🧠:[<think>Okay, so I need to solve this problem about the sequence {a_n} where a₁ = 1 and the sum of the first n terms Sₙ = (n + 2)/3 * aₙ. There are two parts: first, find a₂, a₃, and the general formula; second, for bₙ = 1/aₙ, find the sum of the first n terms Tₙ. Hmm, let me start with part (I).First, I remember that the sum of the first n terms Sₙ is related to the terms of the sequence. Specifically, Sₙ = a₁ + a₂ + ... + aₙ. Also, the problem gives a formula for Sₙ in terms of aₙ: Sₙ = (n + 2)/3 * aₙ. So, if I can express Sₙ in terms of aₙ, maybe I can find a recursive relation or something.Given that a₁ = 1, let's compute S₁ first. S₁ is just the first term, so S₁ = a₁ = 1. According to the formula, S₁ should also equal (1 + 2)/3 * a₁ = 3/3 * 1 = 1. Okay, that checks out. Good.Now, let's find a₂. For n = 2, S₂ = a₁ + a₂ = 1 + a₂. According to the formula, S₂ should be (2 + 2)/3 * a₂ = 4/3 * a₂. Therefore, we have the equation 1 + a₂ = (4/3) * a₂. Let's solve for a₂.Subtract a₂ from both sides: 1 = (4/3)a₂ - a₂ = (4/3 - 3/3)a₂ = (1/3)a₂. Therefore, a₂ = 1 / (1/3) = 3. So a₂ is 3. That seems straightforward.Now, moving on to a₃. Let's compute S₃. S₃ = a₁ + a₂ + a₃ = 1 + 3 + a₃ = 4 + a₃. According to the formula, S₃ should be (3 + 2)/3 * a₃ = 5/3 * a₃. So, 4 + a₃ = (5/3)a₃. Let's solve for a₃.Subtract a₃ from both sides: 4 = (5/3)a₃ - a₃ = (5/3 - 3/3)a₃ = (2/3)a₃. Therefore, a₃ = 4 / (2/3) = 4 * (3/2) = 6. So a₃ is 6. Okay, so a₁=1, a₂=3, a₃=6. Hmm, I wonder if there's a pattern here. Let's see: 1, 3, 6... those look like triangular numbers? Wait, triangular numbers are 1, 3, 6, 10, 15,... but let me check if that's the case here. But maybe not, let's see the general formula.To find the general formula for aₙ, I need to find a relation between Sₙ and aₙ. Since Sₙ = (n + 2)/3 * aₙ, and also Sₙ = Sₙ₋₁ + aₙ. So we can set up an equation: Sₙ = Sₙ₋₁ + aₙ = [(n - 1 + 2)/3] * aₙ₋₁ + aₙ = [(n + 1)/3] * aₙ₋₁ + aₙ. But according to the formula, Sₙ = (n + 2)/3 * aₙ. Therefore:[(n + 1)/3] * aₙ₋₁ + aₙ = (n + 2)/3 * aₙLet me write that again:[(n + 1)/3] * aₙ₋₁ + aₙ = [(n + 2)/3] * aₙSubtract [(n + 2)/3] * aₙ from both sides:[(n + 1)/3] * aₙ₋₁ + aₙ - [(n + 2)/3] * aₙ = 0Simplify the terms with aₙ:[(n + 1)/3] * aₙ₋₁ + [1 - (n + 2)/3] * aₙ = 0Compute 1 - (n + 2)/3:1 = 3/3, so 3/3 - (n + 2)/3 = (3 - n - 2)/3 = (1 - n)/3So the equation becomes:[(n + 1)/3] * aₙ₋₁ + [(1 - n)/3] * aₙ = 0Multiply both sides by 3 to eliminate denominators:(n + 1)aₙ₋₁ + (1 - n)aₙ = 0Rearranged:(n + 1)aₙ₋₁ = (n - 1)aₙTherefore, solving for aₙ:aₙ = [(n + 1)/(n - 1)] * aₙ₋₁Hmm, so this is a recursive formula. So each term is [(n + 1)/(n - 1)] times the previous term. Let's see if this works with our known terms.For n=2: a₂ = [(2 + 1)/(2 - 1)] * a₁ = 3/1 * 1 = 3. Correct.For n=3: a₃ = [(3 + 1)/(3 - 1)] * a₂ = 4/2 * 3 = 2 * 3 = 6. Correct.Okay, so the recursive formula is valid. Now, to find the general formula, we can write aₙ in terms of a₁ by expanding the recursion.So, let's write out the recursion:aₙ = [(n + 1)/(n - 1)] * aₙ₋₁Similarly, aₙ₋₁ = [n/(n - 2)] * aₙ₋₂And so on, until a₂ = 3/1 * a₁.Therefore, multiplying all these together:aₙ = [(n + 1)/(n - 1)] * [n/(n - 2)] * [(n - 1)/(n - 3)] * ... * [4/2] * [3/1] * a₁Notice that there's a telescoping product here. Let me write this out:aₙ = [ (n + 1)/1 ] * [ n / 2 ] * [ (n - 1)/3 ] * ... * [4/(n - 2)] * [3/(n - 1)] * a₁Wait, maybe I need to track the terms more carefully.Wait, starting from aₙ:aₙ = [(n + 1)/(n - 1)] * aₙ₋₁But aₙ₋₁ = [n/(n - 2)] * aₙ₋₂So substitute:aₙ = [(n + 1)/(n - 1)] * [n/(n - 2)] * aₙ₋₂Similarly, aₙ₋₂ = [(n - 1)/(n - 3)] * aₙ₋₃So:aₙ = [(n + 1)/(n - 1)] * [n/(n - 2)] * [(n - 1)/(n - 3)] * aₙ₋₃Continuing this pattern until we get to a₁.Let me try writing the product:aₙ = a₁ * product from k=2 to n of [(k + 1)/(k - 1)]Because each term aₖ = [(k + 1)/(k - 1)] * aₖ₋₁So multiplying from k=2 to n:aₙ = a₁ * Π_{k=2}^n [(k + 1)/(k - 1)]Let me compute this product.Notice that [(k + 1)/(k - 1)] can be written as [(k - 1 + 2)/(k - 1)] = 1 + 2/(k - 1). But maybe it's better to look for telescoping.The product Π_{k=2}^n [(k + 1)/(k - 1)] can be expanded as:(3/1) * (4/2) * (5/3) * (6/4) * ... * [(n + 1)/(n - 1)]Each numerator cancels with the denominator two terms ahead. Let's see:3/1 * 4/2 * 5/3 * 6/4 * ... * (n + 1)/(n - 1)So, the 3 in the numerator cancels with the 3 in the denominator of the third term. The 4 in the numerator cancels with the 4 in the denominator of the fourth term, and so on. What's left is the last two numerators and the first two denominators.Specifically, after cancellation, the product becomes:(n * (n + 1)) / (1 * 2) )Wait, let's check with small n.For n=2: product is (3/1) = 3. Then (2 * 3)/(1 * 2) = 6/2 = 3. Correct.For n=3: product is (3/1)*(4/2) = 3*2=6. Then (3*4)/(1*2)=12/2=6. Correct.For n=4: product is (3/1)*(4/2)*(5/3)= (4/2)*(5/1)=2*5=10. Then (4*5)/(1*2)=20/2=10. Correct.So in general, the product Π_{k=2}^n [(k + 1)/(k - 1)] = [n(n + 1)] / [1 * 2] = n(n + 1)/2Therefore, aₙ = a₁ * [n(n + 1)/2]. Since a₁ = 1, then aₙ = n(n + 1)/2.Wait, but hold on. For n=1, this formula gives a₁ = 1(2)/2 = 1, which is correct. For n=2, 2(3)/2=3, correct. For n=3, 3(4)/2=6, correct. So the general formula is aₙ = n(n + 1)/2. That's the formula for triangular numbers, but starting from n=1: 1, 3, 6, 10,... So yes, that's correct. So part (I) is solved: a₂=3, a₃=6, and aₙ = n(n + 1)/2.But wait, let me verify this with the original condition Sₙ = (n + 2)/3 * aₙ. If aₙ = n(n + 1)/2, then Sₙ should be the sum of the first n terms of this sequence. Let's check if Sₙ = (n + 2)/3 * aₙ.First, compute Sₙ for aₙ = n(n + 1)/2. The sum Sₙ = Σ_{k=1}^n [k(k + 1)/2] = 1/2 Σ_{k=1}^n (k² + k) = 1/2 [ Σk² + Σk ].We know that Σk from 1 to n is n(n + 1)/2, and Σk² from 1 to n is n(n + 1)(2n + 1)/6.Therefore, Sₙ = 1/2 [n(n + 1)(2n + 1)/6 + n(n + 1)/2] = 1/2 [ n(n + 1)(2n + 1)/6 + 3n(n + 1)/6 ] = 1/2 [ n(n + 1)(2n + 1 + 3)/6 ] = 1/2 [n(n + 1)(2n + 4)/6] = 1/2 [n(n + 1)(2)(n + 2)/6] = 1/2 [2n(n + 1)(n + 2)/6] = [2n(n + 1)(n + 2)/12] = n(n + 1)(n + 2)/6.Now, according to the problem statement, Sₙ should be equal to (n + 2)/3 * aₙ. Let's compute (n + 2)/3 * aₙ:(n + 2)/3 * [n(n + 1)/2] = n(n + 1)(n + 2)/6. Which matches the sum we just calculated. Perfect. So the general formula aₙ = n(n + 1)/2 is correct.Therefore, part (I) is done: a₂=3, a₃=6, general formula aₙ = n(n + 1)/2.Now, part (II): Let bₙ = 1/aₙ, find the sum of the first n terms Tₙ for {bₙ}.Given that aₙ = n(n + 1)/2, so bₙ = 1/[n(n + 1)/2] = 2/[n(n + 1)].So, Tₙ = Σ_{k=1}^n bₙ = Σ_{k=1}^n [2/(k(k + 1))].Hmm, this seems like a telescoping series. Let's recall that 1/(k(k + 1)) can be written as 1/k - 1/(k + 1). Let me verify:1/k - 1/(k + 1) = (k + 1 - k)/[k(k + 1)] = 1/[k(k + 1)]. Exactly. So 2/(k(k + 1)) = 2[1/k - 1/(k + 1)].Therefore, Tₙ = Σ_{k=1}^n 2[1/k - 1/(k + 1)] = 2 Σ_{k=1}^n [1/k - 1/(k + 1)].This is a telescoping sum. Let's write out the terms:For k=1: 2[1/1 - 1/2]For k=2: 2[1/2 - 1/3]For k=3: 2[1/3 - 1/4]...For k=n: 2[1/n - 1/(n + 1)]Adding all these up:2[(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n + 1))]Notice that all intermediate terms cancel out: -1/2 cancels with +1/2, -1/3 cancels with +1/3, etc., leaving:2[1 - 1/(n + 1)] = 2[ (n + 1 - 1)/(n + 1) ] = 2[ n/(n + 1) ] = 2n/(n + 1)Therefore, Tₙ = 2n/(n + 1)Let me check this with n=1,2,3 to make sure.For n=1: T₁ = b₁ = 2/(1*2) = 1. According to formula, 2*1/(1 + 1)=2/2=1. Correct.For n=2: T₂ = b₁ + b₂ = 1 + 2/(2*3) = 1 + 1/3 = 4/3. Formula: 2*2/(2 + 1)=4/3. Correct.For n=3: T₃ = 1 + 1/3 + 2/(3*4) = 1 + 1/3 + 1/6 = (6 + 2 + 1)/6 = 9/6 = 3/2. Formula: 2*3/(3 + 1) = 6/4 = 3/2. Correct.Good, so the formula holds. Therefore, the sum of the first n terms Tₙ is 2n/(n + 1).So, part (II) answer is Tₙ = 2n/(n + 1).Final Answer(I) (a_2 = boxed{3}), (a_3 = boxed{6}), and the general formula is (a_n = boxed{dfrac{n(n+1)}{2}});(II) The sum of the first (n) terms (T_n = boxed{dfrac{2n}{n+1}}).</think>Given the sequence ({a_n}) where (a_1 = 1) and the sum of the first (n) terms (S_n = frac{n+2}{3}a_n):(I) To find (a_2), (a_3), and the general formula for ({a_n}):1. Calculate (a_2): [ S_2 = a_1 + a_2 = 1 + a_2 quad text{and} quad S_2 = frac{4}{3}a_2 ] Solving (1 + a_2 = frac{4}{3}a_2) gives (a_2 = 3).2. Calculate (a_3): [ S_3 = a_1 + a_2 + a_3 = 4 + a_3 quad text{and} quad S_3 = frac{5}{3}a_3 ] Solving (4 + a_3 = frac{5}{3}a_3) gives (a_3 = 6).3. Derive the general formula using the recursive relation: [ a_n = frac{n+1}{n-1} a_{n-1} ] This telescopes to: [ a_n = a_1 prod_{k=2}^n frac{k+1}{k-1} = frac{n(n+1)}{2} ]Thus, the values are (a_2 = boxed{3}), (a_3 = boxed{6}), and the general formula is (a_n = boxed{dfrac{n(n+1)}{2}}).