=10101010101. So, 10101010101. Let me see if this is composite. Sum of digits: 1+0+1+0+1+0+1+0+1+0+1=6. Divisible by 3. Therefore, 10101010101 ÷3=3367003367. So, composite. Therefore, A_5 is composite.Similarly, for n=4, we saw it's composite. n=6 is composite. n=10 would also be composite. Therefore, so far, the only prime in the sequence is A_1=101. Let me check n=0, but the sequence starts at n=1, with 101. Therefore, maybe 101 is the only prime in the sequence.But let me check n=5 (n+1=6). Wait, n=5 corresponds to n+1=6, which is composite, so A_5 is composite. Therefore, up to here, only A_1 is prime. Let me check n=12 (n+1=13). A_12=(100^{13}-1)/99. This is a very large number, and checking its primality would be difficult manually. However, considering the previous pattern, it's highly probable that all terms for n>1 are composite.Another approach: suppose that for n>1, A_n can be written as a product of smaller numbers. For example, when n is even, say n=2m, then 100^{2m+1}-1 factors as (100-1)(100^{2m} +100^{2m-1} +...+1). Wait, but 100^{n+1}-1 factors as (100-1)(100^{n} +100^{n-1} +...+1). Therefore, (100^{n+1}-1)/99= (100^{n} +100^{n-1} +...+1). Which is the same as the sum we had before. However, unless n+1 is prime, but even then, as seen in previous examples, this sum can be factored further.Alternatively, perhaps there's a theorem related to repunit numbers. Wait, repunit numbers are numbers consisting of all 1s. Their primality is studied. Our numbers are similar but in base 100. That is, A_n is a repunit number in base 100 with n+1 digits. The repunit primes are primes in the repunit numbers. It's known that in base 10, repunit primes are rare and their primality depends on the number of digits being prime. Similarly, in base 100, the same logic applies. But since 100 is a square, 100=10^2, maybe there's more structure.In any case, given that for n>1, A_n can be factored when n+1 is composite, but even when n+1 is prime, A_n tends to be composite (as shown for n+1=3,5,7,11), it's plausible that the only prime in the sequence is the first term, 101. Therefore, the answer is that only the first term, 101, is prime.Final AnswerThe only prime number in the sequence is boxed{101}.</think>To determine which members of the sequence 101, 10101, 1010101, ... are prime, we first express the general form of the nth term. Each term in the sequence can be written as a geometric series:[ A_n = frac{100^{n+1} - 1}{99} ]We need to check the primality of these terms for different values of ( n ).1. For ( n = 1 ): [ A_1 = frac{100^2 - 1}{99} = frac{9999}{99} = 101 ] 101 is a prime number.2. For ( n = 2 ): [ A_2 = frac{100^3 - 1}{99} = frac{999999}{99} = 10101 ] 10101 is not prime (divisible by 3).3. For ( n = 3 ): [ A_3 = frac{100^4 - 1}{99} = frac{99999999}{99} = 1010101 ] 1010101 is not prime (factors into 101 and 10001).4. For ( n = 4 ): [ A_4 = frac{100^5 - 1}{99} = frac{9999999999}{99} = 101010101 ] 101010101 is not prime (factors into 11111 and 9091).5. General Analysis: - If ( n+1 ) is composite, ( A_n ) can be factored and is therefore composite. - Even if ( n+1 ) is prime, ( A_n ) might still be composite. For example, ( n+1 = 5 ) (prime) results in ( A_4 = 101010101 ), which is composite.Through detailed factorization and primality checks, it is found that all terms for ( n > 1 ) are composite. The only prime number in the sequence is the first term.[boxed{101}],<think>Alright, let me try to figure out which numbers in the sequence 101, 10101, 1010101, ... are prime. So first, I need to understand the pattern here. Each term in the sequence seems to be made by adding "10" alternately, but starting with 101. Let me write out the first few terms:1st term: 1012nd term: 101013rd term: 10101014th term: 101010101And so on. So each subsequent term has two more digits, adding "10" each time. So the first term is three digits, the second is five digits, third is seven digits, etc. So the nth term has 2n + 1 digits. Now, the question is which of these numbers are prime. Let me recall that primes are numbers greater than 1 that have no divisors other than 1 and themselves. So we need to check if each of these numbers is prime. But since these numbers get large very quickly, factoring them directly might be challenging. So maybe there's a pattern or a formula that can help us generalize whether these numbers can be prime or not.First, let me try to express these numbers mathematically. Let's take the first term: 101. That's straightforward. The second term is 10101. Let's break that down. 10101 can be thought of as 1*10^4 + 0*10^3 + 1*10^2 + 0*10 + 1 = 10^4 + 10^2 + 1. Similarly, 1010101 is 10^6 + 10^4 + 10^2 + 1. So the general form for the nth term seems to be the sum of 10^{2k} for k from 0 to n. Wait, let me check:For n=1: sum from k=0 to 1 of 10^{2k} = 10^0 + 10^2 = 1 + 100 = 101. Correct.For n=2: sum from k=0 to 2 of 10^{2k} = 10^0 + 10^2 + 10^4 = 1 + 100 + 10000 = 10101. Correct.So in general, the nth term is the sum_{k=0}^{n} 10^{2k}. That is a geometric series. The sum of a geometric series is (r^{n+1} - 1)/(r - 1), where r is the common ratio. Here, r is 100, since each term is 10^2, which is 100. So the sum would be (100^{n+1} - 1)/(100 - 1) = (100^{n+1} - 1)/99. Therefore, each term in the sequence can be written as (100^{n+1} - 1)/99. Let me verify that with n=1: (100^2 -1)/99 = (10000 -1)/99 = 9999/99 = 101. Correct. For n=2: (100^3 -1)/99 = 1000000 -1 = 999999/99 = 10101. Correct. So that formula works.So the problem reduces to determining for which n >=1, the number (100^{n+1} -1)/99 is prime.So we need to check if this number is prime. Let's note that (100^{n+1} -1)/99 is a number consisting of n+1 1's in base 100, which translates to alternating 1's and 0's in base 10. For example, in base 10, 101 is 1*100 + 0*10 +1, which is 1*10^2 + 0*10 +1, which is the same as 100 + 0 +1 = 101. Similarly, 10101 is 1*10^4 + 0*10^3 +1*10^2 +0*10 +1 = 10000 + 100 +1 = 10101. So in base 100, each digit is represented by two digits in base 10, hence the alternating 1's and 0's.But maybe there's another way to express this. Let's see, (100^{n+1} -1)/99. Let's write 100 as 10^2. So 100^{n+1} is 10^{2(n+1)}. Then, (10^{2(n+1)} -1)/99. Hmm. Alternatively, 99 is 9*11, so this number can be written as (10^{2(n+1)} -1)/(9*11). Wait, but (10^{2(n+1)} -1) is a number like 999...9 with 2(n+1) digits. So dividing by 99 gives us a number with n+1 digits, each being 10101...01. For example, 9999/99 = 101, 999999/99 = 10101, etc. So that makes sense.Now, the key is to find n such that this number is prime. Let's check small n first.n=1: (100^2 -1)/99 = 101. 101 is a prime number. So that's prime.n=2: (100^3 -1)/99 = 10101. Let's check if 10101 is prime. Hmm. Let me test divisibility by small primes. First, 2: it's odd. 3: sum the digits: 1+0+1+0+1 = 3. Since 3 divides 3, then 3 divides 10101. So 10101 is divisible by 3. Therefore, 10101 is not prime.n=3: (100^4 -1)/99 = 1010101. Let's check if 1010101 is prime. Again, check divisibility by small primes. 2: odd. 3: sum the digits: 1+0+1+0+1+0+1 = 4. 4 is not divisible by 3. 5: ends with 1, so no. 7: Let's try dividing 1010101 by 7. 7*144300 = 1010100. Then 1010101 - 1010100 =1, so remainder 1. Not divisible by 7. 11: Alternating sum: (1 +1 +1 +1) - (0 +0 +0) =4 -0=4. Not divisible by 11. 13: Let's try 13*77623 = 1009099. 1010101 - 1009099 = 1002. 1002/13 is 77.07... Not integer. So maybe 13 doesn't divide. Hmm, maybe 101? Wait, 101*10000 = 1010000. 1010101 - 1010000 = 101. So 1010101 = 101*10000 + 101 = 101*(10000 +1) = 101*10001. Therefore, 1010101 factors as 101*10001. So 101 is prime, but 10001 is 100^2 +1. Is 10001 prime? Let's check. 10001 divided by 17: 17*588 = 10000 - 17*588=10000-9996=4. 17*588 +4=10000. So 10001 is 17*588 +4 +1=17*588 +5. Not divisible by 17. Let's try 73: 73*136= 73*(100+36)=7300 + 2628=9928. 10001-9928=73. So 10001=73*137. Let me check 73*137: 70*137=9590, 3*137=411, so total 9590+411=99901? Wait, no, 73*137. Let me compute 73*100=7300, 73*30=2190, 73*7=511. So 7300+2190=9490, 9490+511=99901? Wait, no. Wait, 73*137: 70*137 +3*137 =9590 +411=9590+411=10001. Yes, correct. So 10001=73*137. Therefore, 1010101=101*73*137. So that's composite. Therefore, n=3 gives a composite number.n=4: (100^5 -1)/99 =101010101. Let me check if this is prime. Again, check divisibility. Sum of digits: 1+0+1+0+1+0+1+0+1=5. Not divisible by 3. 5: ends with 1, no. 7: Let's see. 7*14430014=101010098. Then 101010101-101010098=3. Not divisible by 7. 11: Alternating sum: (1+1+1+1+1)-(0+0+0+0)=5-0=5. Not divisible by 11. Maybe 101 is a factor? Let's check 101*1000000=101000000. 101010101 -101000000=10101. Then 10101 divided by 101 is 100.0099... Wait, 101*100=10100, so 10101-10100=1. So 10101=101*100 +1. Therefore, 101010101=101*1000000 +101*100 +1=101*(1000000 +100) +1. Not a multiple of 101. Alternatively, maybe another factor. Let's try 73: 73*1383001=73*(1380000 +3001)=73*1380000=100,740,000; 73*3001=219,073. So total 100,740,000 +219,073=100,959,073. Which is less than 101,010,101. Difference is 51,028. Not helpful. Maybe 11? Wait, already checked. Alternatively, perhaps 101010101 is prime. Wait, maybe there's a pattern here. Let's try to see if 101010101 can be written as a product. Alternatively, since the previous term had factors 101, 73, 137, maybe similar here. Wait, 101010101 divided by 101: 101010101 ÷ 101. Let's do the division: 101 into 101010101. 101 into 101 is 1. Bring down 01: 101 into 101 is 1 again. Then bring down 01: another 1. Then bring down 01: another 1. Then bring down 01: another 1. So it's 101*100010001. Wait, let's check 101*100010001. 100010001*100=10001000100, plus 100010001*1=100010001. So total 10001000100 +100010001=10101010101. Wait, that's 10101010101, which is longer. Wait, perhaps my approach is wrong. Let me try a better division. Let's compute 101 into 101010101.Start with 101 into 101 (first three digits): 1. 1*101=101. Subtract, get 0. Bring down next digit 0: 00. Bring down next digit 1: 001. 101 into 1: 0. Bring down next digit 0: 0010. 101 into 10: 0. Bring down next digit 1: 00101. 101 into 101:1. Multiply 1*101=101. Subtract, remainder 0. Bring down next digit 0: 0. Bring down next digit 1:01. 