(II) For the sequence ({b_n}) where (b_n = frac{1}{a_n}), find the sum of the first (n) terms (T_n):1. Express (b_n): [ b_n = frac{2}{n(n+1)} ]2. Sum the telescoping series: [ T_n = sum_{k=1}^n frac{2}{k(k+1)} = 2 sum_{k=1}^n left( frac{1}{k} - frac{1}{k+1} right) ] This simplifies to: [ T_n = 2 left( 1 - frac{1}{n+1} right) = frac{2n}{n+1} ]Thus, the sum of the first (n) terms (T_n = boxed{dfrac{2n}{n+1}}).,<think>Okay, so I need to solve this problem about the sequence {a_n} where a_1 = 1 and the sum of the first n terms S_n is given by (n + 2)/3 times a_n. Part I asks me to find a_2, a_3, and the general formula for {a_n}. Then part II is about finding the sum of the first n terms for the sequence {b_n} where each b_n is 1/a_n. Let me start with part I.First, I know that S_n is the sum of the first n terms of the sequence. So S_n = a_1 + a_2 + ... + a_n. The problem says that S_n = [(n + 2)/3] * a_n. Given that a_1 is 1, let me compute S_1 first. For n=1, S_1 should be equal to (1 + 2)/3 * a_1, which is (3/3)*1 = 1. That checks out since S_1 is just a_1 = 1.Now, moving on to n=2. S_2 should be the sum of a_1 and a_2, which is 1 + a_2. According to the formula, S_2 = (2 + 2)/3 * a_2 = (4/3) * a_2. Therefore, we can set up the equation: 1 + a_2 = (4/3) * a_2. Let me solve for a_2.Subtracting a_2 from both sides: 1 = (4/3)a_2 - a_2. That simplifies to 1 = (1/3)a_2, so a_2 = 3. Okay, that seems straightforward.Next, for n=3, S_3 = a_1 + a_2 + a_3 = 1 + 3 + a_3 = 4 + a_3. According to the formula, S_3 = (3 + 2)/3 * a_3 = (5/3)a_3. So 4 + a_3 = (5/3)a_3. Let's solve this equation.Subtract a_3 from both sides: 4 = (5/3)a_3 - a_3 = (2/3)a_3. Then, multiplying both sides by 3/2 gives a_3 = 4 * (3/2) = 6. So a_3 is 6. Hmm, interesting. So a_1=1, a_2=3, a_3=6. Let me check if there's a pattern here. 1, 3, 6... these look like triangular numbers, but let's see if that's the case or if there's another pattern.Wait, triangular numbers are 1, 3, 6, 10, 15,... but here we have a_3=6, which is the third triangular number. Let me see if a_4 follows. Let's compute a_4 to check. For n=4, S_4 = S_3 + a_4 = 4 + a_3 + a_4 = 4 + 6 + a_4 = 10 + a_4. According to the formula, S_4 = (4 + 2)/3 * a_4 = 6/3 * a_4 = 2 * a_4. So 10 + a_4 = 2a_4. Then, subtract a_4: 10 = a_4. So a_4=10. Hmm, which is the 4th triangular number. Wait, triangular numbers are T_n = n(n+1)/2. Let's check: T_1=1, T_2=3, T_3=6, T_4=10. So yes, this seems like a_n is the nth triangular number. But let's confirm this with the general formula.Alternatively, maybe there's a different pattern here. Let's try to derive the general formula.Given that S_n = [(n + 2)/3] * a_n. But S_n is also equal to S_{n-1} + a_n. So, we can write:S_n = S_{n-1} + a_n.But according to the problem's formula, S_n = [(n + 2)/3] * a_n and S_{n-1} = [(n -1 + 2)/3] * a_{n-1} = [(n +1)/3] * a_{n-1}.Therefore, substituting into the first equation:[(n + 2)/3] * a_n = [(n +1)/3] * a_{n-1} + a_n.Let me rearrange this equation. Subtract a_n from both sides:[(n + 2)/3] * a_n - a_n = [(n +1)/3] * a_{n-1}.Factor out a_n on the left:a_n [ (n + 2)/3 - 1 ] = [(n +1)/3] * a_{n-1}.Simplify the left side:(n + 2)/3 - 3/3 = (n + 2 - 3)/3 = (n -1)/3.Therefore:a_n * (n -1)/3 = (n +1)/3 * a_{n-1}.Multiply both sides by 3 to eliminate denominators:a_n * (n -1) = (n +1) * a_{n-1}.So, we have a recurrence relation:a_n = [(n +1)/(n -1)] * a_{n-1}.This is a recursive formula. Let's see if we can express a_n in terms of a_1.Let me compute a few terms:We have a_1=1.a_2 = [(2 +1)/(2 -1)] * a_1 = 3/1 * 1 = 3. Which matches our previous result.a_3 = [(3 +1)/(3 -1)] * a_2 = 4/2 * 3 = 2 * 3 = 6. Correct.a_4 = [(4 +1)/(4 -1)] * a_3 = 5/3 * 6 = 10. Correct.a_5 = [(5 +1)/(5 -1)] * a_4 = 6/4 * 10 = 15. Which would be the next triangular number. So yes, the pattern continues.So, to find a general formula, let's write out the recursion:a_n = [(n +1)/(n -1)] * a_{n-1}.If we expand this recursion:a_n = [(n +1)/(n -1)] * a_{n-1}a_{n-1} = [n/(n -2)] * a_{n-2}...a_2 = [3/1] * a_1Multiplying all these together:a_n = [(n +1)/(n -1)] * [n/(n -2)] * [ (n -1)/(n -3) ] * ... * [3/1] * a_1Notice that this is a telescoping product. Let's see:The numerators are (n +1), n, (n -1), ..., 3.The denominators are (n -1), (n -2), (n -3), ..., 1.So when we multiply them, most terms cancel out. Let's write the product step by step.Numerators: (n +1) * n * (n -1) * ... * 3Denominators: (n -1) * (n -2) * ... * 1So, the product is [ (n +1) * n * (n -1)! ] / [ (n -1)! * 1 ]Wait, let me check.Wait, the numerators from a_n down to a_2 are:(n +1)/ (n -1) * n/(n -2) * (n -1)/(n -3) * ... * 3/1Let me write the first few terms:Term for a_n: (n +1)/(n -1)Term for a_{n-1}: n/(n -2)Term for a_{n-2}: (n -1)/(n -3)...Term for a_3: 4/2Term for a_2: 3/1So, multiplying all these terms:[(n +1)/ (n -1)] * [n / (n -2)] * [(n -1)/ (n -3)] * ... * [4/2] * [3/1]Let me write this as a product:Product from k=2 to k=n of [(k +1)/(k -1)]Wait, for k from 2 to n, when k=2: (2 +1)/(2 -1) = 3/1When k=3: 4/2...When k=n: (n +1)/(n -1)So indeed, the product is from k=2 to k=n of (k +1)/(k -1)But if I write this product out:(3/1) * (4/2) * (5/3) * ... * [(n +1)/(n -1)]Notice that this telescopes. The denominator of each term cancels with the numerator of the previous term.For example, the 3 in the numerator of the first term cancels with the 3 in the denominator of the third term. Wait, actually, let's see:Wait, first term: 3/1Second term: 4/2Third term: 5/3Fourth term: 6/4...Last term: (n +1)/(n -1)So, if I write all numerators and denominators:Numerators: 3, 4, 5, 6, ..., (n +1)Denominators: 1, 2, 3, 4, ..., (n -1)So, the product is (3 * 4 * 5 * ... * (n +1)) / (1 * 2 * 3 * ... * (n -1))This simplifies to [ (n +1)! / (2!) ] / [ (n -1)! ) ]Wait, because the numerator is 3*4*5*...*(n+1). Let's see:3*4*5*...*(n+1) = (n +1)! / (2!) )Similarly, the denominator is 1*2*3*...*(n -1) = (n -1)! )Therefore, the product is [ (n +1)! / 2 ] / (n -1)! ) = [ (n +1)! / (n -1)! ) ] / 2But (n +1)! / (n -1)! ) = (n +1)*n*(n -1)! ) / (n -1)! ) = n(n +1)Therefore, the product is n(n +1)/2Therefore, the product of the terms from k=2 to n of (k +1)/(k -1) is n(n +1)/2Therefore, a_n = [n(n +1)/2] * a_1But since a_1=1, then a_n = n(n +1)/2Wait, that's the formula for the nth triangular number. So, that's consistent with our earlier terms: a_1=1, a_2=3, a_3=6, a_4=10, etc. So the general formula is a_n = n(n +1)/2. Let me verify this.Let me check if with this formula, S_n = (n + 2)/3 * a_n.Given that a_n = n(n +1)/2, then (n +2)/3 * a_n = (n +2)/3 * [n(n +1)/2] = [n(n +1)(n +2)] /6.But S_n is the sum of the first n terms of the sequence {a_n}. Since a_n = n(n +1)/2, then S_n = sum_{k=1}^n [k(k +1)/2] = (1/2) sum_{k=1}^n [k^2 +k] = (1/2)[ sum_{k=1}^n k^2 + sum_{k=1}^n k ]We know that sum_{k=1}^n k = n(n +1)/2 and sum_{k=1}^n k^2 = n(n +1)(2n +1)/6.Therefore, S_n = (1/2)[ n(n +1)(2n +1)/6 + n(n +1)/2 ] = (1/2)[ n(n +1)(2n +1 + 3)/6 ] = (1/2)[ n(n +1)(2n +4)/6 ] = (1/2)[ n(n +1)(2)(n +2)/6 ] = (1/2)[ n(n +1)(n +2)/3 ] = n(n +1)(n +2)/6Which matches the expression we had from the given formula: S_n = (n +2)/3 * a_n = (n +2)/3 * [n(n +1)/2] = n(n +1)(n +2)/6. So yes, it's correct. Therefore, the general formula for a_n is indeed n(n +1)/2.So, for part I, the answers are:a_2 = 3, a_3 = 6, and the general formula is a_n = n(n +1)/2.Moving on to part II: Let b_n = 1/a_n, find the sum of the first n terms T_n for the sequence {b_n}.Given that a_n = n(n +1)/2, so b_n = 1/a_n = 2/[n(n +1)]. So we need to compute T_n = sum_{k=1}^n b_k = sum_{k=1}^n [2/(k(k +1))].This is a telescoping series. Let's recall that 2/(k(k +1)) can be written as 2[1/k - 1/(k +1)]. Let's check:1/k - 1/(k +1) = (k +1 - k)/[k(k +1)] = 1/[k(k +1)], so 2[1/k -1/(k +1)] = 2/[k(k +1)], which matches b_n. Therefore, each term b_k is 2[1/k -1/(k +1)].Therefore, the sum T_n = sum_{k=1}^n 2[1/k -1/(k +1)] = 2 sum_{k=1}^n [1/k -1/(k +1)]This telescopes: the sum becomes 2[ (1/1 -1/2) + (1/2 -1/3) + (1/3 -1/4) + ... + (1/n -1/(n +1)) ) ]All the intermediate terms cancel out:-1/2 cancels with +1/2, -1/3 cancels with +1/3, etc., leaving:2[1 - 1/(n +1)] = 2[ (n +1 -1)/(n +1) ) ] = 2[ n/(n +1) ) ] = 2n/(n +1)Therefore, T_n = 2n/(n +1)Let me verify this with the first few terms.For n=1: T_1 = b_1 = 2/(1*2) = 1. According to the formula, 2*1/(1 +1) = 2/2 =1. Correct.For n=2: T_2 = b_1 + b_2 = 1 + 2/(2*3) = 1 + 1/3 = 4/3. According to the formula, 2*2/(2 +1) =4/3. Correct.For n=3: T_3 = 4/3 + 2/(3*4) =4/3 +1/6= (8 +1)/6=9/6=3/2. Formula: 2*3/(3 +1)=6/4=3/2. Correct.So the formula works. Therefore, the sum T_n is 2n/(n +1).Thus, part II's answer is T_n = 2n/(n +1).Final Answer(I) (a_2 = boxed{3}), (a_3 = boxed{6}), and (a_n = boxed{dfrac{n(n+1)}{2}});(II) The sum of the first (n) terms (T_n = boxed{dfrac{2n}{n+1}}).</think>Given the sequence ({a_n}) where (a_1 = 1) and the sum of the first (n) terms (S_n = frac{n+2}{3}a_n):(I) To find (a_2), (a_3), and the general formula for ({a_n}):1. For (n = 2): [ S_2 = a_1 + a_2 = 1 + a_2 ] Given (S_2 = frac{4}{3}a_2): [ 1 + a_2 = frac{4}{3}a_2 implies 1 = frac{1}{3}a_2 implies a_2 = 3 ]2. For (n = 3): [ S_3 = a_1 + a_2 + a_3 = 1 + 3 + a_3 = 4 + a_3 ] Given (S_3 = frac{5}{3}a_3): [ 4 + a_3 = frac{5}{3}a_3 implies 4 = frac{2}{3}a_3 implies a_3 = 6 ]3. General formula derivation: Using the recurrence relation derived from the given sum formula: [ a_n = frac{(n+1)}{(n-1)}a_{n-1} ] This telescopes to: [ a_n = frac{n(n+1)}{2} ] Verification shows this formula satisfies the given sum condition.Thus, the answers are:[a_2 = boxed{3}, quad a_3 = boxed{6}, quad text{and} quad a_n = boxed{dfrac{n(n+1)}{2}}](II) For the sequence ({b_n}) where (b_n = frac{1}{a_n}), find the sum of the first (n) terms (T_n):1. Given (a_n = frac{n(n+1)}{2}), we have: [ b_n = frac{2}{n(n+1)} ] This can be written as a telescoping series: [ b_n = 2left(frac{1}{n} - frac{1}{n+1}right) ]2. Summing the first (n) terms: [ T_n = sum_{k=1}^n b_k = 2 sum_{k=1}^n left(frac{1}{k} - frac{1}{k+1}right) ] This telescopes to: [ T_n = 2left(1 - frac{1}{n+1}right) = frac{2n}{n+1} ]Thus, the answer is:[T_n = boxed{dfrac{2n}{n+1}}]]

🔑:(I) Using the given information, a_1=1 and S_n=frac{n+2}{3}a_n, we can find:S_2=frac{2+2}{3}a_2=a_1+a_2 Rightarrow a_2=3a_1=3,Similarly, frac{2}{3}a_3=a_1+a_2=4, thus a_3=6;Now let's derive the general formula for {a_n}. For ngeq 2, we have two equations:S_n=frac{n+2}{3}a_n (1), andS_{n-1}=frac{n+1}{3}a_{n-1} (2).Subtracting equation (2) from (1), we get:a_n=frac{n+2}{3}a_n-frac{n+1}{3}a_{n-1}, which simplifies to:frac{n-1}{3}a_n=frac{n+1}{3}a_{n-1}, and finally:frac{a_n}{a_{n-1}}=frac{n+1}{n-1}.Using this recurrence relation, we can derive the general formula:a_n=frac{a_n}{a_{n-1}}cdotfrac{a_{n-1}}{a_{n-2}}dotsfrac{a_2}{a_1}cdot a_1=frac{n+1}{n-1}cdotfrac{n}{n-2}dotsfrac{3}{1}cdot1=frac{n^2+n}{2} (ngeq 2).As frac{1^2+1}{2}=1=a_1, the general formula for a_n is:a_n=frac{n^2+n}{2}.(II) Since b_n=frac{1}{a_n}=frac{2}{n^2+n}=2(frac{1}{n}-frac{1}{n+1}), we can find the sum of the first n terms, T_n, as follows:T_n=2[(1-frac{1}{2})+(frac{1}{2}-frac{1}{3})+dots+(frac{1}{n}-frac{1}{n+1})].Simplifying, we get:T_n=2(1-frac{1}{n+1})=boxed{frac{2n}{n+1}}.

🧠:[<think>Okay, so I need to figure out the minimum number of weighings required to order four stones from lightest to heaviest using a balanced scale. The options are from A to E, with numbers 4 to 8. Hmm. Let me think about how this process works.First, I remember that a balanced scale can compare two groups of items and tell which side is heavier or if they're equal. Since all stones have distinct weights, each weighing will give a definite result. The goal is to determine the exact order of the four stones with the fewest weighings possible.Let me start by recalling that sorting algorithms often use comparisons, and this is similar. In computer science, the minimum number of comparisons needed to sort n elements is known to be on the order of n log n. For four elements, that would be around 4 log 4 = 4 * 2 = 8 comparisons, but this is just a general case. However, in reality, the exact minimum number can be lower. Wait, but the answer choices here go up to 8, but maybe there's a smarter way.Wait, actually, I remember that the exact minimal number of comparisons needed to sort 4 elements is 5. Let me try to recall. For example, using a decision tree model, the height of the decision tree must be at least log2(4!) which is log2(24) ≈ 4.58, so you need at least 5 weighings. So the answer should be 5? But let me verify this step by step.First, the number of possible permutations of 4 stones is 4! = 24. Each weighing provides some information, specifically, each weighing can result in three outcomes: left side heavier, right side heavier, or equal. But since we are dealing with distinct weights, equal might not be possible unless we are weighing the same number of stones on both sides. Wait, but in this problem, we can weigh any number of stones on either side. So, actually, the scale is balanced, so we can compare two groups, say, one stone vs another, or two stones vs two stones, or one vs two, but the scale will tip to the heavier side or balance if equal. But since all stones have distinct weights, if we weigh different numbers of stones, the side with more stones will not necessarily be heavier because the individual stones could be lighter. Hmm, this complicates things.Wait, but maybe we need to use a strategy where we only compare individual stones or pairs of stones. Let me think. If we do one stone against another, then each weighing gives a comparison between two stones. So in that case, each weighing is a comparison between two elements, which can be modeled as a binary decision (which is heavier). Then, the decision tree would have a certain depth. But the number of leaves of the decision tree needs to be at least 24 (the number of permutations). Each internal node has three children (since each weighing has three possible outcomes). Wait, no, actually, if we compare two stones, there are three outcomes: A > B, A < B, or A = B. But in this problem, all stones have distinct weights, so the equal case is impossible. Therefore, each weighing actually only has two possible outcomes. Therefore, each internal node in the decision tree would have two children. So the minimum depth required would be log2(24) ≈ 4.58, so 5 weighings. So that's the same as before.But maybe we can do better by using weighings that compare more than two stones at a time. For example, comparing two stones against two stones. If we do that, then the outcome would tell us which pair is heavier. But how much information does that give us? Let's see. Suppose we have four stones: A, B, C, D. If we compare A + B vs C + D. Then if A + B > C + D, we know that the combined weight of A and B is greater than C and D. But how does that help us order the individual stones? It's not immediately clear. Similarly, if we compare A vs B, that's a direct comparison. So maybe using pairwise comparisons is more straightforward.Alternatively, maybe using a tournament method. For example, first compare A vs B, then C vs D. Then compare the winners of those two. Then, perhaps insert the losers into the hierarchy. But I need to think through this step by step.Let me try to outline a possible sequence of weighings:1. Compare A and B. Suppose A < B.2. Compare C and D. Suppose C < D.3. Now compare the two winners: B and D. Suppose B < D. So now we know that D is the heaviest. Then we have A < B and C < D, and B < D. But we still need to figure out where C stands relative to A and B, and where A and C stand.Alternatively, maybe another approach. Let me recall that with four elements, there is a known optimal sorting network that can sort four elements in five comparisons. For example, the following sequence:1. Compare A and B.2. Compare C and D.3. Compare the two winners (e.g., if A < B and C < D, compare B and D).4. Then compare the losers of the first two comparisons (e.g., A and C).5. Then depending on the results, another comparison.This might result in five weighings. Alternatively, let me check the Ford-Johnson algorithm, which is a comparison sort that is known to be efficient for small numbers of elements. For n=4, the algorithm requires five comparisons. So this seems to confirm that five is the minimal number.But let's try to actually construct such a sequence of weighings. Let's name the stones as W, X, Y, Z for clarity.First, weigh W vs X. Suppose W < X.Second, weigh Y vs Z. Suppose Y < Z.Third, weigh X vs Z. Suppose X < Z. So now we know W < X < Z and Y < Z. But we don't know how Y compares to W and X. So fourth weighing: weigh W vs Y. Suppose W < Y. Then we know W < Y, and since X < Z and Y < Z, but we need to place Y relative to X. So fifth weighing: weigh Y vs X. Suppose Y < X. Then the order is W < Y < X < Z. If Y > X, then W < X < Y < Z.Alternatively, if in the fourth weighing, W vs Y, if Y < W, then Y < W < X < Z.Wait, but depending on the results of each weighing, the required subsequent weighings might differ. This is starting to get complex, but in the worst case, it might take five weighings. Alternatively, maybe there's a smarter way that can do it in four? Let me try.Suppose I start by weighing two pairs:1. Compare A + B vs C + D. Suppose A + B < C + D. This tells us that the total weight of C and D is greater, but individually, we don't know which is which. Maybe then compare A vs B. If A < B, and C vs D, if C < D. Then, as before, compare B vs D. But we still might need more comparisons. This approach might not necessarily reduce the number of required weighings below five.Alternatively, another method: divide the four stones into two groups of two, weigh each group, then compare the winners. Let me see.Weigh W vs X: suppose W < X.Weigh Y vs Z: suppose Y < Z.Then weigh X vs Z: suppose X < Z. Now, the heaviest stone is Z. Now, we need to determine the order among W, X, Y. But Y hasn't been compared to W or X yet. So fourth weighing: compare Y vs X. If Y < X, then fifth weighing: compare Y vs W. Then we can place Y. Alternatively, if Y > X, then we know W < X < Y < Z. Wait, but maybe we can combine some steps.Alternatively, perhaps after the first three weighings (W vs X, Y vs Z, X vs Z), we can proceed as follows:We have W < X < Z and Y < Z. Next, we need to place Y relative to W and X. So weigh Y vs X. If Y < X, then weigh Y vs W. If Y > W, then order is W < Y < X < Z. If Y < W, then Y < W < X < Z. But in that case, since we already know W < X, and Y < Z, but Y is less than W. So this might take five weighings. Alternatively, maybe there's a way to do it in four?Wait, suppose we do a different initial set of weighings.First, compare A and B. Let's say A < B.Second, compare A and C. Now, depending on whether C < A or A < C, we can place C relative to A and B. Suppose C < A. Then we have C < A < B. Then third weighing: compare C and D. If C < D, then fourth weighing: compare D with A. If D < A, then order is C < D < A < B. If D > A, then compare D with B. If D < B, then C < A < D < B. If D > B, then C < A < B < D. This seems to require more than five weighings in the worst case. Hmm.Alternatively, maybe using a merge sort approach. Split the stones into two pairs, sort each pair, then merge them.First, compare W vs X: suppose W < X.Second, compare Y vs Z: suppose Y < Z.Now merge the two sorted pairs. To merge, we need to compare the smallest of each pair. So third weighing: compare W vs Y. Suppose W < Y. Then W is the lightest. Then compare X vs Y. If X < Y, then order is W < X < Y < Z. If X > Y, then order is W < Y < X < Z. That would take four weighings. Wait, does that work?Let's walk through an example:1. Weigh W vs X: W < X.2. Weigh Y vs Z: Y < Z.3. Weigh W vs Y: W < Y.4. Weigh X vs Y: suppose X < Y. Then we have W < X < Y < Z. But how do we know Z is the heaviest? From step 2, we know Y < Z. Since X < Y and Y < Z, so X < Z, and W < X < Y < Z. So in four weighings, maybe?But wait, in this case, after four weighings, we have ordered them. But what if in step 4, X > Y?Then after step 4, we have W < Y and X > Y. So order is W < Y < X. But we still need to place Z. We know Y < Z from step 2, but how does Z compare to X? We haven't compared Z and X yet. So fifth weighing: compare X vs Z. If X < Z, then order is W < Y < X < Z. If X > Z, then order is W < Y < Z < X. Therefore, in this case, we need a fifth weighing.So depending on the results of the fourth weighing, we might need a fifth. So in the worst case, it's five weighings. Therefore, the minimal number of weighings required is five. Therefore, the answer is B) 5.But let me check another approach. Maybe using a different initial weighing. For example, compare one stone against the other three in some way. Wait, but how?Alternatively, think of the problem as a sorting tournament. For four players, the number of rounds needed in a knockout tournament is three, but that's to find the winner. Here, we need the full order. So perhaps the merge sort approach is the right way, which as we saw, can take up to five weighings. Therefore, five is the minimal number required in the worst case.Another way: each weighing can eliminate a certain number of permutations. Since we start with 24 permutations, each weighing can ideally split the remaining permutations into groups. But since each weighing has two possible outcomes (since stones are distinct, so no equality), each weighing can at best halve the number of remaining permutations. Therefore, log2(24) ≈ 4.58, so five weighings are necessary. This information-theoretic lower bound suggests that five is the minimum. Since five is an option here, the answer is B.Wait, but some sources say that four weighings are sufficient. Wait, perhaps there's a better algorithm. Let me check.Wait, here's a possible sequence:1. Compare A vs B. Suppose A < B.2. Compare C vs D. Suppose C < D.3. Compare A vs C. Suppose A < C.4. Compare B vs D. Suppose B < D.Now, what do we know? A < B, A < C, C < D, B < D. But we still don't know the relation between B and C. So we might need a fifth weighing: compare B vs C. Then we can determine the order. So again, fifth weighing is needed.Alternatively, maybe arrange the weighings in a way that more information is gleaned. For example:1. Compare A + B vs C + D. Suppose A + B < C + D. So the total of C and D is greater.2. Compare A vs C. Suppose A < C.3. Compare B vs D. Suppose B < D.4. Compare A vs B. Suppose A < B.5. Compare C vs D. Suppose C < D.But even with this, the interactions might not resolve all ambiguities. Hmm. Alternatively, perhaps there's a way to use transitivity. For example, if A < B and B < C, then A < C without needing to weigh them. But unless you have those comparisons, you can't assume transitivity. So the key is to structure the weighings such that transitivity can be used to infer some relations without explicit weighings.But given that four stones have four elements, there are six pairwise comparisons possible. Each weighing gives one comparison. So if we do five weighings, we cover five out of six possible comparisons, but maybe use transitivity for the sixth. However, in the worst case, you might need five comparisons. Wait, but the minimal number of comparisons required to sort four elements is five. For example, as per the Ford-Johnson algorithm, the number is five for n=4. Therefore, the minimal number of weighings is five.Therefore, the answer is B) 5.Final Answerboxed{B}</think>To determine the minimum number of weighings required to order four stones with distinct weights from lightest to heaviest using a balanced scale, we need to consider the number of permutations and the information each weighing provides. 1. Number of Permutations: There are (4! = 24) possible permutations of four stones.2. Information Theory: Each weighing can result in two outcomes (since all stones have distinct weights, equality is impossible). The minimum number of weighings required is determined by the information-theoretic lower bound, which is (lceil log_2(24) rceil approx 5).3. Comparison with Sorting Algorithms: The Ford-Johnson algorithm, a comparison sort, requires five comparisons to sort four elements. This is confirmed by considering the decision tree model, where the height of the tree must be sufficient to cover all permutations.4. Step-by-Step Weighing Strategy: By dividing the stones into pairs and merging the results, we find that in the worst case, five weighings are necessary. For example: - Compare two pairs, then compare the winners and losers, and finally merge the results with additional weighings.Thus, the minimum number of weighings required to order four stones from lightest to heaviest is (boxed{B}).,<think>Okay, so the problem is asking for the minimum number of weighings needed on a balanced scale to order four stones with distinct weights from lightest to heaviest. The options are from 4 to 8. Hmm, let's think about this step by step.First, I know that a balanced scale can compare two groups of items and tell us which side is heavier or if they're equal. Since all stones have distinct weights, we don't have to worry about ties. Each weighing gives us some information, and we need to figure out how much information we need to sort all four stones.Let me recall that sorting algorithms often use comparisons, and the minimum number of comparisons needed to sort n elements is related to the concept of information theory. The idea is that each comparison gives us a certain amount of information (specifically, log2(3) bits per comparison if we consider ternary decisions, but maybe that's complicating things). Alternatively, the decision tree model says that the number of leaves (possible outcomes) must be at least n!, and each internal node represents a comparison with three possible outcomes. So the minimum height of such a tree would be the ceiling of log3(n!)).Wait, n! is the number of permutations, which is 24 for n=4. So log base 3 of 24. Let me compute that. 3^3=27, so log3(24) is a little less than 3, but since we need an integer number of weighings, it would be 3? But wait, the answer options start at 4. Hmm, maybe this approach is not directly applicable here.Alternatively, maybe the balanced scale can compare two stones at a time, giving a binary result (heavier or lighter). Wait, but a balanced scale can compare two groups, not just two stones. So if you compare two groups, each of which can have one or more stones, the scale tells you which group is heavier. That gives more information than just comparing two stones individually.But how does this affect the number of weighings needed? Let me think. For four stones, we need to determine the order, which has 4! = 24 possible permutations. Each weighing can have three possible outcomes: left side heavier, right side heavier, or balanced. Wait, no, if we use a balanced scale, the two groups compared must have the same number of items? Wait, no. A balanced scale just compares the total weight of the two groups. The groups can be of unequal size, but typically, when people talk about balance scales for comparing weights, they might compare equal-sized groups to make it fair. But in this problem, it's not specified that we have to compare equal-sized groups, so maybe we can compare any two subsets.But in the problem statement, it just says "balanced scale", so maybe it's similar to a regular balance scale where you can put any number of stones on each side. Each weighing has three possible outcomes: left heavier, right heavier, or equal. However, since all stones have distinct weights, equality is only possible if the two sides have the same number of stones and the same total weight, which, given distinct weights, is only possible if they are exact subsets. But with distinct weights, even if you have the same number on each side, their total weights will differ unless they are permutations. So actually, equality is impossible here. Wait, but if you have two different subsets, even with the same number of stones, their total weights will be different because all stones have distinct weights. So, in reality, each weighing can only result in left or right being heavier. So maybe each weighing only gives us one bit of information, not ternary. Wait, that contradicts some previous knowledge.Hold on, perhaps there's confusion here. If we use a balance scale to compare two subsets, and since all stones have distinct weights, any two subsets with the same number of stones will have different total weights. So if we compare, say, one stone against another, we get which one is heavier. If we compare two stones against two stones, we can determine which pair is heavier, but the problem is that the heavier pair might not correspond directly to the order of individual stones.Wait, but maybe in this problem, we can choose how to group the stones for each weighing. So each weighing can be designed to maximize the information gained. The key is to figure out the minimal number of weighings needed to sort four stones.Alternatively, perhaps we can model this as a tournament sort or some other method. Let me recall that in decision tree theory, the minimal number of comparisons needed to sort 4 elements is 5. Wait, but that's in the comparison model where each comparison is between two elements. But here, if we can compare groups, maybe we can get more information each time. For example, comparing two pairs of stones can give more information than just comparing two individual stones. But how does that affect the total number of weighings needed?Wait, but if each weighing can be between any two subsets, then each weighing could potentially give more information. For example, if we weigh two stones against the other two, then depending on the result, we can infer certain relationships. Let's try to work this out.Let me denote the stones as A, B, C, D with unknown distinct weights. Our goal is to sort them from lightest to heaviest with the minimal number of weighings.First, perhaps divide and conquer. Let's split them into two pairs: (A vs B) and (C vs D). But since we can only do one weighing at a time, we need to sequence the weighings.Alternatively, compare two stones first. Let's say we compare A and B. Suppose A is lighter than B. Then we know A < B. Then compare C and D. Suppose C < D. Now we have two ordered pairs: A < B and C < D. Then we need to merge these two pairs into a total order. To merge, we might need to compare the middle elements. For example, compare B and C. If B < C, then the order is A < B < C < D. If B > C, then we have A < C < B < D or similar, but need more information. Wait, maybe this approach requires more weighings.Alternatively, think of it as a sorting network. A sorting network for four elements has a certain number of comparators. Each comparator is a comparison that can be done in parallel. In terms of depth, the minimum number of sequential steps. For four elements, the depth is 3. But each step can have multiple comparisons. However, in our case, each weighing is a single comparison (maybe comparing two individual stones or two groups). So the depth would correspond to the number of weighings if each weighing is a single comparison. Wait, but in a sorting network, the number of comparators is 5 for four elements. But each comparator is a comparison between two elements. So if each weighing can only compare two elements, then you need 5 weighings. But wait, the answer choices include 5 as an option. However, maybe if we use group comparisons, we can reduce the number.Alternatively, maybe the answer is 5, as per the decision tree lower bound. Let's compute log2(4!) = log2(24) ≈ 4.58, so the minimal number of weighings if each weighing gives 1 bit is 5. But if each weighing can give more bits (since comparing groups can result in more information), then maybe fewer weighings. However, if each weighing is a ternary decision (left, right, balance), but in reality, since all stones have distinct weights, balance is impossible unless we have equal subsets, which might not be possible. Wait, but if you compare two stones against two stones, even if all are distinct, the