101 into 1:0. So the quotient is 10001. So 101*10001=1010000 +101=1010101. Wait, but 101010101 divided by 101 is 1000100.01...? Wait, no, let me do this properly.Wait, 101 * 1000000 = 101000000101 * 10000 = 1010000101 * 1 = 101So 101*(1000000 +10000 +1) =101*(1010001)=101*1000000 +101*10000 +101*1=101000000 +1010000 +101=102010000 +101=102010101. Wait, that's not 101010101. Hmm, maybe my approach is wrong. Alternatively, perhaps 101010101 is prime. Let me check with another method.Alternatively, 101010101 can be written as (100^5 -1)/99. Let's factor 100^5 -1. That's a difference of squares: 100^5 = (10^2)^5 = 10^10. So 10^10 -1 = (10^5 -1)(10^5 +1). So 10^10 -1 = (10^5 -1)(10^5 +1) = (99999)(100001). Therefore, (10^10 -1)/99 = (99999/99)(100001) = 10101 * 100001. Therefore, 101010101 = 10101 * 100001. So both 10101 and 100001 are factors. Therefore, 101010101 is composite. Therefore, n=4 gives a composite number.Hmm, so n=1 gives prime, n=2,3,4 composite. Let's check n=5, just to see a pattern. But this might take a long time.Alternatively, maybe there's a general approach. Let me note that (100^{n+1} -1)/99 can be factored if n+1 is composite. Because 100^{ab} -1 is divisible by 100^a -1. So for example, if n+1 is composite, say n+1 = a*b, then 100^{a*b} -1 is divisible by 100^a -1, so (100^{a*b} -1)/99 = (100^a -1)/99 * (100^{a(b-1)} + 100^{a(b-2)} + ... +1). Therefore, unless n+1 is prime, (100^{n+1} -1)/99 is composite. Therefore, if n+1 is composite, the number is composite. So only when n+1 is prime, there is a possibility that (100^{n+1} -1)/99 is prime. But even if n+1 is prime, the number might still be composite. So the necessary condition is that n+1 is prime, but not sufficient.Therefore, for (100^{n+1} -1)/99 to be prime, n+1 must be prime.Therefore, we need to check n where n+1 is prime. For example, n=1: n+1=2 (prime), which gives 101, which is prime. Then n=2: n+1=3 (prime), but 10101 is composite. n=3: n+1=4 (composite), so composite. n=4: n+1=5 (prime), but 101010101 is composite. n=5: n+1=6 (composite). n=6: n+1=7 (prime). So we can check n=6, which would be (100^7 -1)/99. But that's a very large number. Let's see if we can determine whether it's prime without factoring.Alternatively, perhaps all such numbers are composite except the first one. That is, 101 is prime, and all others are composite. Because even when n+1 is prime, the number might not be prime. For example, when n=2 (n+1=3), which is prime, but (100^3 -1)/99=10101=3*3367, which is composite. Similarly, n=4 (n+1=5): 101010101=101*100010001, which is composite. So even if n+1 is prime, the number can be composite. So perhaps the only prime in the sequence is 101. Let's see if there is a mathematical reason for this. The numbers in the sequence are called "permutable primes" or something else? Wait, no. They are numbers consisting of alternating 1s and 0s, starting and ending with 1. These are similar to repunit numbers, which have all 1s. Repunit primes are primes consisting of all 1s, and their indices must be prime. Similarly, here, the numbers are base-10 representations similar to repunits but in base 100. So repunit primes in base 100. However, similar to base 10 repunit primes, even if the number of digits is prime, the number itself might not be prime.But in our case, since (100^{n+1} -1)/99, if n+1 is prime, maybe it's more likely to be prime, but in our examples, even when n+1 is prime, like 3 or 5, it's composite. So maybe 101 is the only prime. Let's confirm for n=6, where n+1=7 (prime). Let's compute (100^7 -1)/99. That's 10101010101. Let me check divisibility. Let's try small primes. Sum of digits: 1+0+1+0+1+0+1+0+1+0+1=5. Not divisible by 3. Ends with 1, so not 2 or 5. Let's check 7: 10101010101 ÷7. 7*1443001443=10101010101. Wait, 7*1443001443=7*(1443000000 +1443)=7*1443000000=10101000000; 7*1443=10101. So total 10101000000 +10101=10101010101. Therefore, 10101010101 is divisible by 7. Therefore, n=6 gives a composite number. So even when n+1 is 7 (prime), the number is composite.Therefore, it seems that except for the first term, all others are composite. Therefore, the only prime in the sequence is 101. But let me check n=0 just to be thorough. Wait, the sequence starts at 101, which is n=1. So n=0 isn't considered here.Alternatively, maybe there is another term. Let me check n=5, which would be n+1=6 (composite). So composite. n=6: n+1=7 (prime), but we saw that's composite. n=7: n+1=8 (composite). n=8: n+1=9 (composite). n=9: n+1=10 (composite). n=10: n+1=11 (prime). Let's see if (100^11 -1)/99 is prime. That's a 23-digit number: 101010101010101010101. Checking primality of such a large number is difficult manually. However, given the pattern, it's likely composite. For example, 100^11 -1 factors as (10^22 -1), which can be factored into (10^11 -1)(10^11 +1). Then (10^11 -1)/99 = (10^11 -1)/(10^2 -1) = (10^11 -1)/((10-1)(10+1)) = ... Alternatively, since 100^11 -1 = (100 -1)(100^10 +100^9 +...+1). Therefore, (100^11 -1)/99 = 100^10 +100^9 +...+1. Which is composite because 11 is prime, but the number could still be composite. Since the number is huge, it's almost certainly composite. Thus, given that all terms beyond the first have been shown to be composite through examples and the general factorization when n+1 is composite, and even when n+1 is prime, the numbers factor into smaller primes, the only prime in the sequence is 101.Final AnswerThe only prime number in the sequence is boxed{101}.</think>The sequence given is 101, 10101, 1010101, ..., where each term can be expressed as (frac{100^{n+1} - 1}{99}) for (n geq 1). 1. First term (n = 1): [ frac{100^{2} - 1}{99} = frac{10000 - 1}{99} = frac{9999}{99} = 101 ] 101 is a prime number.2. Second term (n = 2): [ frac{100^{3} - 1}{99} = frac{1000000 - 1}{99} = frac{999999}{99} = 10101 ] 10101 is divisible by 3 (sum of digits is 3), hence composite.3. Third term (n = 3): [ frac{100^{4} - 1}{99} = frac{100000000 - 1}{99} = frac{99999999}{99} = 1010101 ] 1010101 factors as 101 × 10001, and 10001 further factors into 73 × 137, hence composite.4. Fourth term (n = 4): [ frac{100^{5} - 1}{99} = frac{10000000000 - 1}{99} = frac{9999999999}{99} = 101010101 ] 101010101 factors as 10101 × 100001, both of which are composite.5. General case analysis: - For (n+1) composite, (frac{100^{n+1} - 1}{99}) is composite because it can be factored. - For (n+1) prime, further analysis shows that even if (n+1) is prime, the number can still be composite (e.g., (n = 2) and (n = 4)).After checking several terms and analyzing the structure, it is concluded that the only prime number in the sequence is 101.[boxed{101}]]

🔑:: 101. Fairly easy. Let k n represent the member of the sequence with n 1s. It is obvious that 101 divides k 2n . So we need only consider k 2n+1 . But k 2n+1 = 1 + 10 2 + 10 4 + ... + 10 4n = (10 4n+2 - 1)/99 = (10 2n+1 + 1)/11 (10 2n+1 - 1)/9. Each of these is integral: the first is 1 - 10 + 10 2 - ... + 10 2n , the second is 11...1 (2n+1 1s). 50th Putnam 1989 © John Scholes [email protected] 3 Nov 1999

🧠:[<think>Alright, so I need to add two binary numbers: 101111011 and 1100111011. Hmm, binary addition can be a bit tricky if you're not used to it, but let me think through it step by step. First off, I remember that adding binary numbers is similar to adding decimal numbers, except that you carry over when the sum is 2 instead of 10. So, each column addition can result in 0, 1, or 10 (which is 2 in decimal), right? That means if the sum in a column is 2, we write down 0 and carry over 1 to the next column. If it's 3, which can happen if there's a carry from the previous column, we write down 1 and carry over 1 again. Let me confirm that logic. Yeah, because 1+1=10 in binary, which is 2, so you carry 1. If you have 1+1+1=11 in binary, which is 3, so you write 1 and carry 1. Okay, that makes sense.Now, looking at the two numbers: (101111011)₂ and (1100111011)₂. Let me write them down aligned properly. To add them, they need to be right-aligned, just like with decimal numbers. Let me check their lengths. The first number is 9 digits: 1 0 1 1 1 1 0 1 1. The second is 10 digits: 1 1 0 0 1 1 1 0 1 1. So, the second number is longer by one digit. I need to make sure they are aligned by their least significant bit (rightmost bit). Let me write them one under the other, padding the shorter one with a leading zero to make them the same length. Wait, actually, leading zeros don't affect the value, so padding the shorter number on the left with a zero to make both numbers 10 digits. So, the first number becomes 0101111011, and the second is 1100111011. Let me check:First number padded: 0 1 0 1 1 1 1 0 1 1Second number: 1 1 0 0 1 1 1 0 1 1Yes, that's correct. Now, we can add them column by column starting from the rightmost digit.Let me set up the addition: 0 1 0 1 1 1 1 0 1 1 + 1 1 0 0 1 1 1 0 1 1 ---------------------------Starting from the rightmost bit (position 1):Column 1: 1 + 1 = 10. So, write 0, carry 1.Column 2: 1 + 1 + carry 1 = 11. Write 1, carry 1.Column 3: 0 + 0 + carry 1 = 1. Write 1, carry 0.Column 4: 1 + 1 + carry 0 = 10. Write 0, carry 1.Column 5: 1 + 1 + carry 1 = 11. Write 1, carry 1.Column 6: 1 + 1 + carry 1 = 11. Write 1, carry 1.Column 7: 1 + 1 + carry 1 = 11. Write 1, carry 1.Column 8: 0 + 0 + carry 1 = 1. Write 1, carry 0.Column 9: 1 + 1 + carry 0 = 10. Write 0, carry 1.Column 10: 0 + 1 + carry 1 = 10. Write 0, carry 1.Wait, but since we're at the leftmost digit now, we have a carry of 1 that needs to be added. So, the result will have an additional digit at the beginning. Let me write down the results step by step.Let me list the columns from right to left (positions 1 to 10):Position 1: 1 + 1 = 0, carry 1Position 2: 1 + 1 + 1 = 1, carry 1Position 3: 0 + 0 + 1 = 1, carry 0Position 4: 1 + 1 + 0 = 0, carry 1Position 5: 1 + 1 + 1 = 1, carry 1Position 6: 1 + 1 + 1 = 1, carry 1Position 7: 1 + 1 + 1 = 1, carry 1Position 8: 0 + 0 + 1 = 1, carry 0Position 9: 1 + 1 + 0 = 0, carry 1Position 10: 0 + 1 + 1 = 0, carry 1Then, we have a final carry of 1. So, the result is 1 followed by the results from position 10 to 1. Wait, actually, the order is reversed because when we write the result, the first carry we get is for the next position. Let me make sure.Wait, in positional numbering, the rightmost digit is the least significant bit (position 1). When we add, we start from position 1, go to position 10, and then the final carry is added as position 11. So, the result would be:Carry: 1 1 0 1 1 1 1 0 1 1 1Wait, let me re-examine each position with proper tracking. Maybe I need to write it out step by step.Let me create a table for each column, tracking the two bits being added, the carry in, the sum bit, and the carry out.Let's label the two numbers as A and B. A is the first number (padded to 10 bits): 0 1 0 1 1 1 1 0 1 1B is the second number: 1 1 0 0 1 1 1 0 1 1We'll process each bit from right to left (index 1 to 10):Position 1 (rightmost):A: 1B: 1Carry-in: 0Sum: 1 + 1 + 0 = 10 (binary)Sum bit: 0Carry-out: 1Position 2:A: 1B: 1Carry-in: 1Sum: 1 + 1 + 1 = 11 (binary)Sum bit: 1Carry-out: 1Position 3:A: 0B: 0Carry-in: 1Sum: 0 + 0 + 1 = 01 (binary)Sum bit: 1Carry-out: 0Position 4:A: 1B: 1Carry-in: 0Sum: 1 + 1 + 0 = 10Sum bit: 0Carry-out: 1Position 5:A: 1B: 1Carry-in: 1Sum: 1 + 1 + 1 = 11Sum bit: 1Carry-out: 1Position 6:A: 1B: 1Carry-in: 1Sum: 1 + 1 + 1 = 11Sum bit: 1Carry-out: 1Position 7:A: 1B: 1Carry-in: 1Sum: 1 + 1 + 1 = 11Sum bit: 1Carry-out: 1Position 8:A: 0B: 0Carry-in: 1Sum: 0 + 0 + 1 = 01Sum bit: 1Carry-out: 0Position 9:A: 1B: 1Carry-in: 0Sum: 1 + 1 + 0 = 10Sum bit: 0Carry-out: 1Position 10:A: 0B: 1Carry-in: 1Sum: 0 + 1 + 1 = 10Sum bit: 0Carry-out: 1After Position 10, we still have a carry-out of 1, which becomes the next bit.So compiling all the sum bits from Position 1 to 10 and the final carry:Sum bits (Position 1 to 10):Position 1: 0Position 2: 1Position 3: 1Position 4: 0Position 5: 1Position 6: 1Position 7: 1Position 8: 1Position 9: 0Position 10: 0Final carry: 1So the total result is the final carry followed by the sum bits from Position 10 to 1? Wait, no. Wait, when we add from right to left, the first sum bit we get is the rightmost (Position 1), then Position 2 to the left, etc. So the sum bits should be ordered from Position 1 (LSB) to Position 10, and then the final carry becomes the MSB.But when writing the number, the first bit we get is the LSB, so the order is:Final carry, Position 10, Position 9, ..., Position 1.Wait, no. Let me think.When adding, we start at the rightmost digit (Position 1). The sum bit for Position 1 is the rightmost bit of the result. Then Position 2 is the next bit to the left, and so on. After processing all positions, the carry-out becomes the new leftmost bit.Therefore, the result is:Carry-out (MSB) followed by the sum bits from Position 10 down to Position 1. Wait, actually, no. Wait, when you process Position 1, you get the rightmost sum bit. Then Position 2 gives the next bit to the left, ..., Position 10 gives the 10th bit from the right, and then the final carry is the 11th bit (MSB). So the final result is a 11-bit number where the bits are:[Carry-out][Sum Position 10][Sum Position 9]...[Sum Position 1]But that can't be right because the sum bits from Position 1 to 10 are written right to left. Wait, maybe I need to visualize the actual bits.Let me list the sum bits in the order they were computed:Position 1: 0 (rightmost)Position 2: 1Position 3: 1Position 4: 0Position 5: 1Position 6: 1Position 7: 1Position 8: 1Position 9: 0Position 10: 0Final carry: 1 (leftmost)So when writing the result, it's the final carry followed by the sum bits from Position 10 to Position 1. Wait, but that would be:1 (carry) followed by Position 10 sum bit (0), then Position 9 (0), ..., up to Position 1 (0). Wait, that can't be. Because Position 1 is the rightmost, so in the final number, Position 1's sum bit should be the rightmost, and the carry is the leftmost.Therefore, the correct order is:Final carry (1) followed by Position 10 sum bit (0), Position 9 sum bit (0), ..., Position 1 sum bit (0). Wait, no, that would place the carry as the leftmost bit, and then the sum bits from Position 10 (which was the leftmost original bit) to Position 1 (rightmost original bit). But that's not correct because when we add, the sum bits are computed from right to left, so the first sum bit (Position 1) is the rightmost in the result.Therefore, the result should be:[Final carry][Position 10 sum bit][Position 9 sum bit]...[Position 1 sum bit]But Position 10 sum bit is 0, Position 9 is 0, ..., down to Position 1 is 0. But that seems incorrect because the individual sum bits were 0,1,1,0,1,1,1,1,0,0 (from Position 1 to 10). Wait, no, hold on. Wait, the sum bits are as follows:From Position 1 to 10:Position 1: 0Position 2: 1Position 3: 1Position 4: 0Position 5: 1Position 6: 1Position 7: 1Position 8: 1Position 9: 0Position 10: 0So written in order from Position 1 (LSB) to Position 10 (MSB of the original numbers), the sum bits are:0 1 1 0 1 1 1 1 0 0Then the final carry is 1. So the total result is the final carry followed by the sum bits from Position 10 to Position 1. Wait, but that would be:1 (carry) followed by 0 (Position 10), 0 (Position 9), 1 (Position 8), 1 (Position 7), 1 (Position 6), 1 (Position 5), 0 (Position 4), 1 (Position 3), 1 (Position 2), 0 (Position 1). But that seems reversed. Wait, perhaps the correct way is that after computing all the sum bits from right to left (Position 1 to 10), you have the sum bits as [0,1,1,0,1,1,1,1,0,0], with Position 1 being the rightmost. Then, the carry-out is the next bit to the left. So the full result is:Carry-out (1) followed by the sum bits starting from Position 10 to Position 1. But Position 10 was the leftmost bit of the original numbers, so in the result, it's the 10th bit from the right. But when we add the carry-out, it becomes the 11th bit from the right (i.e., the new leftmost bit). Therefore, the final result is:1 (carry-out) 0 (Position 10 sum bit) 0 (Position 9 sum bit) 1 (Position 8 sum bit) 1 (Position 7 sum bit) 1 (Position 6 sum bit) 1 (Position 5 sum bit) 0 (Position 4 sum bit) 1 (Position 3 sum bit) 1 (Position 2 sum bit) 0 (Position 1 sum bit)Wait, but if we write this out, it would be:1 0 0 1 1 1 1 0 1 1 0But let's check this. Let me write the bits in order from MSB to LSB:Bit 11 (carry): 1Bit 10: 0Bit 9: 0Bit 8: 1Bit 7: 1Bit 6: 1Bit 5: 1Bit 4: 0Bit 3: 1Bit 2: 1Bit 1: 0So the binary number is 1 0 0 1 1 1 1 0 1 1 0.Wait, let me convert this to decimal to check if the addition makes sense.First, let's compute the decimal values of the original numbers to verify.First number: (101111011)₂. Let's convert this to decimal.Breaking it down from right to left (positions 1 to 9):1: 2^0 = 11: 2^1 = 20: 01: 81: 161: 321: 640: 01: 256Adding them up: 256 + 64 + 32 + 16 + 8 + 2 + 1 = 256 + 64=320; 320+32=352; 352+16=368; 368+8=376; 376+2=378; 378+1=379. So the first number is 379.Second number: (1100111011)₂. Let's convert this.From right to left (positions 1 to 10):1:11:20:01:80:01:321:641:1280:01:512Adding them up: 512 + 128 + 64 + 32 + 8 + 2 +1. Let's compute:512 + 128 = 640; 640 +64=704; 704 +32=736; 736 +8=744; 744 +2=746; 746 +1=747. So the second number is 747.Adding 379 + 747: 379 + 700 = 1079; 1079 +47=1126. So the decimal result should be 1126.Now let's convert the binary result we got: (10011110110)₂.Let's convert this to decimal. Breaking it down from right to left:Position 1: 0Position 2: 1*2^1=2Position 3:1*2^2=4Position 4:0*2^3=0Position 5:1*2^4=16Position 6:1*2^5=32Position 7:1*2^6=64Position 8:1*2^7=128Position 9:0*2^8=0Position 10:0*2^9=0Position 11:1*2^10=1024Adding them up: 1024 + 128 + 64 +32 +16 +4 +2 = Let's compute:1024 +128=1152; 1152 +64=1216; 1216 +32=1248; 1248 +16=1264; 1264 +4=1268; 1268 +2=1270.Wait, that's 1270, but we expected 1126. There's a discrepancy here. That means I must have made a mistake in the binary addition. Hmm.So something's wrong. Let's check where the mistake is.First, the decimal conversion of the original numbers: first was 379, second was 747, total 1126. The binary addition gave me 10011110110, which is 1270 in decimal. So clearly, I messed up somewhere.Let me re-examine the addition step by step.Let me write down the two numbers again, properly aligned with the padded zero: 0 1 0 1 1 1 1 0 1 1 + 1 1 0 0 1 1 1 0 1 1 ---------------------------Let's perform the addition column by column, starting from the right (position 1 to 10):Let me create a table for each position:Position | A | B | Carry-in | Sum | Carry-out1 | 1 | 1 | 0 | 0 | 12 | 1 | 1 | 1 | 1 | 13 | 0 | 0 | 1 | 1 | 04 | 1 | 1 | 0 | 0 | 15 | 1 | 1 | 1 | 1 | 16 | 1 | 1 | 1 | 1 | 17 | 1 | 1 | 1 | 1 | 18 | 0 | 0 | 1 | 1 | 09 | 1 | 1 | 0 | 0 | 110 | 0 | 1 | 1 | 0 | 1Final carry-out: 1So the Sum bits from Position 1 to 10 are:Pos1:0, Pos2:1, Pos3:1, Pos4:0, Pos5:1, Pos6:1, Pos7:1, Pos8:1, Pos9:0, Pos10:0And the final carry is 1.So writing the result as 1 (carry) followed by the sum bits from Pos10 to Pos1:1 0 0 1 1 1 1 1 0 1 0Wait, hold on. Wait, if we take the sum bits from Position 1 to 10: 0,1,1,0,1,1,1,1,0,0But when we write the entire result, starting from the final carry, followed by the sum bits from leftmost (Position 10) to rightmost (Position 1). Wait, no, actually, Position 10 is the leftmost bit of the original numbers. But in the sum bits, Position 10 corresponds to the 10th column from the right. So the sum bits, when written as a number, should have Position 1 as the rightmost bit and Position 10 as the 10th bit from the right. Then, the carry-out becomes the 11th bit from the right (the new leftmost bit). So the final result is:Carry-out (1) followed by sum bits from Position 10 to Position 1. But the sum bits from Position 10 to Position 1 are: 0 (Position10), 0 (Position9), 0 (Position8? Wait, no.Wait, confusion arises from how the positions are labeled. Let me think differently.When adding two numbers:A: 0 1 0 1 1 1 1 0 1 1B: 1 1 0 0 1 1 1 0 1 1Let's write them vertically: 0 1 0 1 1 1 1 0 1 1 + 1 1 0 0 1 1 1 0 1 1 ---------------------------Starting from the rightmost bit:Column 1: 1 + 1 = 0, carry 1Column 2: 1 + 1 + 1 = 1, carry 1Column 3: 0 + 0 + 1 = 1, carry 0Column 4: 1 + 1 + 0 = 0, carry 1Column 5: 1 + 1 + 1 = 1, carry 1Column 6: 1 + 1 + 1 = 1, carry 1Column 7: 1 + 1 + 1 = 1, carry 1Column 8: 0 + 0 + 1 = 1, carry 0Column 9: 1 + 1 + 0 = 0, carry 1Column 10: 0 + 1 + 1 = 0, carry 1Final carry: 1So writing the sum bits from right to left (columns 1 to 10):0,1,1,0,1,1,1,1,0,0Then the final carry is 1. So the total result is 1 followed by 0,0,1,1,1,1,0,1,1,0. Wait, no. Wait, the sum bits are in order from right to left as 0,1,1,0,1,1,1,1,0,0. So when written as a number, the rightmost bit is 0, then next is 1, up to the leftmost sum bit which is 0 (from column 10). Then the final carry is added to the left, making it the new leftmost bit.Therefore, the final result is:1 (carry) 0 (column10 sum) 0 (column9) 1 (column8) 1 (column7) 1 (column6) 1 (column5) 0 (column4) 1 (column3) 1 (column2) 0 (column1)So written out: 1 0 0 1 1 1 1 0 1 1 0Which is 10011110110. But converting this to decimal:Breaking it down:From right to left (positions 1 to 11):0: 01: 21: 40: 01: 161: 321: 641: 1280: 00: 01: 1024Adding them up: 1024 + 128 + 64 +32 +16 +4 +2 = 1024+128=1152; 1152+64=1216; 1216+32=1248; 1248+16=1264; 1264+4=1268; 1268+2=1270. Hmm, same as before. So this is 1270 in decimal, but expected was 1126. Therefore, there is a mistake in the binary addition.Wait, but the decimal addition is 379 + 747 = 1126. So where did I go wrong in the binary addition?Let me recheck the step-by-step addition:Original numbers:A (padded): 0 1 0 1 1 1 1 0 1 1B: 1 1 0 0 1 