total weights can differ, so the scale will tip to one side. Therefore, each weighing can be considered as a comparison between two subsets, with the result being which subset is heavier. Each such weighing can partition the possibility space.The information theory approach: the number of possible permutations is 4! = 24. Each weighing can result in 3^k outcomes, where k is the number of weighings. Wait, no. Wait, if each weighing can have three possible outcomes, then the number of distinct outcomes is 3^k. But if in reality, each weighing can only have two possible outcomes (since equality is impossible), then it's 2^k. However, if the subsets compared have different numbers of elements, then equality is possible? Wait, no. If the two subsets have different numbers of elements, then since all stones have distinct weights, the total weight of each subset is unique. So even if you compare 1 vs 3 stones, the side with more stones is more likely to be heavier, but since weights are distinct, it's not certain. However, the problem is that comparing different numbers of stones could result in either side being heavier. So, in that case, the result isn't necessarily trinary. Wait, no. If you compare any two subsets, regardless of size, the scale can tip left, tip right, or balance. But in our problem, since all stones have distinct weights, the total weights of the subsets can never be equal unless the subsets are identical, which they aren't. Wait, but if the two subsets are different, even with the same number of elements, their total weights will be different. Wait, but if subsets have different numbers of elements, their total weights can still be different. So in any case, comparing two different subsets (whether same size or not) will always result in either left or right being heavier. Therefore, each weighing has only two possible outcomes. So the number of possible outcomes with k weighings is 2^k. We need 2^k >= 24. Solving for k, we get k >= log2(24) ≈ 4.58, so k=5. Therefore, 5 weighings. But this contradicts the idea that group weighings can give more information.Wait, maybe the mistake here is assuming that each weighing only gives a binary outcome, but actually, depending on how you structure the weighings, each weighing can give more information. For example, if you compare two stones against two stones, the result can give information about the relative order of four stones. For instance, if you have A+B vs C+D, and A+B is heavier, then you know that the two heavier stones are more likely to be on the left. But how exactly does this help?Alternatively, perhaps using a method similar to merge sort. Let's break down the stones into pairs and sort each pair, then merge them. For four stones, first sort two pairs: let's say compare A vs B, then C vs D. That's two weighings. Suppose A < B and C < D. Then, to merge the two sorted pairs, you would need to compare B vs D. If B < D, then the last stone is D, but you still need to place C and B. Wait, this seems similar to the merge process. But perhaps more weighings are needed. Let me work through an example.Stones: A, B, C, D. Unknown order.First weighing: A vs B. Suppose A < B.Second weighing: C vs D. Suppose C < D.Now we have A < B and C < D.Third weighing: Compare B vs C. If B < C, then we have A < B < C < D. So that's three weighings. But if B > C, then we have A < C < B < D? Wait, not necessarily. Because if B > C, then we know that C is less than B, but how does C compare to A? We already know A < B and C < D, but C could be between A and B or less than A. Wait, no, in the first weighing, A < B, second C < D. Then if in the third weighing, B > C, then C could be less than A or between A and B. Wait, but we don't know C's relation to A yet. Hmm, so that approach might not be sufficient.Wait, let's say after the first two weighings: A < B and C < D. Then, third weighing: A vs C. If A < C, then we know A is the lightest so far. Then we have A < B, A < C < D. Then we need to figure out where B fits in between C and D. So fourth weighing: B vs C. If B < C, then A < B < C < D. If B > C, then A < C < B < D. So that would be four weighings.Alternatively, if in the third weighing, A > C, then we know C < A < B and C < D. Then, we need to sort A, B, D with respect to C. Hmm, maybe fourth weighing: A vs D. If A < D, then we have C < A < B and C < D. Then need to determine relation between D and B. Fifth weighing: B vs D. So total of five weighings.Alternatively, maybe there's a smarter way. Let's see.Another approach is the method used in tournaments. For four players, a single-elimination tournament takes three rounds, but that might not be directly applicable here. Alternatively, a sorting network. According to some references, the optimal comparison network for sorting four elements has five comparisons. For example, the following sequence:1. Compare A and B.2. Compare C and D.3. Compare the winners of the first two comparisons.4. Compare the loser of the first comparison with the winner of the second comparison.5. Compare the remaining two elements.But this is for a comparison network where each comparison is between two elements. If each weighing can compare groups, perhaps we can do better.Wait, but in our problem, the question is about weighings on a balanced scale, which can compare any two subsets. So maybe each weighing can compare two subsets, each of any size, and the result tells us which subset is heavier. So, for example, we can compare two stones against one stone, or three stones against one, etc. However, since all stones have distinct weights, comparing different numbers might not give straightforward information.But perhaps comparing two stones against two stones is more informative. Let's consider this strategy.First weighing: Compare A + B vs C + D.Case 1: A + B < C + D.This tells us that the total weight of A and B is less than the total of C and D. Since all stones have distinct weights, this implies that the two lightest stones might be A and B, or one of them and one of C or D. Hmm, but it's not straightforward.Alternatively, if we have A + B < C + D, then the heaviest stone must be in C or D, and the lightest stone must be in A or B. But is that necessarily true?Wait, suppose the actual order is A < B < C < D. Then A + B = lightest two, C + D = heaviest two. So A + B < C + D.But if the order is A < C < B < D, then A + B vs C + D would be A + B = (A) + (B) vs (C) + (D). Since C > A and D > B, then C + D > A + B. So in this case, the result is same as before. So this weighing might not distinguish between different orderings where the total of two pairs is the same.Hmm, maybe not the best approach.Alternatively, think in terms of the number of weighings required. Since there are 4 stones, and we need to determine the order, perhaps using a decision tree where each internal node is a weighing.The minimum number of weighings required would be the height of the decision tree, which must be sufficient to have at least 24 leaves (the number of permutations). Each weighing can result in multiple outcomes. If each weighing has three possible outcomes (left, right, equal), but in reality, since equality is impossible, each weighing only has two possible outcomes. Wait, but earlier reasoning suggests that if you compare two subsets (even of different sizes), the result is either left or right. So each weighing is a binary decision. Therefore, the number of possible outcomes with k weighings is 2^k. We need 2^k >= 24. Since 2^4=16 <24 and 2^5=32>24, so k=5. Hence, at least 5 weighings. Therefore, the answer is B) 5. But let me verify this with an example.Alternatively, maybe with some weighings, you can get more information. For example, if you compare two stones against two stones, and depending on which side is heavier, you can infer more. Let's try.First, weigh A+B vs C+D.Case 1: A+B < C+D. Then, we know that the two heavier stones are in C and D, and the two lighter ones are in A and B. But is that necessarily true? Suppose the actual order is A < B < C < D. Then A+B < C+D. But if the order is A < C < B < D, then A + B could be less than C + D. Wait, A=1, C=2, B=3, D=4. Then A + B = 4, C + D = 6, so A + B < C + D. But here, the two heaviest stones are B and D, but B is in the first group. So the initial assumption is wrong. Therefore, comparing A + B vs C + D doesn't necessarily put the two heaviest in one group. Therefore, this approach might not work.Alternatively, another strategy: first, compare A vs B. Suppose A < B.Second, compare B vs C. If B < C, then A < B < C. Then compare C vs D. If C < D, then A < B < C < D. So that took 3 weighings. But if in the second weighing, B > C, then we have A < C < B. Then compare A vs C (already done), but then need to place D. Compare D vs B. If D > B, then order is A < C < B < D. If D < B, compare D vs C. If D > C, then A < C < D < B. If D < C, compare D vs A. Wait, this is getting complicated. But in the worst case, it seems like more than 5 weighings might be needed. Hmm.Alternatively, another method: start by comparing two stones, then compare the next two, then merge them as in merge sort. For four stones, merge sort would require two splits and then a merge. Each merge could take up to n-1 comparisons. But for four elements, merge sort would take 3 + 1 = 4 comparisons? Wait, let's think.First, split into two pairs: compare A vs B and C vs D. That's two weighings. Suppose A < B and C < D. Then merge the two pairs. To merge, you need to compare A vs C. Suppose A < C. Then the next comparison is B vs C. If B < C, then the order is A < B < C < D. If B > C, then A < C < B < D. So that took a total of four weighings. But if in the third weighing, A > C, then we have C < A < B and C < D. Then compare A vs D. If A < D, then compare B vs D. So that could take up to five weighings. Hmm. So in the worst case, five weighings. So the answer is 5. Therefore, option B.But the problem says "at least how many weighings are needed". So we need the minimal number that is sufficient in the worst case. Since there exists a strategy that can sort four stones in five weighings, and since information-theoretic lower bound is log2(24) ≈ 4.58, which rounds up to 5, so five is the minimal number.Therefore, the answer is B) 5.Final Answerboxed{B}</think>To determine the minimum number of weighings needed to order four stones with distinct weights from lightest to heaviest using a balanced scale, we need to consider the information each weighing provides and the total number of permutations (4! = 24).1. Information Theory Approach: Each weighing can result in two possible outcomes (left or right side heavier). The number of possible outcomes with ( k ) weighings is ( 2^k ). We need ( 2^k geq 24 ). Solving for ( k ), we get ( k geq log_2(24) approx 4.58 ), so ( k = 5 ).2. Merge Sort Analogy: Using a merge sort approach, we first compare pairs of stones and then merge the results. - Compare two pairs (A vs B and C vs D) in 2 weighings. - Merge the pairs by comparing the winners and losers, which can take up to 3 more weighings in the worst case.3. Decision Tree Model: The decision tree must have at least 24 leaves (permutations). Each internal node (weighing) has 2 outcomes. The height of the tree (number of weighings) must be sufficient to cover all permutations, leading to a minimum of 5 weighings.Thus, the minimum number of weighings required is (boxed{B}).]

🔑:1. Label the stones: Let the four stones be labeled as ( A, B, C, ) and ( D ).2. Initial weighings: - Weigh ( A ) against ( B ). - Weigh ( C ) against ( D ).3. Determine the lightest pair: - Suppose ( A ) is lighter than ( B ) and ( C ) is lighter than ( D ). - Weigh the lighter ones from each pair: ( A ) against ( C ).4. Identify the lightest stone: - If ( A ) is lighter than ( C ), then ( A ) is the lightest. - If ( C ) is lighter than ( A ), then ( C ) is the lightest.5. Order the remaining stones: - Suppose ( A ) is the lightest. We now need to order ( B, C, ) and ( D ). - Weigh ( B ) against ( C ). - If ( B ) is lighter than ( C ), weigh ( B ) against ( D ). - If ( B ) is lighter than ( D ), the order is ( A, B, D, C ). - If ( D ) is lighter than ( B ), the order is ( A, D, B, C ). - If ( C ) is lighter than ( B ), weigh ( C ) against ( D ). - If ( C ) is lighter than ( D ), the order is ( A, C, D, B ). - If ( D ) is lighter than ( C ), the order is ( A, D, C, B ).6. Minimum number of weighings: - Each weighing provides one result, and with ( n ) weighings, there are ( 2^n ) possible outcomes. - To distinguish between ( 4! = 24 ) possible orderings, we need at least ( log_2(24) approx 4.58 ) weighings. - Since the number of weighings must be an integer, we round up to 5.Thus, at least 5 weighings are needed to order the four stones.The final answer is ( boxed{5} )

🧠:[<think>Okay, so I need to find the value of x₀ where the tangent lines to the two curves y = x² - 1 and y = 1 - x³ are parallel. Hmm, parallel tangent lines mean that their slopes are equal at those points, right? So, I guess I need to find the derivatives of both functions and set them equal to each other at x = x₀. Let me write that down step by step.First, let's consider the first curve: y = x² - 1. To find the slope of the tangent line at any point x, I need to take the derivative of y with respect to x. The derivative of x² is 2x, and the derivative of -1 is 0. So, the derivative dy/dx for the first curve is 2x. Therefore, at x = x₀, the slope of the tangent line is 2x₀.Now, moving on to the second curve: y = 1 - x³. Similarly, the derivative here would be the derivative of 1, which is 0, minus the derivative of x³, which is 3x². So, dy/dx for the second curve is -3x². Therefore, at x = x₀, the slope of the tangent line is -3x₀².Since the tangent lines are parallel, their slopes must be equal. That means I can set the two derivatives equal to each other:2x₀ = -3x₀².Okay, now I need to solve this equation for x₀. Let me rearrange it to bring all terms to one side. Adding 3x₀² to both sides:3x₀² + 2x₀ = 0.Hmm, factoring out an x₀ from the left side:x₀(3x₀ + 2) = 0.So, this product equals zero, which means either x₀ = 0 or 3x₀ + 2 = 0. Solving the second equation:3x₀ + 2 = 0 3x₀ = -2 x₀ = -2/3.Therefore, the possible solutions are x₀ = 0 and x₀ = -2/3.But wait, let me verify if both of these are valid. Let's plug them back into the original derivative equations to check.First, for x₀ = 0:Slope for the first curve: 2*0 = 0. Slope for the second curve: -3*(0)² = 0. So, both slopes are 0. That works. The tangent lines are horizontal and hence parallel.Now, for x₀ = -2/3:Slope for the first curve: 2*(-2/3) = -4/3. Slope for the second curve: -3*(-2/3)². Let's compute that. First, square (-2/3): (-2/3)² = 4/9. Then multiply by -3: -3*(4/9) = -12/9 = -4/3. So, both slopes are -4/3. That also checks out. The tangent lines have the same slope and are parallel.Therefore, both x₀ = 0 and x₀ = -2/3 are solutions. Wait, but the problem says "the value of x₀ is ______". Hmm, maybe I need to check if both solutions are acceptable or if there's something I missed.Let me read the problem again: "the tangent line to the curve y = x² - 1 at the point x = x₀ is parallel to the tangent line to the curve y = 1 - x³ at the point x = x₀". So, both tangent lines are at the same x₀? Wait, does that mean that both curves have their tangent lines evaluated at the same x₀ value? Yes, that's what the problem says. So, x₀ is the x-coordinate where both tangent lines are taken, and their slopes must be equal. So, the two solutions we found are both valid because at both x₀ = 0 and x₀ = -2/3, the slopes of the tangent lines on both curves are equal. So, does the problem accept multiple answers? The problem says "the value of x₀ is ______". Maybe I need to check if both are correct. Let me check again.Wait, if x₀ is 0, then on the first curve, the point is (0, 0² -1) = (0, -1), and the tangent line there is horizontal. On the second curve, the point is (0, 1 - 0³) = (0, 1), and the tangent line there is also horizontal. So, both tangent lines are horizontal, hence parallel. Similarly, for x₀ = -2/3, the points are (-2/3, (-2/3)² -1) and (-2/3, 1 - (-2/3)³). Let me compute those:First curve: y = (-2/3)² - 1 = 4/9 - 1 = -5/9. Second curve: y = 1 - (-2/3)³ = 1 - (-8/27) = 1 + 8/27 = 35/27. The tangent lines at these points both have slope -4/3, so they are parallel.Therefore, both x₀ = 0 and x₀ = -2/3 are valid solutions. However, the problem says "the value of x₀ is ______", which might imply a single answer. Did I make a mistake here?Wait, let me check the problem statement again: "the tangent line to the curve y = x² - 1 at the point x = x₀ is parallel to the tangent line to the curve y = 1 - x³ at the point x = x₀". So, they are both at the same x₀. So, in this case, both x₀ = 0 and x₀ = -2/3 satisfy the condition. However, the problem might be expecting both answers, but maybe the original problem expects two answers? Wait, but the user hasn't specified multiple answers. Let me check the original problem again.Wait, the user wrote: "the value of x₀ is ______". The word "value" is singular. Hmm, but we have two solutions. Maybe I did something wrong here.Wait, let me re-examine the equations. The equation I set up was 2x₀ = -3x₀². Then moving all terms to one side: 3x₀² + 2x₀ = 0. Factoring: x₀(3x₀ + 2) = 0. So solutions are x₀ = 0 and x₀ = -2/3. That seems correct. So perhaps the problem is expecting both answers? But how is the answer supposed to be presented? If the user is expecting a single answer, but there are two, maybe there's a mistake in my approach.Wait, let me think again. Maybe the problem is written in a way that x₀ is unique? Let me check the functions. The first function is a parabola opening upwards, and the second is a cubic function that is decreasing for positive x and has an inflection point. The tangent lines could intersect at two points where their slopes are equal. Hmm. Let me graph both functions mentally.For y = x² -1, it's a parabola with vertex at (0, -1). The derivative is 2x, so the slope becomes more negative as x decreases and more positive as x increases. For y = 1 - x³, it's a cubic that is decreasing for all x because the derivative is -3x², which is always non-positive (since x² is non-negative, multiplied by -3 gives non-positive). So, the cubic is decreasing everywhere. The slope of the cubic is zero only at x = 0, where it has a horizontal tangent, and the slope becomes more negative as x moves away from zero in either direction.Wait, but the cubic's slope is -3x², which is always non-positive. The parabola's slope is 2x, which can be positive or negative. So, when do these slopes coincide? When 2x = -3x². So, solving 3x² + 2x = 0 gives x = 0 or x = -2/3. At x = 0, the cubic has slope 0 (horizontal tangent), and the parabola also has slope 0. At x = -2/3, the cubic has slope -3*(-2/3)^2 = -3*(4/9) = -12/9 = -4/3, and the parabola has slope 2*(-2/3) = -4/3. So both slopes are equal there. Therefore, both x₀ = 0 and x₀ = -2/3 are valid.But the problem says "the value of x₀ is ______". If it's expecting multiple answers, maybe separated by commas. But in the original problem statement, the user hasn't indicated that. However, in the initial problem statement, if it's from a textbook or somewhere, perhaps they only expect one answer? Wait, perhaps I made a mistake in interpreting the problem. Let me check again.Wait, the problem says: "the tangent line to the curve y = x² - 1 at the point x = x₀ is parallel to the tangent line to the curve y = 1 - x³ at the point x = x₀". So, both tangent lines are at the same x₀. Therefore, we need to find x₀ such that the slopes of both curves at that x₀ are equal. Which gives us two solutions. Therefore, unless there is a restriction in the problem that I haven't considered, both solutions are valid.But maybe in the original problem, due to some miscalculation, there's only one solution. Wait, let me verify my calculations again.Starting over:For the first curve, y = x² -1. The derivative is 2x. At x = x₀, slope is 2x₀.For the second curve, y = 1 - x³. The derivative is -3x². At x = x₀, slope is -3x₀².Set them equal: 2x₀ = -3x₀².Bring all terms to one side: 3x₀² + 2x₀ = 0.Factor: x₀(3x₀ + 2) = 0.Solutions: x₀ = 0 or 3x₀ + 2 = 0 => x₀ = -2/3.This is straightforward. So, the answer should be both x₀ = 0 and x₀ = -2/3. But the problem says "the value of x₀ is ______". Maybe the problem expects both answers, but the way the user wrote the question here, it's in a singular form. Maybe in the original problem, there's a typo or maybe only one solution is expected. Alternatively, perhaps the user made a mistake in the problem statement.Wait, let me check once more: "the tangent line to the curve y = x² -1 at the point x = x₀ is parallel to the tangent line to the curve y = 1 - x³ at the point x = x₀". So, same x₀ for both curves. So, we have two valid solutions. Unless the problem is from a source where only one solution is correct, but mathematically, both are correct.Alternatively, maybe I need to check if at x₀ = 0, the tangent lines are not only parallel but also coinciding? Let me check the tangent lines.For the first curve at x₀ = 0: the point is (0, -1), slope is 0. So, the tangent line is y = -1.For the second curve at x₀ = 0: the point is (0, 1), slope is 0. So, the tangent line is y = 1.These are two different horizontal lines, both with slope 0, hence parallel. So, they are parallel but not the same line. So, that's valid. So, x₀ = 0 is a valid solution.Similarly, at x₀ = -2/3, both tangent lines have slope -4/3 but pass through different points. So, they are distinct parallel lines. Therefore, both solutions are valid.Therefore, unless the problem has additional constraints, both x₀ = 0 and x₀ = -2/3 are correct. But the problem statement says "the value of x₀ is ______", which is singular. Maybe in the original problem, there was a restriction like x₀ > 0 or something, but the user didn't mention it. Alternatively, maybe it's a mistake in the problem.Alternatively, maybe I made a mistake in interpreting the slopes. Let me check once more.First curve derivative: 2x. At x₀, slope is 2x₀.Second curve derivative: -3x². At x₀, slope is -3x₀².Set equal: 2x₀ = -3x₀².Yes, that's correct. So, solving 2x₀ + 3x₀² = 0. So, x₀(3x₀ + 2) = 0. Hence, solutions x₀ = 0 and x₀ = -2/3. So, the answer should be both.But the problem is presented as if there is a single answer. Maybe the user wants both answers separated by commas. However, in the instructions, the user says "put your final answer within boxed{}". The box is usually for a single answer. But maybe multiple answers can be boxed. Let me check standard practice. In many cases, if there are multiple solutions, they can be presented as boxed{0} and boxed{-dfrac{2}{3}}, but depending on the problem's expectation. But since the problem says "the value of x₀ is", maybe the answer is both. Alternatively, perhaps the problem expects both, but written together.Alternatively, maybe I'm overcomplicating. The problem might have a unique solution, so perhaps I made a mistake. Wait, let me check the arithmetic again.Equation: 2x₀ = -3x₀² => 3x₀² + 2x₀ = 0 => x₀(3x₀ + 2) = 0 Solutions: x₀ = 0 or x₀ = -2/3. Yes, correct.Therefore, the answer is both 0 and -2/3. But since the problem says "the value of x₀ is", maybe the answer is both. However, since the user hasn't specified multiple answers, but the system might accept multiple boxed answers. Alternatively, perhaps the problem is from a source where x₀ is expected to be negative, but that's speculative.Alternatively, maybe when the problem says "parallel", but not "distinct", so x₀ = 0 could be considered, but perhaps in the context, x₀ is non-zero? However, there is no such indication.Alternatively, let's check if x₀ = 0 is a valid solution. The tangent lines there are horizontal, as I saw before. So, they are indeed parallel. So, no reason to discard it. Therefore, mathematically, there are two solutions.But the problem is presented as a fill-in-the-blank with a single box. Maybe in the original problem, the answer is both 0 and -2/3. So, perhaps the answer should be written as two boxed answers. However, in the problem statement, the user wrote "put your final answer within boxed{}", so maybe they expect multiple values inside the box. For example, boxed{0} and boxed{-dfrac{2}{3}}. But I need to check the instructions again.Looking back at the user's message: "put your final answer within boxed{}". So, LaTeX with a single box. Therefore, if there are two solutions, they should be presented together in one box. So, maybe separated by a comma. For example, boxed{0, -dfrac{2}{3}}.Alternatively, since sometimes multiple answers are written as a list inside the box. So, I think that's the correct approach here. Hence, the value of x₀ is 0 and -2/3. Therefore, the answer is both, presented within a single box.Therefore, the final answer is boxed{0} and boxed{-dfrac{2}{3}}. Wait, but can I have two boxed answers in one line? In some contexts, people write multiple boxed answers. But the instruction says "put your final answer within boxed{}", so maybe the user expects a single box. Hmm, this is a bit confusing.Alternatively, maybe I made a mistake in the problem's interpretation. Wait, the problem says "the tangent line to the curve y = x² -1 at the point x = x₀ is parallel to the tangent line to the curve y = 1 - x³ at the point x = x₀". So, same x₀ for both curves. Therefore, the answer is two values of x₀ where this occurs. Therefore, both 0 and -2/3. But maybe the problem expects writing both in the box.But let me check online if similar problems expect multiple answers. For example, if solving a quadratic equation with two solutions, the answer is presented with both solutions in one box, separated by commas. For example, boxed{0, -dfrac{2}{3}}.Alternatively, in some cases, they might use curly braces to denote multiple answers, but I think commas are more standard. Therefore, I will present both solutions separated by a comma inside a single box.Therefore, the value of x₀ is boxed{0} and boxed{-dfrac{2}{3}}, but since the user might expect both answers in one box, maybe boxed{0 text{ and } -dfrac{2}{3}}.But in standard math notation, if multiple answers are required, we can list them separated by commas within the same box. So, boxed{0, -dfrac{2}{3}}.I think that's acceptable. So, even though the problem says "the value", which is singular, the correct answer has two values, so we present both.Final AnswerThe values of ( x_0 ) are boxed{0} and boxed{-dfrac{2}{3}}.Alternatively, if the problem expects both answers in a single box:Final AnswerThe values of ( x_0 ) are boxed{0 text{ and } -dfrac{2}{3}}.But according to standard practice, multiple answers can be boxed together with commas:Final AnswerThe value of ( x_0 ) is boxed{0, -dfrac{2}{3}}.But given the original problem's wording, which uses "the value", it's possible that there's an error, but given the mathematical results, there are two solutions. Therefore, I will present both in the box separated by commas.Final Answerboxed{0} text{ and } boxed{-dfrac{2}{3}}Wait, but in LaTeX, if I put them in separate boxes, they will appear as two boxed answers. But the user's instruction was "put your final answer within boxed{}". So, maybe the user expects two separate boxed answers. However, the system that is going to receive this answer might only accept one box. Hmm.Alternatively, let me check the initial problem statement again. The user wrote: "the value of x₀ is ______". Since "value" is singular, but our solution yields two values, this might imply that the problem expects both answers. But how to reconcile that.Alternatively, perhaps the problem is mistyped, and instead of both tangent lines at x = x₀, maybe they are at different points? But the problem as stated says both are at x = x₀.Alternatively, maybe the problem is from a non-English source where "value" can be plural. But in English, "value" is singular, "values" is plural. So, it's possible that the problem has a typo, and it should be "values".But given the problem as stated, with "value" in singular, but mathematically, there are two solutions. Therefore, maybe the correct answer is both, and the problem has a typo. Therefore, I will present both answers in separate boxes.Final AnswerThe values of ( x_0 ) are boxed{0} and boxed{-dfrac{2}{3}}.But the user might want them in a single box. Hmm. Alternatively, check if the problem is from a specific source. But since I don't have that context, I'll proceed with the two-box approach, since that's the most accurate.However, looking back at the user's exact problem statement: "the value of x₀ is ______". If I have to follow strictly, maybe I should consider that there's a mistake in the problem, but since we derived two solutions, perhaps they expect both. But since in Chinese or other languages sometimes singular is used for multiple answers, but in English, it's a bit conflicting.Alternatively, maybe the problem is designed such that x₀ is unique, and my solution is wrong. Let me double-check.Equation: 2x₀ = -3x₀² Solutions: x₀ = 0 and x₀ = -2/3. This is correct. Therefore, the problem must have two solutions. Therefore, the answer is both. So, to present both in boxes, as per the user's instruction.Final Answerboxed{0} and boxed{-dfrac{2}{3}}But in the guidelines, the user says "put your final answer within boxed{}". So, if there are two answers, maybe separate boxes. However, some systems might not accept multiple boxes. So, in some cases, people list multiple answers separated by commas inside a single box. For example:boxed{0, -dfrac{2}{3}}This is also a common way to present multiple answers. Given that the user's instruction is to put the final answer within a single box, I think that is the way to go. So, I'll adjust my answer to that.Final Answerboxed{0, -dfrac{2}{3}}</think>To find the value of ( x_0 ) where the tangent lines to the curves ( y = x^2 - 1 ) and ( y = 1 - x^3 ) at ( x = x_0 ) are parallel, we need to set their derivatives equal at ( x = x_0 ).1. For the curve ( y = x^2 - 1 ): - The derivative is ( frac{dy}{dx} = 2x ). - At ( x = x_0 ), the slope of the tangent line is ( 2x_0 ).2. For the curve ( y = 1 - x^3 ): - The derivative is ( frac{dy}{dx} = -3x^2 ). - At ( x = x_0 ), the slope of the tangent line is ( -3x_0^2 ).Setting the slopes equal gives the equation:[ 2x_0 = -3x_0^2 ]Rearranging and solving for ( x_0 ):[ 3x_0^2 + 2x_0 = 0 ][ x_0(3x_0 + 2) = 0 ]This gives two solutions:[ x_0 = 0 quad text{or} quad 3x_0 + 2 = 0 implies x_0 = -frac{2}{3} ]Both solutions are valid as they result in equal slopes for the tangent lines of both curves at the respective ( x_0 ) values.Thus, the values of ( x_0 ) are:[boxed{0} text{ and } boxed{-dfrac{2}{3}}]However, if the answer needs to be in a single box, it can be written as:[boxed{0, -dfrac{2}{3}}],<think>Okay, let's see. I need to find the value of x₀ where the tangent lines to the two curves y = x² - 1 and y = 1 - x³ are parallel. Hmm, tangent lines being parallel means their slopes are equal, right? So, first, I should find the derivatives of both functions because the derivative at a point gives the slope of the tangent line there.Starting with the first function, y = x² - 1. The derivative of this with respect to x is dy/dx = 2x. That's straightforward. So, at x = x₀, the slope of the tangent line is 2x₀.Now, the second function is y = 1 - x³. Let's find its derivative. The derivative of 1 is 0, and the derivative of -x³ is -3x². So, dy/dx = -3x². Therefore, at x = x₀, the slope of the tangent line is -3x₀².The problem states that these two tangent lines are parallel, which means their slopes must be equal. So, I can set the two derivatives equal to each other:2x₀ = -3x₀²Now, I need to solve this equation for x₀. Let's rearrange the terms. Bringing all terms to one side:3x₀² + 2x₀ = 0Wait, actually, if I move 2x₀ to the right side, it would be:3x₀² = -2x₀But maybe it's better to factor out x₀. Let's see:x₀(3x₀ + 2) = 0So, this product equals zero when either x₀ = 0 or 3x₀ + 2 = 0. Solving 3x₀ + 2 = 0 gives x₀ = -2/3.Therefore, the solutions are x₀ = 0 or x₀ = -2/3.Wait, but I should check these solutions to make sure they actually work. Let's substitute them back into the original equation for the slopes.First, x₀ = 0:Slope from first function: 2*(0) = 0Slope from second function: -3*(0)² = 0Okay, both slopes are 0, so they are parallel. That works.Now, x₀ = -2/3:Slope from first function: 2*(-2/3) = -4/3Slope from second function: -3*(-2/3)² = -3*(4/9) = -12/9 = -4/3Yep, both slopes are -4/3, so they are equal. That works too.Wait, so there are two solutions here? The problem says "the value of x₀", which suggests a single answer. Maybe I made a mistake in interpreting the problem? Let me check again.The problem states: "the tangent line to the curve y = x² - 1 at the point x = x₀ is parallel to the tangent line to the curve y = 1 - x³ at the point x = x₀". So, both tangent lines are at the same x₀. That is, for the same x₀, the tangent lines on both curves are parallel. So, the x₀ where this happens. So, in this case, both x₀ = 0 and x₀ = -2/3 satisfy the condition. Therefore, are both answers acceptable?But the question says "the value of x₀ is ______". Hmm, maybe there's only one answer? Wait, maybe I made a mistake in solving the equation.Let me re-derive the steps. The slopes must be equal, so 2x₀ = -3x₀². Then, moving everything to one side: 3x₀² + 2x₀ = 0. Factoring: x₀(3x₀ + 2) = 0. So, x₀ = 0 or x₀ = -2/3. So, that's correct. Therefore, there are two solutions. But the problem might be expecting both answers? Or maybe in the original problem, there's some restriction on x₀? Let me check again.Wait, the problem says "the tangent line to the curve y = x² - 1 at the point x = x₀ is parallel to the tangent line to the curve y = 1 - x³ at the point x = x₀". So, same x₀ for both curves. Therefore, since we found two solutions, both x₀ = 0 and x₀ = -2/3, both should be valid. But the way the question is phrased, "the value of x₀ is ______", it might be expecting multiple answers. But in the original problem statement, maybe there's a typo or maybe I misread something?Wait, the problem was presented as: "the tangent line to the curve y=x^2-1 at the point x=x₀ is parallel to the tangent line to the curve y=1-x^3 at the point x=x₀", so same x₀ for both curves. Therefore, solving 2x₀ = -3x₀² gives two solutions. Therefore, both x₀ = 0 and x₀ = -2/3.But perhaps the problem is expecting both answers? Let me check again.Wait, when I substitute x₀ = 0, both tangent lines are horizontal (slope 0). So, for the first curve at x = 0, y = (0)^2 - 1 = -1, so the tangent line is y = 0*(x - 0) -1 = -1. For the second curve at x = 0, y = 1 - 0 = 1, so the tangent line is y = 0*(x - 0) + 1 = 1. These are both horizontal lines, so they are parallel. So that works.For x₀ = -2/3:First curve: y = (-2/3)^2 - 1 = 4/9 - 1 = -5/9. The tangent line has slope -4/3, so the equation is y - (-5/9) = -4/3(x - (-2/3)), which simplifies to y + 5/9 = -4/3(x + 2/3).Second curve: y = 1 - (-2/3)^3 = 1 - (-8/27) = 1 + 8/27 = 35/27. The tangent line has slope -4/3, so equation is y - 35/27 = -4/3(x - (-2/3)), which is y - 35/27 = -4/3(x + 2/3).These two lines both have slope -4/3, so they are indeed parallel. So, both x₀ = 0 and x₀ = -2/3 are valid solutions. Therefore, the answer should be both values. However, the problem asks "the value of x₀ is ______". If the answer requires multiple answers, they might be expected to list both. But the problem might have been designed with only one solution, perhaps I made a mistake.Wait, let me check the algebra again.Given 2x₀ = -3x₀².Moving all terms to left-hand side:3x₀² + 2x₀ = 0.Factoring:x₀(3x₀ + 2) = 0.Solutions: x₀ = 0 or 3x₀ + 2 = 0 => x₀ = -2/3.Yes, that's correct. So, two solutions. But the problem might be expecting both. However, the user is using singular "the value". Maybe there was a typo, or maybe the problem is from a source where only one solution is expected. Wait, perhaps there's a mistake in my calculation? Let me check once again.First derivative: 2x. Second derivative: -3x². Set equal: 2x = -3x². Solutions x=0 and x= -2/3. Yes. Therefore, there are two solutions. So, maybe the answer is both 0 and -2/3. But the problem is presented as "the value of x₀ is ______". If this is from a textbook or a problem set, maybe the answer requires both values. But in the original problem statement, maybe the user made a typo and meant x₁ and x₂, but as given, it's the same x₀ for both curves. So, the answer is two values. However, in the instructions, the user says "put your final answer within boxed{}", which usually indicates a single answer. Maybe I missed something?Wait, perhaps when x₀ = 0, the tangent lines are both horizontal, but they are different lines (y = -1 and y = 1). Similarly, for x₀ = -2/3, the tangent lines are different but have the same slope. So, they are parallel. The problem doesn't require the tangent lines to be the same, just parallel. Therefore, both x₀ = 0 and x₀ = -2/3 are valid. So, unless the problem has a restriction on x₀, both solutions are acceptable. But the problem says "the value of x₀ is ______", which might expect multiple answers. However, in many cases, such problems sometimes have only one solution, so perhaps I need to check again.Wait, no, the algebra is straightforward. The equation 2x₀ = -3x₀² has two real solutions. Therefore, the answer is two values. But how to present them? In the box, if the system expects multiple answers, maybe separated by commas. But the user might expect both. Wait, but maybe the original problem had a different setup? Let me confirm.Wait, the problem is stated exactly as: "the tangent line to the curve y=x^2-1 at the point x=x₀ is parallel to the tangent line to the curve y=1-x^3 at the point x=x₀". So, same x₀ for both curves, so yes, two solutions. Therefore, the answer is x₀ = 0 and x₀ = -2/3. Therefore, maybe the answer is both. However, in the instructions, the user says "put your final answer within boxed{}", which typically is for a single answer. Maybe in the original problem, only one is correct? Wait, maybe I made a mistake in the calculation.Wait, let me check x₀ = 0. For the first curve, derivative is 0, for the second curve derivative is 0. So, yes, equal. For x₀ = -2/3, the first derivative is 2*(-2/3) = -4/3, second derivative is -3*(-2/3)^2 = -3*(4/9) = -12/9 = -4/3. So, equal. Therefore, both are correct.Therefore, the problem has two solutions, x₀ = 0 and x₀ = -2/3. However, the way the question is phrased, "the value of x₀ is ______", which is singular, so maybe it's expecting both answers. But how to present that? Maybe as a comma-separated list inside the box. But I need to check the user's instructions again.The user said: "put your final answer within boxed{}". Typically, boxed{} is used for a single answer. But if there are two answers, we can box them both. For example, boxed{0} and boxed{-dfrac{2}{3}}. But the system might not accept multiple boxes. Alternatively, inside a single box: boxed{0 text{ and } -dfrac{2}{3}}.But the original problem might have a different intended answer. Wait, maybe I need to check if x₀ = 0 is valid. Let's think again. For x₀ = 0, the tangent lines are both horizontal, as we saw. They are parallel, so that's correct. So, there are indeed two solutions.Therefore, perhaps the answer is both x₀ = 0 and x₀ = -2/3. But since the problem says "the value of x₀ is ______", which is singular, maybe it's a mistake in the problem statement. Alternatively, maybe the problem is from a source where only one answer is expected. Alternatively, maybe there's a miscalculation here.Wait, let me check the problem statement again. Maybe I misread the curves? The first curve is y = x² - 1, second is y = 1 - x³. Yes, derivatives are 2x and -3x². Setting equal: 2x = -3x². Yes. So, two solutions. Therefore, the answer should include both. But the user might expect both answers. Therefore, I'll present both solutions.But since the user asked for the final answer within a single box, maybe separated by a comma. In some contexts, multiple answers are boxed together. For example, boxed{0} and boxed{-dfrac{2}{3}}, but maybe written as boxed{0, -dfrac{2}{3}}.I think that's acceptable. So, the final answer is two values: 0 and -2/3.Final AnswerThe values of (x_0) are (boxed{0}) and (boxed{-dfrac{2}{3}}).Alternatively, if the system requires a single box:boxed{0} and boxed{-dfrac{2}{3}}But based on the problem statement asking for "the value", which is singular, but there are two solutions, maybe there was a typo. However, given the mathematical working, two solutions are correct. So, to present both answers properly, perhaps as two separate boxed answers. But if the system allows only one box, then comma-separated.But given that the user hasn't specified, and usually problems like this may expect all correct answers, I think providing both boxed answers is appropriate.Final AnswerThe values of (x_0) are (boxed{0}) and (boxed{-dfrac{2}{3}}).</think>To find the value of (x_0) where the tangent lines to the curves (y = x^2 - 1) and (y = 1 - x^3) at (x = x_0) are parallel, we need to ensure their slopes are equal. 1. The derivative of (y = x^2 - 1) is (2x). At (x = x_0), the slope is (2x_0).2. The derivative of (y = 1 - x^3) is (-3x^2). At (x = x_0), the slope is (-3x_0^2).Setting these slopes equal gives:[2x_0 = -3x_0^2]Rearranging and factoring out (x_0):[3x_0^2 + 2x_0 = 0 implies x_0(3x_0 + 2) = 0]This yields two solutions:[x_0 = 0 quad text{or} quad 3x_0 + 2 = 0 implies x_0 = -frac{2}{3}]Both solutions are verified by substituting back into the slope equations:- For (x_0 = 0), both slopes are 0.- For (x_0 = -frac{2}{3}), both slopes are (-frac{4}{3}).Thus, the values of (x_0) are (boxed{0}) and (boxed{-dfrac{2}{3}}).]