1 1 0 1 1Let me write them down vertically with the carries:Let me attempt to perform the addition step by step, writing down the carries as I go:Starting from the rightmost bit (position 1):C₀ = 0 (initial carry)Position 1:A₁ = 1, B₁ = 1Sum: 1 + 1 + 0 = 2. So sum bit = 0, carry C₁ = 1Position 2:A₂ = 1, B₂ = 1, C₁ = 1Sum: 1 + 1 + 1 = 3. Sum bit = 1, carry C₂ = 1Position 3:A₃ = 0, B₃ = 0, C₂ = 1Sum: 0 + 0 + 1 = 1. Sum bit = 1, carry C₃ = 0Position 4:A₄ = 1, B₄ = 1, C₃ = 0Sum: 1 + 1 + 0 = 2. Sum bit = 0, carry C₄ = 1Position 5:A₅ = 1, B₅ = 1, C₄ = 1Sum: 1 + 1 + 1 = 3. Sum bit = 1, carry C₅ = 1Position 6:A₆ = 1, B₆ = 1, C₅ = 1Sum: 1 + 1 + 1 = 3. Sum bit = 1, carry C₆ = 1Position 7:A₇ = 1, B₇ = 1, C₆ = 1Sum: 1 + 1 + 1 = 3. Sum bit = 1, carry C₇ = 1Position 8:A₈ = 0, B₈ = 0, C₇ = 1Sum: 0 + 0 + 1 = 1. Sum bit = 1, carry C₈ = 0Position 9:A₉ = 1, B₉ = 1, C₈ = 0Sum: 1 + 1 + 0 = 2. Sum bit = 0, carry C₉ = 1Position 10:A₁₀ = 0, B₁₀ = 1, C₉ = 1Sum: 0 + 1 + 1 = 2. Sum bit = 0, carry C₁₀ = 1Final carry: C₁₀ = 1So the sum bits from position 1 to 10 are: 0,1,1,0,1,1,1,1,0,0With final carry 1. So the total result is 1 followed by 0,0,1,1,1,1,0,1,1,0, which is 10011110110. But this is 1270, which is not matching. Hmm.Wait, maybe there was a mistake in padding the original number. Let me check the original numbers again.Original numbers:First number: (101111011)₂. Let's count the bits: 1 0 1 1 1 1 0 1 1. That's 9 bits. So when padding to 10 bits, we add a leading zero: 0101111011.Second number: (1100111011)₂. That's 10 bits: 1 1 0 0 1 1 1 0 1 1. Correct.So the alignment is correct. Then why is the result different? Maybe my conversion is wrong.Wait, let me reconvert the original numbers.First number: 101111011.From right to left:Positions 1-9:1:11:20:01:81:161:321:640:01:256Total: 256 + 64 +32 +16 +8 +2 +1 = 256+64=320, 320+32=352, 352+16=368, 368+8=376, 376+2=378, 378+1=379. Correct.Second number:1100111011.From right to left:Positions 1-10:1:11:20:01:80:01:321:641:1280:01:512Total: 512 + 128 +64 +32 +8 +2 +1 = 512+128=640, 640+64=704, 704+32=736, 736+8=744, 744+2=746, 746+1=747. Correct.379 +747=1126. Correct. So where is the mistake in binary addition?Wait, perhaps I messed up one of the positions. Let's check each position again.Let me write both numbers vertically with the padded zero: 0 1 0 1 1 1 1 0 1 1 + 1 1 0 0 1 1 1 0 1 1 ---------------------------Starting from the right (position 1):Column 1: 1 +1 =0, carry 1.Column 2:1+1+1=1, carry1.Column3:0+0+1=1, carry0.Column4:1+1+0=0, carry1.Column5:1+1+1=1, carry1.Column6:1+1+1=1, carry1.Column7:1+1+1=1, carry1.Column8:0+0+1=1, carry0.Column9:1+1+0=0, carry1.Column10:0+1+1=0, carry1.Final carry1.So writing the sum bits:Positions1-10:0,1,1,0,1,1,1,1,0,0.Adding the final carry:1.So the binary result is1 0 0 1 1 1 1 0 1 1 0.Which is10011110110.Convert this to decimal:From right to left:Position1:0*1=0Position2:1*2=2Position3:1*4=4Position4:0*8=0Position5:1*16=16Position6:1*32=32Position7:1*64=64Position8:1*128=128Position9:0*256=0Position10:0*512=0Position11:1*1024=1024Total:1024+128+64+32+16+4+2=1024+128=1152+64=1216+32=1248+16=1264+4=1268+2=1270.Hmm. So clearly, there's a mistake here. The binary addition gives 1270, but the correct decimal sum is1126. So there must be an error in the addition process.Let me check each column again.Original numbers:A:0 1 0 1 1 1 1 0 1 1B:1 1 0 0 1 1 1 0 1 1Let me label each bit from right (position1) to left (position10):For A:Position1:1Position2:1Position3:0Position4:1Position5:1Position6:1Position7:1Position8:0Position9:1Position10:0For B:Position1:1Position2:1Position3:0Position4:1Position5:1Position6:1Position7:1Position8:0Position9:1Position10:1Wait, hold on! Wait, in the original problem, the second number is (1100111011)₂. Let me confirm its bits:1 1 0 0 1 1 1 0 1 1From left to right, that's position10 to position1:Position10:1Position9:1Position8:0Position7:0Position6:1Position5:1Position4:1Position3:0Position2:1Position1:1Wait, no, this is where the confusion might be. When we write a binary number, the leftmost digit is the highest power (MSB), so in the second number, 1100111011, the bits are:From left to right (MSB to LSB):1 (2^9), 1 (2^8), 0 (2^7), 0 (2^6), 1 (2^5), 1 (2^4), 1 (2^3), 0 (2^2), 1 (2^1), 1 (2^0)So when aligning with the first number (padded to 10 bits):First number (padded): 0 1 0 1 1 1 1 0 1 1Which is:Position10:0 (2^9)Position9:1 (2^8)Position8:0 (2^7)Position7:1 (2^6)Position6:1 (2^5)Position5:1 (2^4)Position4:1 (2^3)Position3:0 (2^2)Position2:1 (2^1)Position1:1 (2^0)Second number:1 1 0 0 1 1 1 0 1 1Position10:1 (2^9)Position9:1 (2^8)Position8:0 (2^7)Position7:0 (2^6)Position6:1 (2^5)Position5:1 (2^4)Position4:1 (2^3)Position3:0 (2^2)Position2:1 (2^1)Position1:1 (2^0)Wait, in this case, when adding vertically, the positions are from 1 (LSB) to 10 (MSB). But when writing the numbers, the leftmost digit is position10 (for 10-bit numbers). Therefore, when we add vertically, we have:A: positions1-10 (right to left):Position1:1, Position2:1, Position3:0, Position4:1, Position5:1, Position6:1, Position7:1, Position8:0, Position9:1, Position10:0B: positions1-10:Position1:1, Position2:1, Position3:0, Position4:1, Position5:1, Position6:1, Position7:0, Position8:0, Position9:1, Position10:1Wait, earlier, I think I mixed up the positions for the second number. Specifically, Position7 for the second number is 0, not 1. Let me check:Original second number:1100111011Breaking it down from left to right (MSB to LSB):1 (2^9), 1 (2^8), 0 (2^7), 0 (2^6), 1 (2^5), 1 (2^4), 1 (2^3), 0 (2^2), 1 (2^1), 1 (2^0)So positions10 to1:Position10:1Position9:1Position8:0Position7:0Position6:1Position5:1Position4:1Position3:0Position2:1Position1:1Therefore, when adding:For Position7 (2^6), A has 1 and B has 0. Previously, I thought both had 1 in Position7, but that's incorrect.Ah! Here's the mistake. Earlier, when I set up the addition, I incorrectly assigned the bits of the second number. Specifically, in position7 (which is 2^6), the second number has a 0, not a 1. But in my previous addition steps, I assumed that in position7, both numbers had 1. This is where the error crept in.Let me correct the addition table.Revised table with correct bits for Position7:Position | A | B | Carry-in | Sum | Carry-out1 | 1 | 1 | 0 | 0 | 12 | 1 | 1 | 1 | 1 | 13 | 0 | 0 | 1 | 1 | 04 | 1 | 1 | 0 | 0 | 15 | 1 | 1 | 1 | 1 | 16 | 1 | 1 | 1 | 1 | 17 | 1 | 0 | 1 | 0 | 1 (Wait, sum is 1+0+1=2, which is sum 0, carry 1)Wait, Position7:A is 1, B is 0, carry-in is 1 (from Position6 carry-out).So sum: 1 + 0 + 1 = 2. Therefore, sum bit 0, carry-out 1.Previously, I had B as 1 in Position7, leading to sum 1+1+1=3, which was incorrect.This is the critical error. Let's correct the addition steps:Starting from Position1 to Position10 with corrected B bits:Position1:A=1, B=1, Cin=0Sum=0, Cout=1Position2:A=1, B=1, Cin=1Sum=1, Cout=1Position3:A=0, B=0, Cin=1Sum=1, Cout=0Position4:A=1, B=1, Cin=0Sum=0, Cout=1Position5:A=1, B=1, Cin=1Sum=1, Cout=1Position6:A=1, B=1, Cin=1Sum=1, Cout=1Position7:A=1, B=0, Cin=1Sum=0, Cout=1Position8:A=0, B=0, Cin=1Sum=1, Cout=0Position9:A=1, B=1, Cin=0Sum=0, Cout=1Position10:A=0, B=1, Cin=1Sum=0, Cout=1Final carry:1Now, the sum bits from Position1 to Position10 are:Position1:0Position2:1Position3:1Position4:0Position5:1Position6:1Position7:0Position8:1Position9:0Position10:0And final carry:1So writing the result as:Carry-out:1 followed by sum bits from Position10 to Position1:0,0,1,0,1,1,0,1,1,0Wait, no. Let's clarify:The sum bits from right to left (Position1 to Position10) are:0,1,1,0,1,1,0,1,0,0Then, the final carry is 1, making the total result:1 followed by the sum bits from Position10 to Position1:0,0,1,0,1,1,0,1,1,0Wait, but Position10 is the leftmost sum bit (from the original numbers), which is 0, and Position1 is the rightmost sum bit, 0.Wait, the confusion arises from the order. Let me structure it properly.After adding, the sum bits are computed from right to left (Position1 to Position10). So the sum bits are:Position1:0 (rightmost)Position2:1Position3:1Position4:0Position5:1Position6:1Position7:0Position8:1Position9:0Position10:0 (leftmost sum bit)Then, the final carry is 1, which is the new leftmost bit.Therefore, the final binary number is:1 (carry) 0 (Position10) 0 (Position9) 1 (Position8) 0 (Position7) 1 (Position6) 1 (Position5) 0 (Position4) 1 (Position3) 1 (Position2) 0 (Position1)Wait, that's:1 0 0 1 0 1 1 0 1 1 0But let's list them in order from MSB to LSB:Bit11:1Bit10:0Bit9:0Bit8:1Bit7:0Bit6:1Bit5:1Bit4:0Bit3:1Bit2:1Bit1:0So the binary result is 1 0 0 1 0 1 1 0 1 1 0, which is 10010110110.Let's convert this to decimal:Breaking down from right to left (positions1-11):0: 01: 21:40:01:161:320:01:1280:00:01:1024Sum:1024 +128 +32 +16 +4 +2=1024+128=1152+32=1184+16=1200+4=1204+2=1206.Wait, 1206 is still not equal to 1126. Hmm, still discrepancy. There must be another mistake.Wait, let's recheck the corrected addition steps.After correcting Position7's B bit to 0:Position7: A=1, B=0, Cin=1 (from Position6's Cout=1)Sum:1 +0 +1=2. Sum bit=0, Cout=1.Position8: A=0, B=0, Cin=1Sum:0 +0 +1=1, Sum bit=1, Cout=0.Position9: A=1, B=1, Cin=0Sum:1+1+0=2. Sum bit=0, Cout=1.Position10: A=0, B=1, Cin=1Sum:0 +1 +1=2. Sum bit=0, Cout=1.Final carry:1.So sum bits:Position1:0Position2:1Position3:1Position4:0Position5:1Position6:1Position7:0Position8:1Position9:0Position10:0Carry:1Thus, the sum is 1 0 0 1 0 1 1 0 1 1 0 (10010110110).Convert this to decimal:From right to left:Position1:0Position2:1*2=2Position3:1*4=4Position4:0*8=0Position5:1*16=16Position6:1*32=32Position7:0*64=0Position8:1*128=128Position9:0*256=0Position10:0*512=0Position11:1*1024=1024Adding up:1024+128+32+16+4+2=1024+128=1152+32=1184+16=1200+4=1204+2=1206.Still not 1126. So where is the error now?Wait, maybe the carry-in for Position9 is incorrect.Looking back at Position8:Position8: A=0, B=0, Cin=1 (carry from Position7, which had Cout=1)Sum=0+0+1=1, Cout=0.Then Position9: A=1, B=1, Cin=0 (from Position8's Cout=0)Sum=1+1+0=2. Sum bit=0, Cout=1.Position10: A=0, B=1, Cin=1 (from Position9's Cout=1)Sum=0+1+1=2. Sum bit=0, Cout=1.Final carry=1.So Position9 sum bit=0, carry=1.Position10 sum bit=0, carry=1.So that part seems correct.Wait, perhaps another bit was misassigned. Let's re-examine the original numbers.Original A (padded):0 1 0 1 1 1 1 0 1 1From left to right (positions10 to1):Position10:0Position9:1Position8:0Position7:1Position6:1Position5:1Position4:1Position3:0Position2:1Position1:1Original B:1 1 0 0 1 1 1 0 1 1From left to right (positions10 to1):Position10:1Position9:1Position8:0Position7:0Position6:1Position5:1Position4:1Position3:0Position2:1Position1:1So when adding Position7:A's bit:1 (from Position7, which is 2^6)B's bit:0 (from Position7, which is 2^6)Correct.Position8:A's bit:0 (Position8:2^7)B's bit:0 (Position8:2^7)Correct.Position9:A's bit:1 (Position9:2^8)B's bit:1 (Position9:2^8)Position10:A's bit:0 (Position10:2^9)B's bit:1 (Position10:2^9)So the corrected addition steps seem accurate.Then why does the binary result translate to 1206 instead of 1126? There must be an error in the sum bits.Wait, let's recompute the sum bits with the corrected steps:Sum bits from Position1 to10:1. 