🔑:Since for the function y=x^2-1, we have y'=2x, and for the function y=1-x^3, we have y'=-3x^2, therefore, according to the problem, we get 2x_0=-3x_0^2, solving this gives x_0=0 or x_0=-frac{2}{3}. The slope of the tangent line to a function at a point equals the derivative of the function at that point, hence the derivatives of the two functions at x_0 are equal. This problem mainly tests the calculation of the derivative of a function at a certain point, utilizing the fact that the slope of the tangent line to a function at a point equals the derivative of the function at that point, and is considered a basic question.Thus, the value of x_0 is boxed{0 text{ or } -frac{2}{3}}.

🧠:[<think>Okay, let me try to figure out how to solve this problem. So, we have a complex number z that satisfies the equation (z)/(1 + 2i) = 1 - 2i. We need to find the imaginary part of z. Hmm, right. Let me recall how complex numbers work. First, I remember that when you have a complex number in the denominator, one common technique is to multiply the numerator and the denominator by the complex conjugate of the denominator to rationalize it. Maybe that will come into play here. Let me see. The equation given is z divided by (1 + 2i) equals 1 - 2i. So, if I can solve for z, then I can find its imaginary part. Let me write down the equation:z / (1 + 2i) = 1 - 2iSo, to solve for z, I can multiply both sides of the equation by (1 + 2i). That should isolate z on the left side. Let me do that step by step.Multiplying both sides by (1 + 2i):z = (1 - 2i)(1 + 2i)Okay, so now I need to compute the product of (1 - 2i) and (1 + 2i). Wait a second, isn't this a product of a binomial and its conjugate? Let me check. The first term is (1 - 2i), and the second is (1 + 2i). Yes, they are conjugates because the imaginary parts have opposite signs. I remember that when you multiply a complex number by its conjugate, you get a real number. Specifically, (a + bi)(a - bi) = a² + b². Let me verify that with these numbers. Here, a is 1 and b is 2. So, (1 + 2i)(1 - 2i) = 1² - (2i)². Wait, is that right? Actually, no. Wait, (a + b)(a - b) = a² - b², right. But in this case, b is 2i. So, (1 + 2i)(1 - 2i) = 1² - (2i)². Let me compute that.1 squared is 1. (2i) squared is 4i². Since i squared is -1, that becomes 4*(-1) = -4. So, substituting back, 1 - (-4) = 1 + 4 = 5. So, the product is 5. Therefore, z = 5. Wait, that's it? So z is just 5? Then the imaginary part of z would be 0. But that seems too straightforward. Let me check my steps again.Original equation: z/(1 + 2i) = 1 - 2i. Multiply both sides by (1 + 2i):z = (1 - 2i)(1 + 2i). That product is 1*1 + 1*2i - 2i*1 - 2i*2i. Let's expand it term by term:First term: 1*1 = 1Second term: 1*2i = 2iThird term: -2i*1 = -2iFourth term: -2i*2i = -4i²So adding all terms together: 1 + 2i - 2i -4i².Simplify the terms: 1 + (2i - 2i) + (-4i²) = 1 + 0 + (-4i²) = 1 - 4i².Since i² = -1, then -4i² = -4*(-1) = 4. So, 1 + 4 = 5. So, z = 5. So, indeed, z is 5. Therefore, the imaginary part of z is 0. But wait, the problem says "Given the complex number z..." so maybe I made a mistake here. Let me double-check.Alternatively, maybe the problem is more complicated. Let me check the original equation again. z/(1 + 2i) = 1 - 2i. Multiply both sides by (1 + 2i): z = (1 - 2i)(1 + 2i). That gives z = 5. So z is 5. Therefore, imaginary part is 0. Hmm. That seems correct, but maybe the problem is expecting me to think differently? Wait, if z is 5, then yes, the imaginary part is 0. Is there another way to approach this problem? Let me try another method.Alternatively, suppose I let z = a + bi, where a and b are real numbers. Then, substitute into the equation (z)/(1 + 2i) = 1 - 2i.So, (a + bi)/(1 + 2i) = 1 - 2i.Multiply both sides by (1 + 2i):a + bi = (1 - 2i)(1 + 2i). Wait, same as before. So, same result. So, a + bi = 5. Therefore, a = 5 and b = 0. Therefore, imaginary part is 0. So, that's consistent.Alternatively, maybe I can compute z by first representing 1/(1 + 2i) and then multiplying by z. Wait, but the equation is z divided by (1 + 2i) equals 1 - 2i. So, z = (1 - 2i)*(1 + 2i). Which again is 5. So, yes, the answer is imaginary part 0. Hmm.Wait, but let me check if maybe I misread the problem. The problem says "the imaginary part of z". So, if z is 5, which is a real number, then its imaginary part is indeed 0. So, the answer is 0. But maybe the problem expects some steps. Let me verify once again.Alternatively, perhaps the user intended z divided by (1 + 2i) to equal 1 - 2i, but maybe there's a different interpretation. Let me parse the equation again. It's written as z/(1 + 2i) = 1 - 2i. So, z over (1 + 2i) equals 1 - 2i. So, solving for z gives z = (1 - 2i)(1 + 2i). That product is 5, as we saw. So, yes, z is 5, imaginary part is 0.Alternatively, maybe I can approach it by rationalizing the denominator. Let me try that.Starting with z/(1 + 2i) = 1 - 2i. To solve for z, multiply both sides by (1 + 2i):z = (1 - 2i)(1 + 2i). As we saw, that product is 5. So, z is 5. Alternatively, if I didn't recognize that (1 - 2i)(1 + 2i) is 5, I could do the multiplication step by step:(1 - 2i)(1 + 2i) = 1*1 + 1*2i - 2i*1 - 2i*2i= 1 + 2i - 2i -4i²Simplify:The 2i and -2i terms cancel out, leaving 1 -4i².Since i² = -1, then -4i² = 4, so total is 1 + 4 = 5. Therefore, z = 5. Therefore, the imaginary part is 0. Wait, so all methods lead to z = 5. So, the answer should be 0. But maybe I need to check if there's any possible mistake here. Let me see.Wait, maybe the problem was written differently. Let me check again: "Given the complex number z that satisfies the equation z/(1 + 2i) = 1 - 2i (where i is the imaginary unit), find the imaginary part of z."Yes, that's exactly what's written. So, solving for z gives 5, imaginary part is 0. Maybe the problem is straightforward. Alternatively, maybe there's a typo, but given the problem as stated, z is 5, so the imaginary part is 0. Alternatively, could the equation be z divided by (1 + 2i) equals 1 - 2i. Let me check with another example. Suppose z/(a + bi) = c + di. Then z = (c + di)(a + bi). So, in our case, c = 1, d = -2, a = 1, b = 2. So z = (1 - 2i)(1 + 2i) = 1*1 + 1*2i -2i*1 -2i*2i. Same steps, same result. So, 5.Alternatively, perhaps the problem is expecting me to find the imaginary part without solving for z. Let me see. Let's suppose I didn't realize that multiplying both sides by the denominator would help. Let's try cross-multiplying. Wait, that's exactly the same as what I did. Alternatively, write z = (1 - 2i)(1 + 2i). Then expand. As before, which is 5. Alternatively, maybe the problem is intended to use substitution. Let me think. For example, let me represent z as a complex number x + yi, then plug into the equation:(x + yi)/(1 + 2i) = 1 - 2i.Multiply both sides by (1 + 2i):x + yi = (1 - 2i)(1 + 2i) = 5. So, x + yi = 5. Therefore, x = 5 and y = 0. So, imaginary part y is 0. Therefore, regardless of the method, the answer is 0. Hmm, so perhaps the answer is 0. But let me check if there's a common mistake here. For example, sometimes when people multiply complex numbers, they might forget that i² is -1. Let me verify that in the calculation. In the expansion of (1 - 2i)(1 + 2i), the cross terms are 2i and -2i, which cancel. The last term is -4i². Since i² is -1, -4*(-1) = 4. So, total is 1 + 4 = 5. That's correct.Alternatively, if someone thought that i² is 1, they would get -4*1 = -4, leading to 1 -4 = -3, which is wrong. But in this case, since we know i² is -1, the calculation is correct. Therefore, the answer is 0. Alternatively, let me try plugging z = 5 back into the original equation to verify. Original equation: z/(1 + 2i) = 1 - 2i. So, substitute z = 5:5/(1 + 2i). Let's compute that. To divide 5 by (1 + 2i), multiply numerator and denominator by (1 - 2i):5*(1 - 2i)/[(1 + 2i)(1 - 2i)] = 5*(1 - 2i)/5 = (1 - 2i). Which is equal to the right-hand side of the equation. So, 1 - 2i. Therefore, the equation holds. So z = 5 is indeed the correct solution, and the imaginary part is 0. Therefore, the answer is 0. I think that's thorough. Maybe I was overcomplicating it, but better safe than sorry. So, final answer: 0.Final AnswerThe imaginary part of ( z ) is boxed{0}.</think>Given the complex number ( z ) that satisfies the equation (frac{z}{1+2i} = 1 - 2i), we need to find the imaginary part of ( z ).First, we solve for ( z ) by multiplying both sides of the equation by ( 1 + 2i ):[z = (1 - 2i)(1 + 2i)]Next, we expand the product on the right-hand side:[(1 - 2i)(1 + 2i) = 1 cdot 1 + 1 cdot 2i - 2i cdot 1 - 2i cdot 2i]Simplifying each term:[= 1 + 2i - 2i - 4i^2]Since ( i^2 = -1 ), we substitute and simplify further:[= 1 + 2i - 2i - 4(-1)= 1 + 0 + 4= 5]Thus, ( z = 5 ), which is a real number. The imaginary part of ( z ) is therefore 0.To verify, we substitute ( z = 5 ) back into the original equation:[frac{5}{1 + 2i} = 1 - 2i]Multiplying the numerator and denominator by the conjugate of the denominator:[frac{5(1 - 2i)}{(1 + 2i)(1 - 2i)} = frac{5(1 - 2i)}{1 + 4} = frac{5(1 - 2i)}{5} = 1 - 2i]The equation holds true, confirming that ( z = 5 ). Therefore, the imaginary part of ( z ) is (boxed{0}).,<think>Okay, let's see. I need to find the imaginary part of the complex number z that satisfies the equation z divided by (1 + 2i) equals 1 - 2i. Hmm, complex numbers can sometimes be tricky, but I think if I take it step by step, I can figure it out.First, let me write down the equation again to make sure I have it right: z / (1 + 2i) = 1 - 2i. So, they're telling me that when z is divided by the complex number (1 + 2i), the result is (1 - 2i). My goal is to find z and then identify its imaginary part.Since z is divided by (1 + 2i), maybe I can solve for z by multiplying both sides of the equation by (1 + 2i). That should isolate z on one side, right? Let me try that.Multiplying both sides by (1 + 2i):z = (1 - 2i) * (1 + 2i)Okay, so z equals the product of (1 - 2i) and (1 + 2i). Now I need to multiply these two complex numbers. Remember, when multiplying complex numbers, I have to use the distributive property, also known as the FOIL method for binomials. Let's apply that here.First, multiply the First terms: 1 * 1 = 1.Next, the Outer terms: 1 * 2i = 2i.Then, the Inner terms: -2i * 1 = -2i.Finally, the Last terms: -2i * 2i = -4i².Now, combine all these results:1 + 2i - 2i - 4i².Hmm, let's simplify this step by step. The 2i and -2i terms should cancel each other out. So, 2i - 2i = 0. That leaves us with 1 - 4i².Wait, now I need to remember that i² is equal to -1. So, substituting that in:1 - 4(-1) = 1 + 4 = 5.Wait, hold on. So the product of (1 - 2i) and (1 + 2i) is 5? That seems like a real number. So z equals 5? But that would mean z is a real number with no imaginary part. But the problem is asking for the imaginary part of z. If z is 5, then the imaginary part is 0. Is that possible?Let me double-check my calculations to make sure I didn't make a mistake. Let's go through the multiplication again:(1 - 2i)(1 + 2i) = 1*1 + 1*2i - 2i*1 - 2i*2i = 1 + 2i - 2i -4i². Yes, that's the same as before. Then combining the like terms: 2i - 2i cancels out, leaving 1 -4i². Then substituting i² = -1: 1 -4(-1) = 1 +4 = 5. So z is indeed 5. Therefore, z = 5 + 0i, so the imaginary part is 0.But wait, the problem says "the complex number z that satisfies the equation". It didn't specify whether z is supposed to have non-zero imaginary part. But according to the calculation, z is 5, which is a real number. So the imaginary part is 0.But let me consider another approach. Maybe I misapplied the multiplication. Alternatively, maybe I can approach the problem differently. For example, sometimes when dealing with complex numbers in denominators, people multiply numerator and denominator by the conjugate to make the denominator real. Wait, in the original equation, z/(1 + 2i) = 1 - 2i. So, if I want to solve for z, I can multiply both sides by (1 + 2i). Which is what I did. So that should be correct.Alternatively, if I wanted to, I could represent z as a complex number a + bi, and then solve for a and b. Let me try that method to confirm.Let z = a + bi, where a and b are real numbers. Then, the equation is (a + bi)/(1 + 2i) = 1 - 2i.To solve for a and b, multiply both sides by (1 + 2i):a + bi = (1 - 2i)(1 + 2i). Wait, that's exactly the same as before. So, again, a + bi = 5. Therefore, a = 5, and b = 0. So the imaginary part is 0. Hmm, so this confirms the same result.Alternatively, maybe the problem was designed to test understanding that when you have a complex number divided by another complex number, sometimes the result can be real. So, in this case, z ends up being real. Therefore, the imaginary part is zero.But let me check once again. If z / (1 + 2i) = 1 - 2i, then z = (1 - 2i)(1 + 2i). Calculating that product:First, 1*1 = 11*2i = 2i-2i*1 = -2i-2i*2i = -4i²Adding them up: 1 + 2i - 2i -4(-1) = 1 +0i +4=5.Yep, same result. So z is 5. Therefore, the imaginary part is 0.Alternatively, maybe I can think in terms of complex conjugates. The denominator is (1 + 2i). The right-hand side is (1 - 2i), which is the conjugate of (1 + 2i). Because the conjugate of a complex number a + bi is a - bi. So here, 1 + 2i conjugate is 1 - 2i. Therefore, z/(1 + 2i) = conjugate(1 + 2i). Then, multiplying both sides by (1 + 2i) gives z = conjugate(1 + 2i) * (1 + 2i). But the product of a complex number and its conjugate is the modulus squared. So modulus of (1 + 2i) squared is (1^2 + (2)^2) = 1 +4=5. Therefore, z=5. So again, z is 5, imaginary part is 0.Therefore, no matter which method I use, I get z=5, so the imaginary part is 0. Therefore, the answer should be 0.But wait, just to be thorough, let me recall that the imaginary part of a complex number a + bi is b. So if z=5, which is 5 + 0i, then the imaginary part is indeed 0.I think this is correct. Maybe the problem is testing the understanding that multiplying a complex number by its conjugate yields a real number, so in this case, z ends up being real, hence no imaginary part. Therefore, the imaginary part is 0.Alternatively, if I had made a mistake in multiplication, perhaps I would have gotten a different result. But after multiple checks, all paths lead to z=5, so the imaginary part is 0.Final AnswerThe imaginary part of ( z ) is boxed{0}.</think>Given the complex number ( z ) that satisfies the equation (frac{z}{1+2i} = 1 - 2i), we need to find the imaginary part of ( z ).First, we solve for ( z ) by multiplying both sides of the equation by ( 1 + 2i ):[z = (1 - 2i)(1 + 2i)]Next, we use the distributive property (FOIL method) to multiply the complex numbers:[(1 - 2i)(1 + 2i) = 1 cdot 1 + 1 cdot 2i - 2i cdot 1 - 2i cdot 2i]Calculating each term:- First terms: (1 cdot 1 = 1)- Outer terms: (1 cdot 2i = 2i)- Inner terms: (-2i cdot 1 = -2i)- Last terms: (-2i cdot 2i = -4i^2)Combining these terms, we get:[1 + 2i - 2i - 4i^2]Simplifying the expression:- The terms (2i) and (-2i) cancel each other out.- Using (i^2 = -1), we substitute (-4i^2) with (4):[1 + 4 = 5]Thus, ( z = 5 ), which is a real number. Therefore, the imaginary part of ( z ) is (0).[boxed{0}]]

🔑:To find z, we multiply both sides of the equation by 1+2i:z = (1+2i)(1-2i)We use the difference of squares formula (a-b)(a+b) = a^2 - b^2:z = 1^2 - (2i)^2Simplifying, we have:z = 1 - 4i^2Since i^2 = -1, we substitute the value:z = 1 - 4(-1)Thus,z = 5The imaginary part of z is boxed{0}.This problem tests your understanding of complex number operations and the definition of the imaginary part.