02. 13. 14. 05. 16. 17. 08. 19. 010. 0With carry1.Writing these bits from LSB to MSB:Positions1-10:0,1,1,0,1,1,0,1,0,0Plus carry1.So the full binary number is1 0 0 1 0 1 1 0 1 1 0.But converting this:Break it into sections:From left to right (MSB to LSB):1 0 0 1 0 1 1 0 1 1 0Which is:1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1Bits:1:10240:00:01:1280:01:321:160:01:41:20:0Sum:1024 +128 +32 +16 +4 +2 = 1024+128=1152+32=1184+16=1200+4=1204+2=1206.Wait, the issue is that the binary result is 10010110110, which is 1206, but the correct sum is 1126. Therefore, there's still a mistake.Wait, maybe I messed up another position. Let's check Position5:Position5: A=1, B=1, Cin=1.Sum=1+1+1=3, which is sum1, carry1. Correct.Position6: A=1, B=1, Cin=1. Sum=3, sum1, carry1. Correct.Position7: A=1, B=0, Cin=1. Sum=0, carry1. Correct.Position8: A=0, B=0, Cin=1. Sum=1, carry0. Correct.Position9: A=1, B=1, Cin=0. Sum=0, carry1. Correct.Position10: A=0, B=1, Cin=1. Sum=0, carry1. Correct.Hmm. All steps seem correct, but the result is wrong. Wait, maybe the initial decimal conversions are wrong?Wait, double-checking decimal conversions.First number:101111011.Which is 1*2^8 + 0*2^7 +1*2^6 +1*2^5 +1*2^4 +1*2^3 +0*2^2 +1*2^1 +1*2^0.=256 +0 +64 +32 +16 +8 +0 +2 +1=256+64=320+32=352+16=368+8=376+2=378+1=379. Correct.Second number:1100111011.Break it down:1*2^9 +1*2^8 +0*2^7 +0*2^6 +1*2^5 +1*2^4 +1*2^3 +0*2^2 +1*2^1 +1*2^0.=512 +256 +0 +0 +32 +16 +8 +0 +2 +1=512+256=768+32=800+16=816+8=824+2=826+1=827.Wait, hold on! This is different from before. Earlier, I thought the second number was 747, but actually it's 827.Wait, original problem states the second number is (1100111011)₂. Converting this:From right to left:Position1:1*1=1Position2:1*2=2Position3:0*4=0Position4:1*8=8Position5:1*16=16Position6:1*32=32Position7:0*64=0Position8:0*128=0Position9:1*256=256Position10:1*512=512Total:512 +256 +32 +16 +8 +2 +1=512+256=768+32=800+16=816+8=824+2=826+1=827.So the second number is 827, not 747. There was a mistake in the initial decimal conversion.Oh my goodness! That was the critical error. Initially, I thought the second number was 747, but it's actually 827.So 379 +827=1206, which matches the binary result of10010110110 (1206). Therefore, the binary addition is correct, and my initial decimal conversion was wrong.So where did the 747 come from? Let me see.Original second number:1100111011.Breakdown:If I convert it as 10 bits:But perhaps I miscounted the bits.Wait, 1100111011 has 10 bits:1st bit:1 (MSB, 2^9)2nd:13rd:04th:05th:16th:17th:18th:09th:110th:1 (LSB, 2^0)So the breakdown:2^9:5122^8:2562^7:02^6:02^5:322^4:162^3:82^2:02^1:22^0:1Sum:512+256=768+32=800+16=816+8=824+2=826+1=827.Yes, correct. So my initial conversion was wrong, thinking it was 747, but it's actually 827. Then 379 +827=1206, which matches the binary result. So no error in the binary addition. The mistake was in the initial conversion of the second number.Therefore, the correct sum is10010110110₂, which is 1206 in decimal.Therefore, the correct final answer is boxed{10010110110}_2.Final AnswerThe sum of the two binary numbers is boxed{10010110110}_2.</think>To add the binary numbers ((101111011)_2) and ((1100111011)_2), we first align them properly by padding the shorter number with a leading zero:[begin{array}{r} texttt{0101111011} + texttt{1100111011} hlineend{array}]Starting from the rightmost bit (position 1) and proceeding to the left, we perform the addition column by column, keeping track of the carry:1. Position 1: (1 + 1 = 10) (sum bit 0, carry 1)2. Position 2: (1 + 1 + 1 = 11) (sum bit 1, carry 1)3. Position 3: (0 + 0 + 1 = 1) (sum bit 1, carry 0)4. Position 4: (1 + 1 + 0 = 10) (sum bit 0, carry 1)5. Position 5: (1 + 1 + 1 = 11) (sum bit 1, carry 1)6. Position 6: (1 + 1 + 1 = 11) (sum bit 1, carry 1)7. Position 7: (1 + 0 + 1 = 10) (sum bit 0, carry 1)8. Position 8: (0 + 0 + 1 = 1) (sum bit 1, carry 0)9. Position 9: (1 + 1 + 0 = 10) (sum bit 0, carry 1)10. Position 10: (0 + 1 + 1 = 10) (sum bit 0, carry 1)After processing all columns, we have a final carry of 1. The sum bits from right to left (positions 1 to 10) are (0, 1, 1, 0, 1, 1, 0, 1, 0, 0). Adding the final carry, the result is:[boxed{10010110110}_2]]

🔑:1. (10010110110)_{2}

🧠:[<think>Okay, so I have this differential equation to solve: y''' - 7y'' + 15y' - 9y = 0. Hmm, it's a third-order linear homogeneous differential equation with constant coefficients. I remember that for equations like this, the standard approach is to find the characteristic equation and solve for its roots. Once we have the roots, we can write the general solution based on whether the roots are real and distinct, repeated, or complex.Alright, let's start by writing down the characteristic equation. The original DE is y''' -7y'' +15y' -9y =0. Replacing y''' with r^3, y'' with r^2, y' with r, and y with 1, the characteristic equation becomes:r^3 -7r^2 +15r -9 = 0.Now, I need to solve this cubic equation. Cubic equations can sometimes be tricky, but maybe there are rational roots. The Rational Root Theorem says that any possible rational root p/q is such that p divides the constant term (-9) and q divides the leading coefficient (1). So possible rational roots are ±1, ±3, ±9.Let me test r=1 first. Plugging into the equation:1 -7 +15 -9 = (1 -7) + (15 -9) = (-6) + (6) = 0. Oh, so r=1 is a root. Great!Since r=1 is a root, we can factor out (r -1) from the cubic polynomial. Let's perform polynomial division or use synthetic division.Using synthetic division with root 1:Coefficients of the polynomial: 1 | -7 | 15 | -9Bring down the 1.Multiply by 1: 1*1=1. Add to next coefficient: -7 +1= -6.Multiply by 1: -6*1= -6. Add to next coefficient: 15 + (-6)=9.Multiply by 1: 9*1=9. Add to last coefficient: -9 +9=0. Perfect, no remainder.So the cubic factors as (r -1)(r^2 -6r +9). Now, let's factor the quadratic: r^2 -6r +9.Looking at that quadratic, discriminant is (-6)^2 -4*1*9 =36 -36=0. So discriminant is zero, which means a repeated root. The root is [6 ±0]/2 =3. So the quadratic factors as (r -3)^2.Therefore, the characteristic equation factors as (r -1)(r -3)^2=0. So the roots are r=1, r=3 (with multiplicity 2).Now, for the general solution of the differential equation. When we have distinct real roots, each contributes a term of the form e^(rt). For repeated roots, say root r with multiplicity m, we have terms e^(rt), t e^(rt), ..., t^(m-1) e^(rt).So here, we have roots 1, 3, and 3 (since 3 is a double root). Therefore, the general solution is:y(t) = C1 e^(1*t) + C2 e^(3*t) + C3 t e^(3*t).Simplifying, that's:y(t) = C1 e^t + (C2 + C3 t) e^(3t).I should check if this makes sense. For a third-order DE, we should have three linearly independent solutions, which we do here: e^t, e^(3t), and t e^(3t). The Wronskian should be non-zero, but since they are solutions corresponding to distinct roots or repeated roots, they are linearly independent. So this should be the correct general solution.Let me just verify by plugging in a solution into the DE. Let's take y = e^(3t). Then y' = 3 e^(3t), y'' =9 e^(3t), y'''=27 e^(3t). Plug into DE:27 e^(3t) -7*9 e^(3t) +15*3 e^(3t) -9 e^(3t) = (27 -63 +45 -9) e^(3t) = 0. Yep, that works.Similarly, take y = t e^(3t). Compute y', y'', y''':y = t e^(3t)y' = e^(3t) + 3t e^(3t)y'' = 3 e^(3t) + 3 e^(3t) + 9t e^(3t) = 6 e^(3t) + 9t e^(3t)y''' = 18 e^(3t) + 27t e^(3t)Plug into DE:y''' -7y'' +15y' -9y = [18 e^(3t) +27t e^(3t)] -7[6 e^(3t) +9t e^(3t)] +15[e^(3t) +3t e^(3t)] -9[t e^(3t)]Calculate term by term:18 e^(3t) +27t e^(3t)-42 e^(3t) -63t e^(3t)+15 e^(3t) +45t e^(3t)-9t e^(3t)Combine like terms:For e^(3t): 18 -42 +15 = (18 +15) -42 = 33 -42 = -9For t e^(3t): 27t -63t +45t -9t = (27 -63 +45 -9)t = 0tSo total: -9 e^(3t) +0 = -9 e^(3t). Wait, that's not zero. Hmm, that's a problem. Did I compute something wrong?Wait, y''' -7y'' +15y' -9y should equal zero. But according to my calculation, plugging in y = t e^(3t) gives -9 e^(3t). That's a mistake. Which means I must have miscalculated somewhere.Let me check again:Compute y''':y = t e^(3t)y' = e^(3t) + 3 t e^(3t)y'' = 3 e^(3t) + 3 e^(3t) + 9 t e^(3t) = 6 e^(3t) + 9 t e^(3t)y''' = 6*3 e^(3t) + 9 e^(3t) + 9*3 t e^(3t) = 18 e^(3t) +9 e^(3t) +27 t e^(3t) = 27 e^(3t) +27 t e^(3t)Ah, there's a mistake here. Previously, I thought y''' was 18 e^(3t) +27t e^(3t), but actually, differentiating y'':y'' = 6 e^(3t) + 9t e^(3t)First term derivative: 6*3 e^(3t) =18 e^(3t)Second term derivative: 9*e^(3t) +9t*3 e^(3t) =9 e^(3t) +27 t e^(3t)Therefore, y''' =18 e^(3t) +9 e^(3t) +27 t e^(3t) =27 e^(3t) +27 t e^(3t)So correct y''' is 27 e^(3t) +27 t e^(3t)Now plug into DE:y''' -7y'' +15y' -9y= [27 e^(3t) +27 t e^(3t)] -7[6 e^(3t) +9 t e^(3t)] +15[e^(3t) +3 t e^(3t)] -9[t e^(3t)]Compute each term:First term: 27 e^(3t) +27 t e^(3t)Second term: -7*(6 e^(3t) +9 t e^(3t)) = -42 e^(3t) -63 t e^(3t)Third term: 15*(e^(3t) +3 t e^(3t)) =15 e^(3t) +45 t e^(3t)Fourth term: -9 t e^(3t)Now combine all terms:For e^(3t):27 -42 +15 = (27 +15) -42 =42 -42=0For t e^(3t):27 -63 +45 -9 = (27 -63) + (45 -9)= (-36) +36=0So total: 0 e^(3t) +0 t e^(3t)=0. Perfect, now it works. So that was a calculation mistake earlier. So y = t e^(3t) is indeed a solution.Similarly, check y = e^t:y = e^ty' = e^ty'' = e^ty''' = e^tPlug into DE:e^t -7 e^t +15 e^t -9 e^t = (1 -7 +15 -9) e^t =0. So that works.Thus, all three solutions are valid. Therefore, the general solution is as above.I think that's it. So the answer should be y(t) = C1 e^t + (C2 + C3 t) e^(3t).Final AnswerThe general solution is boxed{y = C_1 e^{t} + left(C_2 + C_3 tright) e^{3t}}.</think>To solve the differential equation ( y''' - 7y'' + 15y' - 9y = 0 ), we start by finding the characteristic equation:[r^3 - 7r^2 + 15r - 9 = 0]Testing for rational roots, we find that ( r = 1 ) is a root. Using synthetic division, we factor the cubic polynomial:[(r - 1)(r^2 - 6r + 9) = 0]The quadratic ( r^2 - 6r + 9 ) factors further as ( (r - 3)^2 ), giving the roots ( r = 1 ) and ( r = 3 ) (with multiplicity 2).For the root ( r = 1 ), the corresponding solution is ( e^t ). For the repeated root ( r = 3 ), the corresponding solutions are ( e^{3t} ) and ( t e^{3t} ).Combining these solutions, the general solution to the differential equation is:[y(t) = C_1 e^t + (C_2 + C_3 t) e^{3t}]Thus, the final answer is:[boxed{y = C_1 e^{t} + left(C_2 + C_3 tright) e^{3t}}],<think>Okay, so I need to solve this third-order linear differential equation: y''' - 7y'' + 15y' -9y = 0. Hmm, let me think. I remember that for linear homogeneous differential equations with constant coefficients, the general approach is to find the characteristic equation and solve for its roots. Then, based on the roots, we write the general solution. First, let me write down the characteristic equation. For a differential equation like this, we replace y''' with r^3, y'' with r^2, y' with r, and y with 1. So the characteristic equation should be r^3 -7r^2 +15r -9 = 0. Now, I need to solve this cubic equation. Cubic equations can be tricky, but maybe I can factor it. Let me try to find rational roots using the Rational Root Theorem. The possible rational roots are the factors of the constant term divided by the factors of the leading coefficient. Here, the constant term is -9 and the leading coefficient is 1, so possible roots are ±1, ±3, ±9.Let me test r=1 first. Plugging into the equation: 1 -7 +15 -9 = (1 -7) + (15 -9) = (-6) + (6) = 0. Oh, so r=1 is a root. That means (r -1) is a factor. Now I can perform polynomial division or use synthetic division to factor out (r -1) from the cubic.Let me do synthetic division with root 1:Coefficients: 1 (r^3), -7 (r^2), 15 (r), -9 (constant)Bring down the 1. Multiply by 1: 1. Add to next coefficient: -7 +1 = -6. Multiply by 1: -6. Add to next coefficient: 15 + (-6) = 9. Multiply by 1: 9. Add to last coefficient: -9 +9=0. Perfect, no remainder.So after factoring out (r -1), the quadratic is r^2 -6r +9. Now, let's factor the quadratic. r^2 -6r +9. Hmm, discriminant is (-6)^2 -4*1*9 = 36 -36 =0. So discriminant is zero, which means there's a repeated root. The root is (6)/2 = 3. So the quadratic factors as (r -3)^2.Therefore, the characteristic equation factors as (r -1)(r -3)^2 =0. So the roots are r=1 and r=3 (with multiplicity 2). For the general solution of the differential equation, each distinct root contributes a term. For a single root r, we have e^(rt). For a repeated root r with multiplicity m, we have e^(rt) multiplied by polynomials of degree m-1. So in this case:- The root r=1 is a simple root, so contributes e^(1*t) = e^t.- The root r=3 has multiplicity 2, so contributes e^(3*t) multiplied by a linear polynomial: (C2 + C3 t) e^(3t).Therefore, the general solution is the sum of these contributions:y(t) = C1 e^t + (C2 + C3 t) e^(3t)Where C1, C2, C3 are constants determined by initial conditions.Wait, let me verify if I did everything correctly. Let me check the factoring again. Original equation: r^3 -7r^2 +15r -9. Factored as (r -1)(r^2 -6r +9). Then (r^2 -6r +9) is (r -3)^2. So roots at 1 and 3 (double). That seems correct. To check if these are correct, maybe plug in r=3 into the original equation. Wait, no, r=3 is a root of the characteristic equation. Let me check the quadratic part. If I plug r=3 into r^2 -6r +9: 9 -18 +9=0, correct. Then r=3 is a root. So the factorization is correct.Therefore, the solution should be as above. Let me recall the standard form for repeated roots. For example, if we have a root r repeated twice, the solutions are e^(rt) and t e^(rt). So here, for r=3, multiplicity 2, we have e^(3t) and t e^(3t). So combining with the simple root r=1, which gives e^t. Hence, the general solution is indeed C1 e^t + C2 e^(3t) + C3 t e^(3t). Alternatively, written as y(t) = C1 e^t + (C2 + C3 t) e^(3t). Either way is correct. So I think that's the answer. Let me check if I made any mistakes in the steps.First, characteristic equation correct: r^3 -7r^2 +15r -9=0. Then factoring with r=1, synthetic division correct. Quadratic factor r^2 -6r +9, which is (r -3)^2. So the roots are 1, 3, 3. Then the solution accordingly. Yes, that seems right. Alternatively, maybe I can plug the solution back into the differential equation to verify. Let me take a simple case where C1=1, C2=0, C3=0. Then y = e^t. Compute y''' -7y'' +15y' -9y. Since y = e^t, all derivatives are e^t. So:y''' = e^t, y'' = e^t, y' = e^t, y = e^t. Plug into the equation: e^t -7e^t +15e^t -9e^t = (1 -7 +15 -9)e^t = 0. Correct.Now take y = e^(3t). Then y' = 3 e^(3t), y'' = 9 e^(3t), y''' = 27 e^(3t). Plug into equation: 27 e^(3t) -7*9 e^(3t) +15*3 e^(3t) -9 e^(3t) = (27 -63 +45 -9) e^(3t) = (27 -63 is -36, then -36 +45 is 9, 9 -9=0). Correct.Then take y = t e^(3t). Compute y', y'', y''':First derivative: y' = e^(3t) + 3t e^(3t) = e^(3t)(1 + 3t)Second derivative: y'' = 3 e^(3t)(1 + 3t) + e^(3t)*3 = e^(3t)(3 +9t +3) = e^(3t)(6 +9t)Third derivative: y''' = 3 e^(3t)(6 +9t) + e^(3t)*9 = e^(3t)(18 +27t +9) = e^(3t)(27 +27t)Now plug into the equation: y''' -7y'' +15y' -9y= e^(3t)(27 +27t) -7*e^(3t)(6 +9t) +15*e^(3t)(1 +3t) -9*t e^(3t)Factor out e^(3t):e^(3t)[ (27 +27t) -7*(6 +9t) +15*(1 +3t) -9t ]Compute the coefficients inside:First term: 27 +27tSecond term: -7*6 = -42; -7*9t = -63tThird term: 15*1 =15; 15*3t=45tFourth term: -9tCombine all terms:27 +27t -42 -63t +15 +45t -9tCombine constants: 27 -42 +15 = (27 +15) -42 = 42 -42=0Combine t terms: 27t -63t +45t -9t = (27 -63 +45 -9)t = (27 -63 is -36; -36 +45 is 9; 9 -9=0) So total coefficient is 0.Therefore, the entire expression is 0, which checks out. So y = t e^(3t) is indeed a solution, as expected.Therefore, the general solution is correct. So I think that's all. The answer should be y(t) = C1 e^t + (C2 + C3 t) e^(3t). Final AnswerThe general solution is boxed{y = C_1 e^{x} + (C_2 + C_3 x) e^{3x}}.</think>To solve the differential equation ( y''' - 7y'' + 15y' - 9y = 0 ), we start by finding the characteristic equation:[r^3 - 7r^2 + 15r - 9 = 0]We test possible rational roots and find that ( r = 1 ) is a root. Using synthetic division, we factor the characteristic equation:[(r - 1)(r^2 - 6r + 9) = 0]The quadratic factor ( r^2 - 6r + 9 ) can be factored further as ( (r - 3)^2 ). Thus, the characteristic equation has roots ( r = 1 ) and ( r = 3 ) (with multiplicity 2).For the root ( r = 1 ), the corresponding solution is ( e^{x} ). For the repeated root ( r = 3 ), the corresponding solutions are ( e^{3x} ) and ( xe^{3x} ).Combining these solutions, the general solution to the differential equation is:[y = C_1 e^{x} + (C_2 + C_3 x) e^{3x}][boxed{y = C_1 e^{x} + (C_2 + C_3 x) e^{3x}}]]

🔑:Solution. Let's form the characteristic equationr^{3}-7 r^{2}+15 r-9=0one of the roots of which r_{1}=1 can be obtained by trial and error.Sinceleft(r^{3}-7 r^{2}+15 r-9right):(r-1)=r^{2}-6 r+9the equation becomes(r-1)left(r^{2}-6 r+9right)=0from which r_{2}=r_{3}=3. Thus, the characteristic equation has one simple (single) root r_{1}=1 and a double root r_{2}=3(k=2).According to formulas (11.12) and (11.8), we obtain the general solutiony=C_{1} e^{x}+C_{2} e^{3 x}+C_{3} x e^{3 x}ory=C_{1} e^{x}+left(C_{2}+C_{3} xright) e^{3 x}

🧠:[<think>Okay, so I have this problem here where t is given as sqrt(2) minus 1, which is approximately 0.4142. The equation is (a t + m)(b t + m) = 17 m, and I need to find the value of ab where a, b, and m are positive integers. Alright, let's try to break this down.First, I know that expanding the left side of the equation might help. Let me try that. So, expanding (a t + m)(b t + m) would give me:ab t² + (a m + b m) t + m². And this is equal to 17 m. So, putting it all together:ab t² + m(a + b) t + m² = 17 m.Hmm, maybe I can rearrange terms to group like terms. Let me subtract 17 m from both sides to set the equation to zero:ab t² + m(a + b) t + m² - 17 m = 0.Alternatively, maybe factor out m from the terms that have it. Let's see:ab t² + m(a + b) t + m(m - 17) = 0.But I'm not sure if that helps yet. Let's think about t being sqrt(2) - 1. Since t is irrational, but a, b, m are integers, perhaps the coefficients of the irrational parts and the rational parts must separately equal zero? Because if you have an equation where a rational number plus an irrational number equals zero, both the rational and irrational coefficients must be zero.So, let's consider that. Let me expand t² first. Since t = sqrt(2) - 1, then t² = (sqrt(2) - 1)² = 2 - 2 sqrt(2) + 1 = 3 - 2 sqrt(2). So t² is 3 - 2 sqrt(2). So substituting back into the equation:ab(3 - 2 sqrt(2)) + m(a + b)(sqrt(2) - 1) + m² = 17 m.Let me distribute the terms:3 ab - 2 ab sqrt(2) + m(a + b) sqrt(2) - m(a + b) + m² = 17 m.Now, let's collect the terms with sqrt(2) and the terms without:Terms with sqrt(2): [-2 ab + m(a + b)] sqrt(2)Terms without sqrt(2): 3 ab - m(a + b) + m²And this equals 17 m. So, since sqrt(2) is irrational, the coefficients of sqrt(2) must be zero, and the constant terms must equal 17 m. Therefore, we can set up two equations:1. Coefficient of sqrt(2): -2 ab + m(a + b) = 02. Constant term: 3 ab - m(a + b) + m² = 17 mOkay, so now we have a system of two equations:Equation 1: -2 ab + m(a + b) = 0Equation 2: 3 ab - m(a + b) + m² = 17 mLet me write Equation 1 as:m(a + b) = 2 abSo, m(a + b) = 2 ab. Let's note that. Then, Equation 2 can be rewritten using this.Equation 2: 3 ab - [m(a + b)] + m² = 17 mBut since m(a + b) = 2 ab, substitute that into Equation 2:3 ab - 2 ab + m² = 17 mSimplify:(3 ab - 2 ab) + m² = 17 mWhich is:ab + m² = 17 mTherefore, ab = 17 m - m²Or, ab = m(17 - m)So now we have two equations:1. m(a + b) = 2 ab2. ab = m(17 - m)We need to find positive integers a, b, m such that these hold.Since a, b, m are positive integers, and m divides ab (from equation 2: ab = m(17 - m)), so m is a divisor of ab. Also, since ab = m(17 - m), m must be a positive integer less than 17 because 17 - m must be positive (since ab is positive). So m can be from 1 to 16.But let's see. Let's express ab from equation 2 as ab = m(17 - m). Then substitute ab into equation 1:m(a + b) = 2 ab = 2 m(17 - m)Divide both sides by m (since m is positive, m ≠ 0):a + b = 2(17 - m) = 34 - 2mTherefore, we have:a + b = 34 - 2mandab = m(17 - m)So now, we can consider a and b as positive integers such that their sum is 34 - 2m and their product is m(17 - m). Since a and b are positive integers, the quadratic equation x² - (34 - 2m)x + m(17 - m) = 0 must have integer roots.The discriminant of this quadratic is:D = (34 - 2m)² - 4 * 1 * m(17 - m)Let me compute D:D = (34 - 2m)² - 4 m(17 - m)First expand (34 - 2m)²:= 34² - 2*34*2m + (2m)²= 1156 - 136m + 4m²Then subtract 4 m(17 - m):= 1156 - 136m + 4m² - 68m + 4m²Wait, wait, 4 m(17 - m) is 68m - 4m². So subtracting that is:1156 - 136m + 4m² - (68m - 4m²) = 1156 - 136m + 4m² - 68m + 4m²Combine like terms:1156 - (136m + 68m) + (4m² + 4m²)= 1156 - 204m + 8m²So discriminant D = 8m² - 204m + 1156We need D to be a perfect square for a and b to be integers. Let's see if we can factor D or simplify it.First, factor out 4:D = 4*(2m² - 51m + 289)Hmm, 2m² -51m +289. Let me check if this quadratic can be factored or if the discriminant becomes a perfect square.Compute discriminant of 2m² -51m +289:Discriminant Δ = (-51)² - 4*2*289 = 2601 - 2312 = 289Ah! 289 is 17². So the discriminant is 17². Therefore, the roots of 2m² -51m +289 are:m = [51 ± 17]/(2*2) = (51 ±17)/4So, m = (51 +17)/4 = 68/4 =17, or m=(51 -17)/4=34/4=8.5But m must be an integer between 1 and 16. 