🧠:[<think>Okay, so I need to find the value of the algebraic expression (3x + 2)(3x - 2) + x(x - 2) given that 5x² - x - 1 = 0. Hmm, let me think about how to approach this step by step.First, maybe I should simplify the expression (3x + 2)(3x - 2) + x(x - 2). Let me start by expanding the terms. The first part is (3x + 2)(3x - 2). I remember that when you multiply two binomials that are conjugates, like (a + b)(a - b), it becomes a² - b². So applying that here, (3x)² - (2)². Let's compute that:(3x)² is 9x², and 2² is 4, so 9x² - 4. That simplifies the first part of the expression.Now, the second part is x(x - 2). Distribute the x into the parentheses:x * x is x², and x * (-2) is -2x. So that part becomes x² - 2x.Now, combining both parts together, the entire expression is:9x² - 4 + x² - 2x.Let me combine like terms. The 9x² and x² terms can be added together: 9x² + x² = 10x².Then the -2x remains as it is, and the constant term is -4. So putting it all together, the expression simplifies to:10x² - 2x - 4.Alright, so the original expression simplifies to 10x² - 2x - 4. Now, the problem gives me the quadratic equation 5x² - x - 1 = 0. Maybe I can use this equation to find the value of 10x² - 2x - 4 without solving for x explicitly.Let me see. If I can express 10x² - 2x - 4 in terms of the given equation, that might work. Let's look at the given equation: 5x² - x - 1 = 0. Let me multiply both sides of this equation by 2 to see if that helps:2*(5x² - x - 1) = 2*0 => 10x² - 2x - 2 = 0.Wait, that's interesting! So 10x² - 2x - 2 = 0. But our simplified expression is 10x² - 2x - 4. Hmm, how does this relate?If I subtract 2 from both sides of the equation 10x² - 2x - 2 = 0, I get 10x² - 2x = 2. Then, if I subtract 4 more from both sides, that would be 10x² - 2x - 4 = 2 - 4 = -2. Wait, is that right? Let me check. Given 10x² - 2x - 2 = 0, then 10x² - 2x = 2. Then subtract 4 from both sides: 10x² - 2x - 4 = 2 - 4 = -2. So that means the expression 10x² - 2x - 4 equals -2. Therefore, the value of (3x + 2)(3x - 2) + x(x - 2) is -2.Wait, let me verify this another way to make sure I didn't make a mistake. Let's substitute x from the original equation into the expression.But solving the quadratic equation 5x² - x - 1 = 0 would involve using the quadratic formula. Let's find the roots:x = [1 ± sqrt(1 + 20)] / 10 = [1 ± sqrt(21)] / 10.So x is (1 + sqrt(21))/10 or (1 - sqrt(21))/10. Let me take one of these roots and plug into the expression 10x² - 2x - 4 and see if it equals -2.Take x = [1 + sqrt(21)] / 10.First, compute x²:x² = [(1 + sqrt(21))/10]^2 = (1 + 2sqrt(21) + 21) / 100 = (22 + 2sqrt(21)) / 100 = (11 + sqrt(21)) / 50.Now, compute 10x²:10x² = 10*(11 + sqrt(21))/50 = (11 + sqrt(21))/5.Compute -2x:-2x = -2*(1 + sqrt(21))/10 = -(1 + sqrt(21))/5.Now, sum all parts:10x² - 2x - 4 = (11 + sqrt(21))/5 - (1 + sqrt(21))/5 - 4.Combine the first two terms:[(11 + sqrt(21)) - (1 + sqrt(21))]/5 - 4 = (10)/5 - 4 = 2 - 4 = -2.Okay, that checks out. Let me check the other root as well, x = [1 - sqrt(21)] / 10.Compute x²:x² = [(1 - sqrt(21))/10]^2 = (1 - 2sqrt(21) + 21)/100 = (22 - 2sqrt(21))/100 = (11 - sqrt(21))/50.10x² = 10*(11 - sqrt(21))/50 = (11 - sqrt(21))/5.-2x = -2*(1 - sqrt(21))/10 = -(1 - sqrt(21))/5 = (-1 + sqrt(21))/5.Now, 10x² - 2x -4:(11 - sqrt(21))/5 + (-1 + sqrt(21))/5 -4.Combine first two terms:[(11 - sqrt(21)) + (-1 + sqrt(21))]/5 -4 = (10)/5 -4 = 2 -4 = -2.So either way, substituting the roots into the expression gives -2. Therefore, my answer is correct.Alternatively, another approach: since the original equation is 5x² - x -1 = 0, then 5x² = x + 1. Multiply both sides by 2: 10x² = 2x + 2. Then substitute 10x² in the expression 10x² -2x -4:(2x + 2) -2x -4 = 2x + 2 -2x -4 = (2x -2x) + (2 -4) = 0 -2 = -2.Yes, that's a much quicker way. Since 10x² can be replaced by 2x + 2 from the original equation, substituting that in directly gives the result. So that's another way to see it. Either method works, and both confirm that the value is -2.Therefore, the value of the algebraic expression is -2.Final AnswerThe value of the expression is boxed{-2}.</think>Given the quadratic equation (5x^2 - x - 1 = 0), we need to find the value of the algebraic expression ((3x + 2)(3x - 2) + x(x - 2)).First, we simplify the expression:[(3x + 2)(3x - 2) + x(x - 2)]Expanding the terms, we get:[(3x + 2)(3x - 2) = (3x)^2 - (2)^2 = 9x^2 - 4]and[x(x - 2) = x^2 - 2x]Combining these results, we have:[9x^2 - 4 + x^2 - 2x = 10x^2 - 2x - 4]Next, we use the given quadratic equation (5x^2 - x - 1 = 0). Multiplying both sides by 2, we get:[10x^2 - 2x - 2 = 0]This implies:[10x^2 - 2x = 2]Substituting this into our simplified expression (10x^2 - 2x - 4), we get:[(10x^2 - 2x) - 4 = 2 - 4 = -2]Thus, the value of the algebraic expression is (boxed{-2}).,<think>Okay, so I need to solve this problem where I'm given the quadratic equation 5x² - x - 1 = 0, and I have to find the value of the algebraic expression (3x + 2)(3x - 2) + x(x - 2). Hmm, let me think about how to approach this.First, maybe I should try to simplify the expression they want me to find. Let's see, expanding the products might help. The first part is (3x + 2)(3x - 2). That looks like a difference of squares. Remember, (a + b)(a - b) = a² - b². So applying that formula here, where a is 3x and b is 2, this should be (3x)² - (2)². Let me calculate that: (3x)² is 9x² and 2 squared is 4, so that part simplifies to 9x² - 4.Then the next part is x(x - 2). That's a simple distribution. Multiply x by both terms inside the parentheses: x * x is x², and x * (-2) is -2x. So that part becomes x² - 2x.Now, putting both parts together: (9x² - 4) + (x² - 2x). Combine like terms here. The 9x² and x² add up to 10x². Then there's a -2x, and the constant term is -4. So the entire expression simplifies to 10x² - 2x - 4.Okay, so the original expression simplifies to 10x² - 2x - 4. Now, the problem wants me to find the value of this expression given that 5x² - x - 1 = 0. Maybe I can use the given equation to substitute for parts of the expression. Let me see.Looking at the expression 10x² - 2x - 4, I notice that 10x² is 2 times 5x². The given equation is 5x² - x - 1 = 0, so perhaps I can solve for 5x² in terms of x and constants. Let me try that.From 5x² - x - 1 = 0, I can rearrange to solve for 5x²:5x² = x + 1.So 5x² is equal to x + 1. Therefore, 10x² would be 2*(5x²) = 2*(x + 1) = 2x + 2.So substituting back into the expression 10x² - 2x - 4, replacing 10x² with 2x + 2:(2x + 2) - 2x - 4.Now, let's simplify this. The 2x and -2x cancel out, leaving 2 - 4, which is -2.Wait, so that means the entire expression simplifies to -2? Let me check my steps again to make sure I didn't make a mistake.Original expression: (3x + 2)(3x - 2) + x(x - 2)First step: Expanding (3x + 2)(3x - 2) to 9x² - 4. That seems right.Second part: x(x - 2) becomes x² - 2x. Correct.Adding them together: 9x² - 4 + x² - 2x = 10x² - 2x - 4. That's correct.Then, using the equation 5x² - x - 1 = 0 to express 5x² as x + 1, so 10x² is 2x + 2. Then substituting into 10x² - 2x - 4 gives 2x + 2 - 2x - 4 = (2x - 2x) + (2 - 4) = 0 - 2 = -2. Yep, that seems to check out.So the value of the expression is -2. Let me verify this another way, just to be sure. Maybe plug in the roots of the equation into the expression and see if it equals -2.But wait, the quadratic equation 5x² - x - 1 = 0 has two roots. Let me compute them using the quadratic formula. The roots are x = [1 ± sqrt(1 + 20)] / 10 = [1 ± sqrt(21)] / 10. So two roots: (1 + sqrt(21))/10 and (1 - sqrt(21))/10.Let me take one of them and substitute into the original expression. Let's pick x = [1 + sqrt(21)] / 10.First, compute (3x + 2)(3x - 2). Let's compute 3x first. 3x = 3*(1 + sqrt(21))/10 = (3 + 3sqrt(21))/10. Then 3x + 2 = (3 + 3sqrt(21))/10 + 20/10 = (23 + 3sqrt(21))/10. Similarly, 3x - 2 = (3 + 3sqrt(21))/10 - 20/10 = (-17 + 3sqrt(21))/10.Multiplying these two: [(23 + 3sqrt(21))( -17 + 3sqrt(21))]/100. Let me compute the numerator.23*(-17) = -39123*(3sqrt(21)) = 69sqrt(21)3sqrt(21)*(-17) = -51sqrt(21)3sqrt(21)*3sqrt(21) = 9*(21) = 189Adding all these terms:-391 + 69sqrt(21) -51sqrt(21) + 189Combine like terms:(-391 + 189) + (69sqrt(21) -51sqrt(21)) = (-202) + (18sqrt(21))So the numerator is -202 + 18sqrt(21). Divided by 100, so the first part is (-202 + 18sqrt(21))/100.Now compute the second part, x(x - 2). Let's compute x first: x = (1 + sqrt(21))/10. Then x - 2 = (1 + sqrt(21))/10 - 20/10 = (-19 + sqrt(21))/10.Multiply x and (x - 2): [(1 + sqrt(21))/10] * [(-19 + sqrt(21))/10] = [ (1)(-19) + 1*sqrt(21) + sqrt(21)*(-19) + sqrt(21)*sqrt(21) ] / 100Calculating numerator:-19 + sqrt(21) -19sqrt(21) + 21Combine terms:(-19 +21) + (sqrt(21) -19sqrt(21)) = 2 -18sqrt(21)So the second part is (2 -18sqrt(21))/100.Now add the two parts together:First part: (-202 + 18sqrt(21))/100Second part: (2 -18sqrt(21))/100Total: [(-202 + 18sqrt(21)) + (2 -18sqrt(21))]/100 = (-200)/100 = -2. Exactly! So that works out.Alternatively, the other root would also give the same result, since the expression we're evaluating is quadratic in x and the original equation is also quadratic. But substituting the other root would follow the same steps and also result in -2. So that confirms that the value is indeed -2 regardless of which root we plug in.Therefore, even though we didn't need to find the actual roots of the equation, we used substitution based on the given equation to simplify the expression, leading to the answer -2.Final Answerboxed{-2}</think>Given the quadratic equation (5x^2 - x - 1 = 0), we need to find the value of the algebraic expression ((3x + 2)(3x - 2) + x(x - 2)).First, we simplify the expression:1. Expand ((3x + 2)(3x - 2)) using the difference of squares: [ (3x + 2)(3x - 2) = (3x)^2 - 2^2 = 9x^2 - 4 ]2. Expand (x(x - 2)) by distribution: [ x(x - 2) = x^2 - 2x ]3. Combine the results: [ 9x^2 - 4 + x^2 - 2x = 10x^2 - 2x - 4 ]Next, we use the given quadratic equation (5x^2 - x - 1 = 0) to substitute for (5x^2):1. Rearrange the equation to solve for (5x^2): [ 5x^2 = x + 1 ]2. Multiply both sides by 2 to find (10x^2): [ 10x^2 = 2(x + 1) = 2x + 2 ]3. Substitute (10x^2) in the simplified expression: [ 10x^2 - 2x - 4 = (2x + 2) - 2x - 4 = 2 - 4 = -2 ]Thus, the value of the algebraic expression is (boxed{-2}).]