8.5 is not an integer, so perhaps this is not helpful. Wait, but since D =4*(2m² -51m +289) and 2m² -51m +289 has roots at m=17 and m=8.5, so for m between 1 and 16, 2m² -51m +289 is positive or negative?Let me check for m=1: 2 -51 +289=240, positiveFor m=16: 2*256 -51*16 +289=512 -816 +289= (512+289)-816=801-816=-15, negativeSo the quadratic 2m² -51m +289 is positive from m=1 up to some m where it becomes zero. Since one root is at m=8.5 and m=17, between m=8.5 and m=17 it is negative. So for m=1 to 8, it's positive, m=9 to16, negative.But D=4*(quadratic), so for m=1 to8, D is positive, and for m=9 to16, D is negative. But D must be a perfect square, which can't be negative, so m must be between1 and8.Therefore, possible m values are 1 to8.So, now, since we have D=8m² -204m +1156=4*(2m² -51m +289). Since 2m² -51m +289 has discriminant 17², then D=4*(quadratic with roots 17 and8.5). Hmm, perhaps not directly helpful.Alternatively, since D must be a perfect square. Let me denote D=8m² -204m +1156= k² for some integer k. Then, 8m² -204m +1156=k². This seems a bit complicated, but maybe we can look for m from1 to8 and check if D is a perfect square.Alternatively, note that from earlier, we have a + b =34 -2m and ab= m(17 -m). Since a and b are positive integers, 34 -2m must be positive, so 34 -2m >0 => m <17, which is already satisfied as m<=8.But also, m(17 -m) must be positive, which is true for m<17, which is already the case. So let's compute possible m from1 to8 and check if ab= m(17 -m) and a + b=34 -2m lead to integers a and b.Let me tabulate possible m values:m: 1 to8For each m:Compute ab = m(17 -m)Compute a + b =34 -2mThen check if there are integers a,b with that sum and product.Let's start with m=1:ab=1*16=16a + b=34 -2=32So two numbers that add to32 and multiply to16. The quadratic is x² -32x +16=0. The discriminant is 1024 -64=960, which is not a perfect square. So no integer solutions.m=2:ab=2*15=30a + b=34 -4=30So numbers adding to30, product30. The factors of30:1+29, 2+28, 3+27, 5+25, 6+24, etc. 30=5*6, but 5+6=11≠30. So no integer solutions. Quadratic equation x² -30x +30=0, discriminant 900 -120=780, not a square.m=3:ab=3*14=42a + b=34 -6=28Looking for two numbers adding to28, product42. Factors of42:1&42 (sum43), 2&21 (23), 3&14 (17), 6&7 (13). None sum to28. Discriminant 28² -4*42=784 -168=616, not a square.m=4:ab=4*13=52a + b=34 -8=26Find a and b such that a +b=26, ab=52. Factors of52:1&52 (sum53), 2&26 (28), 4&13 (17). None sum to26. Discriminant 26² -4*52=676 -208=468, not a square.m=5:ab=5*12=60a + b=34 -10=24Find a, b such that a + b=24, ab=60. Factors of60:1&60, 2&30, 3&20, 4&15, 5&12, 6&10. Checking sums: 6+10=16, 5+12=17, 4+15=19, 3+20=23, 2+30=32, 1+60=61. None sum to24. Discriminant 24² -4*60=576-240=336, not square.m=6:ab=6*11=66a + b=34 -12=22Looking for a +b=22, ab=66. Factors of66:1&66 (67), 2&33 (35), 3&22 (25), 6&11 (17). None sum to22. Discriminant 22² -4*66=484 -264=220, not square.m=7:ab=7*10=70a + b=34 -14=20Looking for a +b=20, ab=70. Factors of70:1&70, 2&35, 5&14, 7&10. Sums:7+10=17, 5+14=19, 2+35=37, 1+70=71. No sum to20. Discriminant 400 -280=120, not square.m=8:ab=8*9=72a + b=34 -16=18Looking for a +b=18, ab=72. Factors of72:1&72, 2&36, 3&24, 4&18, 6&12, 8&9. Check sums:8+9=17, 6+12=18. Wait, 6 and12: sum18, product72. Yes! So a=6, b=12 or vice versa.Therefore, for m=8, we have a=6, b=12 (or a=12, b=6). So ab=6*12=72.Wait, but according to equation 2, ab= m(17 - m)=8*9=72, which matches. So this works.So the solution is m=8, a=6, b=12. Therefore, ab=72. Hence, the answer should be 72. Let me verify this.Let me check if (a t + m)(b t + m) equals 17 m. Given a=6, b=12, m=8, t=sqrt(2)-1.Compute (6t +8)(12t +8). Let's first compute 6t +8 and 12t +8.6t +8=6(sqrt(2)-1) +8=6 sqrt(2) -6 +8=6 sqrt(2) +212t +8=12(sqrt(2)-1)+8=12 sqrt(2) -12 +8=12 sqrt(2) -4Now multiply them:(6 sqrt(2) +2)(12 sqrt(2) -4)Multiply term by term:6 sqrt(2)*12 sqrt(2) = 72*(2)=1446 sqrt(2)*(-4) = -24 sqrt(2)2*12 sqrt(2)=24 sqrt(2)2*(-4)= -8So adding all together:144 -24 sqrt(2) +24 sqrt(2) -8=144 -8=136Now, 17 m=17*8=136. So yes, 136=136. Therefore, it checks out.Therefore, the correct value of ab is72.Final Answerboxed{72}</think>Given ( t = sqrt{2} - 1 ), we need to find the value of ( ab ) where positive integers ( a ), ( b ), and ( m ) satisfy the equation:[(a t + m)(b t + m) = 17 m]First, we expand the left-hand side of the equation:[(ab t^2 + m(a + b) t + m^2) = 17 m]Substituting ( t = sqrt{2} - 1 ) and ( t^2 = 3 - 2sqrt{2} ), we rewrite the equation:[ab(3 - 2sqrt{2}) + m(a + b)(sqrt{2} - 1) + m^2 = 17 m]Separating the rational and irrational parts, we get two equations:1. Coefficient of ( sqrt{2} ): [ -2ab + m(a + b) = 0 ]2. Constant term: [ 3ab - m(a + b) + m^2 = 17 m ]From the first equation, we solve for ( m(a + b) ):[m(a + b) = 2ab]Substituting ( m(a + b) = 2ab ) into the second equation:[3ab - 2ab + m^2 = 17m]Simplifying, we get:[ab + m^2 = 17m]Thus, ( ab = m(17 - m) ). We also know ( a + b = 34 - 2m ). We need to find positive integers ( a ), ( b ), and ( m ) such that these conditions hold.By testing values of ( m ) from 1 to 8 (since ( m ) must be a positive integer less than 17), we find:- For ( m = 8 ): - ( ab = 8(17 - 8) = 72 ) - ( a + b = 34 - 2(8) = 18 )The quadratic equation ( x^2 - 18x + 72 = 0 ) has roots ( a = 6 ) and ( b = 12 ) (or vice versa), giving ( ab = 6 times 12 = 72 ).Verification shows that ( (6t + 8)(12t + 8) = 17 times 8 ), confirming the solution.Thus, the value of ( ab ) is:[boxed{72}],<think>Okay, let's see. The problem says that t is equal to sqrt(2) minus 1, and we have positive integers a, b, and m such that when we multiply (a*t + m) by (b*t + m), we get 17 times m. We need to find the value of a*b. Alright, let's break this down step by step.First, let's write down the equation again to make sure I have it right:(a*t + m)(b*t + m) = 17*mGiven that t = sqrt(2) - 1, and a, b, m are positive integers. So, we need to find integers a, b, m such that this equation holds, and then compute a*b.Hmm. Let me think. Since t is irrational (because sqrt(2) is irrational), but a, b, m are integers, the left-hand side of the equation will involve terms with sqrt(2), and the right-hand side is a multiple of m, which is an integer. So, for the equation to hold, the coefficients of sqrt(2) on the left-hand side must cancel out, leaving a rational number (specifically, an integer since m is an integer). Therefore, the irrational parts must sum to zero, and the rational parts must equal 17*m.So, perhaps we can expand the left-hand side and then separate the terms with sqrt(2) and those without. Then set the coefficient of sqrt(2) to zero and the constant term equal to 17*m. That seems like a plan.Let's try expanding (a*t + m)(b*t + m). Using the distributive property:= a*t*b*t + a*t*m + m*b*t + m*mSimplify each term:First term: a*b*t^2Second term: a*m*tThird term: b*m*tFourth term: m^2So, combining terms:= a*b*t^2 + (a*m + b*m)*t + m^2Now, substitute t = sqrt(2) - 1 into this expression.First, compute t^2:t = sqrt(2) - 1t^2 = (sqrt(2) - 1)^2 = (sqrt(2))^2 - 2*sqrt(2)*1 + 1^2 = 2 - 2*sqrt(2) + 1 = 3 - 2*sqrt(2)So, t^2 = 3 - 2*sqrt(2)Therefore, substituting back into the expanded left-hand side:a*b*(3 - 2*sqrt(2)) + (a*m + b*m)*(sqrt(2) - 1) + m^2Now, let's expand this further:First term: 3*a*b - 2*a*b*sqrt(2)Second term: (a*m + b*m)*sqrt(2) - (a*m + b*m)Third term: m^2Now, combine like terms. Let's group the terms with sqrt(2) and those without.Terms with sqrt(2):-2*a*b*sqrt(2) + (a*m + b*m)*sqrt(2)= [ -2*a*b + (a*m + b*m) ] * sqrt(2)Terms without sqrt(2):3*a*b - (a*m + b*m) + m^2Therefore, the entire left-hand side becomes:[ -2*a*b + (a*m + b*m) ] * sqrt(2) + [3*a*b - (a*m + b*m) + m^2]This must equal 17*m, which is an integer. Since sqrt(2) is irrational, the coefficient of sqrt(2) must be zero. Otherwise, the left-hand side would be irrational, which cannot equal the right-hand side, which is an integer. Therefore, we can set up two equations:1. Coefficient of sqrt(2) is zero:-2*a*b + a*m + b*m = 02. Constant term equals 17*m:3*a*b - (a*m + b*m) + m^2 = 17*mSo, now we have a system of two equations:Equation (1): -2*a*b + m*(a + b) = 0Equation (2): 3*a*b - m*(a + b) + m^2 = 17*mLet me write them more clearly:1) m*(a + b) = 2*a*b2) 3*a*b - m*(a + b) + m^2 = 17*mNotice that in equation (1), we can solve for m*(a + b):m*(a + b) = 2*a*bSo, let's substitute m*(a + b) from equation (1) into equation (2). From equation (1), m*(a + b) = 2*a*b, so substitute that into equation (2):3*a*b - (2*a*b) + m^2 = 17*mSimplify:3*a*b - 2*a*b + m^2 = 17*mWhich simplifies to:a*b + m^2 = 17*mSo, equation (3): a*b + m^2 = 17*mTherefore, we now have two equations:From equation (1): m*(a + b) = 2*a*bFrom equation (3): a*b = 17*m - m^2So, substituting equation (3) into equation (1):m*(a + b) = 2*(17*m - m^2)Let's compute the right-hand side:2*(17*m - m^2) = 34*m - 2*m^2Therefore, equation (1) becomes:m*(a + b) = 34*m - 2*m^2Assuming m ≠ 0 (which it can't be, since m is a positive integer), we can divide both sides by m:a + b = 34 - 2*mSo, equation (4): a + b = 34 - 2*mNow, from equation (3): a*b = 17*m - m^2So, now we have:a + b = 34 - 2*manda*b = 17*m - m^2So, now we can consider that a and b are positive integers that satisfy these two equations. Since a and b are positive integers, their sum and product are also integers, and we need to find m such that 34 - 2*m and 17*m - m^2 are positive integers, since a and b are positive integers.So, let's analyze the possible values of m.First, note that m is a positive integer. Let's find possible m such that:1. a + b = 34 - 2*m must be positive. So, 34 - 2*m > 0 => 2*m < 34 => m < 17. So m ≤ 16.2. a*b = 17*m - m^2 must also be positive. So, 17*m - m^2 > 0 => m*(17 - m) > 0. Since m is positive, this requires 17 - m > 0 => m < 17. Which is the same as above.Therefore, m can be from 1 to 16 inclusive.But also, a and b must be positive integers. So, given that a + b = 34 - 2*m and a*b = 17*m - m^2, then for each m from 1 to 16, we can check if 17*m - m^2 can be factored into two positive integers a and b such that their sum is 34 - 2*m.Alternatively, since a and b are roots of the quadratic equation x^2 - (a + b)x + a*b = 0. So, for given m, the quadratic equation would be:x^2 - (34 - 2*m)x + (17*m - m^2) = 0We can check if this quadratic has integer roots. The discriminant D must be a perfect square.Compute discriminant D:D = [34 - 2*m]^2 - 4*(17*m - m^2)Let me compute that:First, expand [34 - 2*m]^2:= 34^2 - 2*34*2*m + (2*m)^2= 1156 - 136*m + 4*m^2Then subtract 4*(17*m - m^2):= 1156 - 136*m + 4*m^2 - 68*m + 4*m^2Wait, wait, hold on. Let me redo:D = (34 - 2m)^2 - 4*(17m - m^2)= (34^2 - 4*34*m + 4*m^2) - 68*m + 4*m^2Wait, no. Let's compute each part:First, (34 - 2m)^2 = 34^2 - 2*34*2m + (2m)^2 = 1156 - 136m + 4m^2Second, 4*(17m - m^2) = 68m - 4m^2Therefore, D = (1156 - 136m + 4m^2) - (68m - 4m^2) = 1156 -136m +4m^2 -68m +4m^2Combine like terms:= 1156 - (136m +68m) + (4m^2 +4m^2)= 1156 -204m +8m^2So, discriminant D =8m^2 -204m +1156We can factor out a 4:D =4*(2m^2 -51m +289)Hmm. Let's check if 2m^2 -51m +289 is a perfect square. If D is a perfect square, then 2m^2 -51m +289 must be a perfect square divided by 4, but since m is integer, 2m^2 -51m +289 must be a perfect square. Wait, no, since D is 4*(2m^2 -51m +289), then for D to be a perfect square, 2m^2 -51m +289 must be a perfect square as well. Because if 4*(something) is a perfect square, then (something) must be a perfect square divided by 4, but since (something) is 2m^2 -51m +289, which is an integer (since m is integer), then 2m^2 -51m +289 must be a perfect square. Let me check:Let me denote K = 2m^2 -51m +289We need K to be a perfect square.So, the discriminant D =4*K, and if K is a perfect square, then D is 4*(perfect square), which is a perfect square. Therefore, for the quadratic equation to have integer roots, K must be a perfect square.Therefore, we need to find integer m between 1 and 16 such that 2m^2 -51m +289 is a perfect square.This seems a bit complicated, but maybe we can test m from 1 to 16 and see which ones give perfect squares.Alternatively, perhaps there's a smarter way.Alternatively, since we have a + b = 34 - 2m and ab =17m -m^2, perhaps we can express a and b in terms of m.Alternatively, since a and b are positive integers, given their sum and product, so for each m from 1 to16, compute sum =34 -2m and product=17m -m^2, and check if there exist integers a and b such that a + b = sum and a*b=product.Alternatively, note that product =17m -m^2 =m*(17 -m). So, product is m*(17 -m). Also, sum is 34 -2m.So, if we set variables:Let sum S =34 -2mProduct P =m*(17 -m)Then, the possible pairs (a,b) are positive integers such that a + b = S and a*b = P.So, perhaps m*(17 -m) must be equal to the product of two numbers summing to 34 -2m.Alternatively, maybe we can express this as follows:Let’s set m*(17 -m) = a*band 34 -2m =a + bTherefore, given that a and b are positive integers, let's try to find m such that these two equations hold.Alternatively, maybe express 34 -2m as S, so S =a + b, and P =a*b.Then, since S and P relate to m, perhaps we can express m in terms of S or P.But maybe it's easier to iterate through m from 1 to16 and check for each m whether 17m -m^2 can be written as a product of two integers a and b such that a + b =34 -2m.Alternatively, maybe find m such that 17m -m^2 is a positive integer, and 34 -2m is also positive, which we already know m is from1 to16.Alternatively, maybe m is a factor of something. Let me think.Alternatively, note that from equation (3): a*b =17m -m^2 =m(17 -m). So, since a and b are positive integers, and m and (17 -m) are integers (since m is integer), m must divide the product a*b. But since m is positive integer, and a and b are positive integers, perhaps m divides a or m divides b, but since m is part of the expression, maybe not directly. Alternatively, since a and b are variables, perhaps m is a factor in the product a*b. But this may not necessarily hold. Hmm.Alternatively, perhaps we can look for m such that 17m -m^2 is a positive integer, which it is for m from1 to16, and then see if 17m -m^2 can be expressed as a product of two integers a and b with sum 34 -2m.Alternatively, let's consider that a and b are roots of the quadratic equation x^2 - Sx + P =0 where S=34 -2m and P=17m -m^2. So, for each m, compute S and P, and check if the quadratic factors into integer roots.This might be tedious, but since m is from1 to16, perhaps we can find m such that this is possible.Alternatively, maybe there is a specific m that works. Let me try plugging in some values of m.Starting with m=1:m=1:S=34 -2*1=32P=17*1 -1^2=17 -1=16So, a + b=32, a*b=16. The solutions would be a and b such that they are factors of16 adding to32. The factors of16 are (1,16),(2,8),(4,4). None of these pairs add to32. So m=1 is invalid.m=2:S=34 -4=30P=34 -4=30? Wait, no. P=17*2 -4=34 -4=30.So, a + b=30, a*b=30. The factors of30: 1,30 (sum31); 2,15 (sum17);3,10(sum13);5,6(sum11). None sum to30. So m=2 invalid.m=3:S=34 -6=28P=51 -9=42So, a + b=28, a*b=42. Factors of42:1,42 (43);2,21(23);3,14(17);6,7(13). None sum to28. No.m=4:S=34 -8=26P=68 -16=52a + b=26, a*b=52. Factors of52:1,52(53);2,26(28);4,13(17). Not 26.m=5:S=34 -10=24P=85 -25=60a + b=24, a*b=60. Factors:1,60;2,30;3,20;4,15;5,12;6,10. Sums:61,32,23,19,17,16. None 24.m=6:S=34 -12=22P=102 -36=66a + b=22, a*b=66. Factors:1,66;2,33;3,22;6,11. Sums:67,35,25,17. None 22.m=7:S=34 -14=20P=119 -49=70a + b=20, a*b=70. Factors:1,70;2,35;5,14;7,10. Sums:71,37,19,17. No.m=8:S=34 -16=18P=136 -64=72a + b=18, a*b=72. Factors:1,72;2,36;3,24;4,18;6,12;8,9. Sums:73,38,27,22,18,17. So 6 and 12 sum to18. So a=6, b=12 or vice versa. Wait, 6*12=72. Yes, so m=8 gives a=6, b=12. So this works! Therefore, m=8 is a solution. Let's check.So, if m=8, then a and b are 6 and12. Therefore, a*b=6*12=72. So the answer is 72? Wait, but let me verify.Wait, hold on. Let's check if m=8 satisfies the original equation. Let me substitute m=8, a=6, b=12 into (a*t + m)(b*t + m) and see if it equals 17*8=136.Compute each term:a*t + m =6*(sqrt(2)-1) +8=6*sqrt(2)-6 +8=6*sqrt(2) +2b*t + m=12*(sqrt(2)-1)+8=12*sqrt(2)-12 +8=12*sqrt(2)-4Multiply them:(6*sqrt(2)+2)(12*sqrt(2)-4)Let's compute this:First, expand:6*sqrt(2)*12*sqrt(2) +6*sqrt(2)*(-4) +2*12*sqrt(2) +2*(-4)Calculate each term:First term:6*12*(sqrt(2))^2=72*2=144Second term:-24*sqrt(2)Third term:24*sqrt(2)Fourth term:-8Combine terms:144 -24*sqrt(2)+24*sqrt(2)-8=144 -8=136Which is indeed 17*8=136. So yes, m=8, a=6, b=12 is a valid solution, giving a*b=72.But let me check other possible m values to see if there are other solutions.m=9:S=34 -18=16P=17*9 -81=153 -81=72a + b=16, a*b=72. Factors:1,72;2,36;3,24;4,18;6,12;8,9. Sums:73,38,27,22,18,17. None sum to16. No.m=10:S=34 -20=14P=170 -100=70a + b=14, a*b=70. Factors:1,70;2,35;5,14;7,10. Sums:71,37,19,17. No.m=11:S=34 -22=12P=187 -121=66a + b=12, a*b=66. Factors:1,66;2,33;3,22;6,11. Sums:67,35,25,17. No.m=12:S=34 -24=10P=204 -144=60a + b=10, a*b=60. Factors:1,60;2,30;3,20;4,15;5,12;6,10. Sums:61,32,23,19,17,16. No.m=13:S=34 -26=8P=221 -169=52a + b=8, a*b=52. Factors:1,52;2,26;4,13. Sums:53,28,17. No.m=14:S=34 -28=6P=238 -196=42a + b=6, a*b=42. Factors:1,42;2,21;3,14;6,7. Sums:43,23,17,13. No.m=15:S=34 -30=4P=255 -225=30a + b=4, a*b=30. Factors:1,30;2,15;3,10;5,6. Sums:31,17,13,11. No.m=16:S=34 -32=2P=272 -256=16a + b=2, a*b=16. Factors:1,16;2,8;4,4. Sums:17,10,8. No.So, the only valid m is m=8, leading to a=6, b=12, giving a*b=72.Wait, but before finalizing, let me check m=8 again. From equation (1): m*(a + b)=2*a*b. So, m=8, a=6, b=12:Left-hand side:8*(6 +12)=8*18=144Right-hand side:2*6*12=144. So it's equal. Also, equation (3):a*b +m^2=72 +64=136=17*8=136. So, that's correct. Therefore, m=8 is valid.Therefore, the answer is a*b=72.But wait, let me check if there's another solution. For example, maybe m= something else. Wait, when m=8, we get a=6 and b=12, but also, since a and b are interchangeable, a=12 and b=6 would also work, but since the problem asks for a*b, which is the same either way. So, 6*12=72, 12*6=72. So, the answer is unique.Therefore, the value of a*b is 72. Therefore, boxed answer: boxed{72}Final Answerboxed{72}</think>Given ( t = sqrt{2} - 1 ), we need to find positive integers ( a ), ( b ), and ( m ) such that:[(a t + m)(b t + m) = 17 m]First, we expand the left-hand side:[(a t + m)(b t + m) = a b t^2 + (a m + b m) t + m^2]Substituting ( t = sqrt{2} - 1 ) and ( t^2 = 3 - 2sqrt{2} ):[a b (3 - 2sqrt{2}) + (a m + b m)(sqrt{2} - 1) + m^2]Separating the terms with (sqrt{2}) and the constant terms:[[-2 a b + (a m + b m)] sqrt{2} + [3 a b - (a m + b m) + m^2]]Since the right-hand side ( 17 m ) is an integer, the coefficient of (sqrt{2}) must be zero:1. (-2 a b + (a m + b m) = 0)2. (3 a b - (a m + b m) + m^2 = 17 m)From the first equation:[m (a + b) = 2 a b]From the second equation:[3 a b - m (a + b) + m^2 = 17 m]Substituting ( m (a + b) = 2 a b ) into the second equation:[3 a b - 2 a b + m^2 = 17 m implies a b + m^2 = 17 m]We now have two equations:1. ( m (a + b) = 2 a b )2. ( a b = 17 m - m^2 )Substituting ( a b = 17 m - m^2 ) into the first equation:[m (a + b) = 2 (17 m - m^2) implies a + b = 34 - 2 m]We need ( a ) and ( b ) such that their sum is ( 34 - 2m ) and their product is ( 17m - m^2 ). Testing values of ( m ) from 1 to 16, we find:For ( m = 8 ):- ( a + b = 34 - 2 cdot 8 = 18 )- ( a b = 17 cdot 8 - 8^2 = 72 )The integers ( a = 6 ) and ( b = 12 ) (or vice versa) satisfy these conditions. Therefore, the value of ( a b ) is:[boxed{72}]]

🔑:Given that t=sqrt{2}-1, we havet^{2}=3-2 sqrt{2} text {. }From the problem, we knowbegin{array}{l}a b t^{2}+m(a+b) t+m^{2}=17 m Rightarrow a b(3-2 sqrt{2})+m(a+b)(sqrt{2}-1)+m^{2}=17 m Rightarrow sqrt{2}[m(a+b)-2 a b]+ quadleft[3 a b-m(a+b)+m^{2}-17 mright]=0 .end{array}Since a, b, m are positive integers, we havebegin{array}{l}left{begin{array}{l}m(a+b)-2 a b=0, 3 a b-m(a+b)+m^{2}-17 m=0end{array}right. Rightarrowleft{begin{array}{l}a+b=2(17-m), a b=17 m-m^{2} .end{array}right.end{array}Therefore, a, b are the two integer roots of the quadratic equation in x:x^{2}+2(m-17) x+17 m-m^{2}=0The discriminant of equation (1) isbegin{array}{l}Delta=4(m-17)^{2}-4left(17 m-m^{2}right) =4(17-m)(17-2 m) geqslant 0 .end{array}Since a, b, m are positive integers, we havea+b=2(17-m)>0 text {. }Thus, 0<m leqslant frac{17}{2}.Since the discriminant Delta is a perfect square, upon verification, only m=8 meets the requirement.Substituting m=8 into the equation, we geta b=17 m-m^{2}=72 text {. }