🔑:To solve the given problem, we start by simplifying the given algebraic expression left(3x+2right)left(3x-2right)+xleft(x-2right) step by step:1. First, we expand the products:begin{align*}left(3x+2right)left(3x-2right)+xleft(x-2right) &= (9x^2 - 4) + (x^2 - 2x) &= 9x^2 + x^2 - 2x - 4 &= 10x^2 - 2x - 4.end{align*}2. We are given the equation 5x^2 - x - 1 = 0. We can manipulate this equation to express it in a form that will help us substitute values into our simplified expression. We add 1 to both sides to isolate the terms involving x on one side:begin{align*}5x^2 - x &= 1.end{align*}3. Now, we notice that the simplified expression 10x^2 - 2x - 4 can be related to the given equation. Specifically, 10x^2 - 2x is twice the left-hand side of the given equation. So, we substitute 2(5x^2 - x) for 10x^2 - 2x in our expression:begin{align*}10x^2 - 2x - 4 &= 2(5x^2 - x) - 4 &= 2(1) - 4 &= 2 - 4 &= -2.end{align*}Therefore, the value of the algebraic expression left(3x+2right)left(3x-2right)+xleft(x-2right) is boxed{-2}.

🧠:[<think>Okay, so I need to find the range of values for z = x + y, where x and y satisfy the equation (x²)/16 + (y²)/9 = 1. Hmm, that equation looks familiar. Oh right, it's the standard form of an ellipse centered at the origin with semi-major axis 4 along the x-axis and semi-minor axis 3 along the y-axis. So, x can range from -4 to 4, and y can range from -3 to 3. But z is the sum of x and y, so the maximum and minimum values of z won't just be the sum of the individual maxes and mins because x and y can't both be at their extremes at the same time. I need a better approach.Maybe I can parameterize x and y using trigonometric functions since it's an ellipse. The standard parametrization for an ellipse (x²)/a² + (y²)/b² = 1 is x = a cosθ and y = b sinθ. So here, a = 4 and b = 3, so x = 4 cosθ and y = 3 sinθ. Then z = x + y = 4 cosθ + 3 sinθ. Now I need to find the maximum and minimum of this expression. That seems manageable.I remember that expressions like A cosθ + B sinθ can be rewritten using a single sine or cosine function with a phase shift. The formula is A cosθ + B sinθ = C cos(θ - φ) or C sin(θ + φ), where C is the amplitude, which is sqrt(A² + B²). Let me verify that. Suppose we have 4 cosθ + 3 sinθ. Let me see, if we write this as R cos(θ - φ), then expanding R cos(θ - φ) = R cosθ cosφ + R sinθ sinφ. So equating coefficients, R cosφ = 4 and R sinφ = 3. Therefore, R² = 4² + 3² = 16 + 9 = 25, so R = 5. Then φ = arctan(3/4). So, 4 cosθ + 3 sinθ = 5 cos(θ - φ), where φ = arctan(3/4). Therefore, the maximum value of z = 5 cos(θ - φ) would be 5, since the maximum of cosine is 1, and the minimum would be -5. But wait, is that correct? Wait, if it's 5 cos(θ - φ), then yes, the range would be [-5, 5]. But let me check if my parametrization is correct. Because sometimes when dealing with ellipses, the parametrization is a bit different? Wait, the parametrization x = a cosθ and y = b sinθ does cover the entire ellipse, right? As θ goes from 0 to 2π, the point (4 cosθ, 3 sinθ) moves around the ellipse. So then, z = 4 cosθ + 3 sinθ is indeed expressible as 5 cos(θ - φ), which has amplitude 5, so the maximum is 5 and minimum is -5. So the range of z is from -5 to 5. Therefore, the answer is [-5, 5]. Hmm, but let me verify this with another method to be sure.Alternatively, maybe I can use the Cauchy-Schwarz inequality. For real numbers, (a1*b1 + a2*b2)^2 ≤ (a1² + a2²)(b1² + b2²). Let me see. Let me think of z = x + y. So, z = x + y = 1*x + 1*y. So, using Cauchy-Schwarz, (x*1 + y*1)^2 ≤ (x² + y²)(1² + 1²). But wait, in our case, the constraint is (x²)/16 + (y²)/9 = 1, which is different from x² + y². So maybe I need to adjust the inequality.Alternatively, write z = x + y. Let me express x in terms of y or vice versa. From the ellipse equation, (x²)/16 + (y²)/9 = 1. Let's solve for x in terms of y: x² = 16*(1 - (y²)/9) = 16 - (16/9)y². Then x = ±√(16 - (16/9)y²). Then z = x + y = ±√(16 - (16/9)y²) + y. Hmm, but taking the square root makes this a bit complicated. Then to find the maximum and minimum of z, we would need to take the derivative with respect to y and set it to zero. But this might get messy with the square roots. Let's see.Suppose we take the positive square root first (for the upper half of the ellipse). Then z = √(16 - (16/9)y²) + y. To find the maximum, take derivative dz/dy = (1/2)(16 - (16/9)y²)^(-1/2)*(-32/9)y + 1. Set this equal to zero. So:(1/2)(16 - (16/9)y²)^(-1/2)*(-32/9)y + 1 = 0Multiply both sides by (16 - (16/9)y²)^(1/2):(1/2)*(-32/9)y + (16 - (16/9)y²)^(1/2) = 0Hmm, this is getting complicated. Let me compute:(16 - (16/9)y²)^(1/2) = (32/18)yWait, let's rearrange:(16 - (16/9)y²)^(1/2) = (32/18)ySquare both sides:16 - (16/9)y² = (1024/324)y²Simplify the right-hand side: 1024/324 = 256/81So:16 = (256/81)y² + (16/9)y²Convert 16/9 to 144/81:16 = (256/81 + 144/81)y² = (400/81)y²Multiply both sides by 81:16*81 = 400y²So y² = (16*81)/400 = (1296)/400 = 324/100 = 81/25Therefore, y = ±9/5. Hmm, so y = 9/5 or y = -9/5. But we were considering the positive square root, so x = √(16 - (16/9)y²). Let's compute x when y = 9/5:x = √(16 - (16/9)*(81/25)) = √(16 - (16*9)/25) = √(16 - 144/25) = √(400/25 - 144/25) = √(256/25) = 16/5. So z = x + y = 16/5 + 9/5 = 25/5 = 5. Similarly, if we take the negative square root, x = -16/5, and y = -9/5, then z = -16/5 -9/5 = -25/5 = -5. So that gives the same result as before. So using calculus, we also get z ranging from -5 to 5. So that seems to confirm the previous answer.Alternatively, another method: Lagrange multipliers. Let's set up the optimization problem. We want to maximize/minimize z = x + y subject to the constraint (x²)/16 + (y²)/9 = 1. The Lagrangian is L = x + y - λ[(x²)/16 + (y²)/9 - 1]. Taking partial derivatives:∂L/∂x = 1 - λ*(2x)/16 = 1 - λ*(x/8) = 0∂L/∂y = 1 - λ*(2y)/9 = 1 - λ*(2y/9) = 0∂L/∂λ = -(x²)/16 - (y²)/9 +1 = 0From the first equation: 1 = λx/8 ⇒ λ = 8/xFrom the second equation: 1 = (2λ y)/9 ⇒ 1 = (2*(8/x)*y)/9 ⇒ 1 = (16 y)/(9x) ⇒ 9x = 16 y ⇒ y = (9/16)xSo substitute y = (9/16)x into the constraint equation:(x²)/16 + [(9/16 x)²]/9 = 1Compute each term:First term: x²/16Second term: (81/256 x²)/9 = (81/256)/9 x² = 9/256 x²So total:x²/16 + 9x²/256 = (16x² + 9x²)/256 = 25x²/256 = 1Thus, 25x² = 256 ⇒ x² = 256/25 ⇒ x = ±16/5Then y = (9/16)x = (9/16)(±16/5) = ±9/5Thus, when x = 16/5, y = 9/5, z = 16/5 +9/5 =25/5=5When x = -16/5, y = -9/5, z = -25/5= -5Therefore, the maximum value of z is 5 and the minimum is -5. So the range is [-5,5]. All methods seem to converge to this answer.Wait, but let me check with another approach. Maybe using the method of auxiliary variables or parametric equations. Let's think of z = x + y. We can imagine the line z = x + y moving along the ellipse. The extreme values of z occur where this line is tangent to the ellipse. So, substituting y = z - x into the ellipse equation:(x²)/16 + ( (z - x)² ) /9 =1Multiply through by 144 (the least common multiple of 16 and 9):9x² + 16(z - x)² = 144Expand the second term:16(z² - 2zx +x²) = 16z² -32zx +16x²So total equation:9x² +16z² -32zx +16x² =144Combine like terms:(9x² +16x²) -32zx +16z² =14425x² -32zx +16z² -144=0This is a quadratic in x. For the line to be tangent to the ellipse, this quadratic equation must have exactly one solution, so discriminant is zero.The quadratic equation is: 25x² -32zx + (16z² -144)=0Discriminant D = ( -32z )² -4*25*(16z² -144)Compute D:D = 1024z² -100*(16z² -144)= 1024z² -1600z² +14400= (1024 -1600)z² +14400= (-576)z² +14400Set D =0:-576z² +14400=0→ 576z²=14400→ z²=14400/576=25→ z=±5So again, z can be 5 or -5, confirming the previous results. Therefore, the range of z is from -5 to 5. All methods lead to the same conclusion, so I think this is solid.Final AnswerThe range of values for ( z = x + y ) is boxed{[-5, 5]}.</think>To determine the range of values for ( z = x + y ) where ( x ) and ( y ) satisfy the equation (frac{x^2}{16} + frac{y^2}{9} = 1), we can use several methods:1. Parametrization Method: - The given equation represents an ellipse. We parametrize ( x ) and ( y ) as ( x = 4 cos theta ) and ( y = 3 sin theta ). - Substituting these into ( z = x + y ), we get ( z = 4 cos theta + 3 sin theta ). - This expression can be rewritten using the amplitude form: ( 4 cos theta + 3 sin theta = 5 cos(theta - phi) ) where ( phi = arctanleft(frac{3}{4}right) ). - The range of ( 5 cos(theta - phi) ) is ([-5, 5]).2. Calculus Method: - Solving for ( x ) in terms of ( y ) from the ellipse equation and substituting into ( z ), we derive the expression for ( z ) and take its derivative with respect to ( y ). - Setting the derivative to zero and solving, we find critical points ( y = pm frac{9}{5} ) and corresponding ( x = pm frac{16}{5} ). - Substituting these back into ( z ), we get ( z = pm 5 ).3. Lagrange Multipliers Method: - Setting up the Lagrangian ( L = x + y - lambdaleft(frac{x^2}{16} + frac{y^2}{9} - 1right) ), we solve the system of equations derived from partial derivatives. - Solving, we find ( x = pm frac{16}{5} ) and ( y = pm frac{9}{5} ), leading to ( z = pm 5 ).4. Quadratic Equation Method: - Substituting ( y = z - x ) into the ellipse equation and simplifying, we get a quadratic in ( x ). - For the line to be tangent to the ellipse, the discriminant of this quadratic must be zero, leading to ( z = pm 5 ).All methods confirm that the range of ( z = x + y ) is (boxed{[-5, 5]}).,<think>Okay, so I need to find the range of values for z = x + y, given that x and y satisfy the equation (x²)/16 + (y²)/9 = 1. Hmm, let's see. That equation looks familiar—it's the standard form of an ellipse, right? The major axis is along the x-axis because 16 is larger than 9. The semi-major axis is 4 (since sqrt(16) is 4) and the semi-minor axis is 3 (sqrt(9)). So, the problem is asking for the maximum and minimum possible values of z = x + y on this ellipse. That sounds like an optimization problem with a constraint. I remember that for such problems, we can use the method of Lagrange multipliers, but maybe there's a simpler way using trigonometric parameterization since it's an ellipse. Let me think.First, parametric equations for an ellipse. For the standard ellipse (x²/a²) + (y²/b²) = 1, the parametric coordinates are (a cosθ, b sinθ), where θ is the parameter. So in this case, a² = 16, so a = 4, and b² = 9, so b = 3. Therefore, the parametric equations would be x = 4 cosθ and y = 3 sinθ. Then z = x + y = 4 cosθ + 3 sinθ. Okay, so now I need to find the maximum and minimum values of z = 4 cosθ + 3 sinθ. This is a linear combination of sine and cosine functions. I remember that any expression of the form A cosθ + B sinθ can be rewritten as C cos(θ - φ) or C sin(θ + φ), where C is the amplitude, sqrt(A² + B²). Then the maximum value would be C and the minimum would be -C. Let me verify that.So, if we have 4 cosθ + 3 sinθ, then the amplitude C would be sqrt(4² + 3²) = sqrt(16 + 9) = sqrt(25) = 5. Therefore, z can be written as 5 cos(θ - φ), where φ is some phase shift. Therefore, the maximum value of z would be 5 and the minimum would be -5. Is that correct?Wait, let me check with an example. Let's take θ where cosθ is 4/5 and sinθ is 3/5. Then 4*(4/5) + 3*(3/5) = 16/5 + 9/5 = 25/5 = 5. So yes, that's the maximum. Similarly, if θ is such that cosθ is -4/5 and sinθ is -3/5, then z would be -16/5 -9/5 = -25/5 = -5. So that seems correct.Therefore, the range of z = x + y is from -5 to 5. So the answer should be that z can take any value between -5 and 5, inclusive. But let me think again if there's a different way to approach this, maybe using Cauchy-Schwarz inequality or something else.Alternatively, using the method of Lagrange multipliers. Let's set up the optimization problem: maximize and minimize z = x + y subject to (x²)/16 + (y²)/9 = 1. The Lagrangian is L = x + y - λ[(x²)/16 + (y²)/9 - 1]. Taking partial derivatives:∂L/∂x = 1 - λ*(2x)/16 = 1 - λx/8 = 0∂L/∂y = 1 - λ*(2y)/9 = 1 - (2λy)/9 = 0∂L/∂λ = -(x²)/16 - (y²)/9 + 1 = 0So from the first equation, 1 = λx/8 → λ = 8/xFrom the second equation, 1 = (2λy)/9 → substitute λ here: 1 = (2*(8/x)*y)/9 = (16 y)/(9 x)So 1 = (16 y)/(9 x) → 9x = 16 y → y = (9/16) xNow substitute y = (9/16)x into the ellipse equation:(x²)/16 + [(9/16 x)^2]/9 = 1Compute each term:First term: x²/16Second term: (81/256 x²)/9 = (81/256)/9 x² = 9/256 x²Therefore, x²/16 + 9x²/256 = 1Convert to same denominator. 16 is 256/16, so x²/16 = 16x²/256Thus, 16x²/256 + 9x²/256 = (16x² + 9x²)/256 = 25x²/256 = 1So, 25x² = 256 → x² = 256/25 → x = ±16/5Then y = (9/16)x = (9/16)*(±16/5) = ±9/5Therefore, the critical points are (16/5, 9/5) and (-16/5, -9/5)Compute z = x + y at these points:For (16/5, 9/5): z = 16/5 + 9/5 = 25/5 = 5For (-16/5, -9/5): z = -16/5 -9/5 = -25/5 = -5So that's consistent with the previous method. Therefore, the maximum is 5 and the minimum is -5.Therefore, the range of z is from -5 to 5. So any value between -5 and 5, inclusive, is attainable. Wait, but let me check if there isn't any mistake here. Let me see. Suppose x and y are on the ellipse, then z = x + y. So to confirm, if I plug in x = 4 cosθ and y = 3 sinθ, then z = 4 cosθ + 3 sinθ. The maximum of this is indeed 5 because sqrt(4² + 3²) = 5, as per the trigonometric identity. So yes, that's correct. So the maximum is 5 and the minimum is -5.Alternatively, using Cauchy-Schwarz: For any real numbers a, b, c, d, (ac + bd)^2 ≤ (a² + b²)(c² + d²). Let's apply that here.Let me think of z = x + y as 1*x + 1*y. Then by Cauchy-Schwarz, (x + y)^2 ≤ (1² + 1²)(x² + y²). Wait, but the constraint is (x²)/16 + (y²)/9 = 1, so x² + y² is not directly given. Maybe not the best approach. Alternatively, we can use weighted Cauchy-Schwarz.Alternatively, write the constraint as (x²)/16 + (y²)/9 = 1. Let me consider vectors u = (x/4, y/3) and v = (4, 3). Then z = x + y = 4*(x/4) + 3*(y/3) = u · v. The vector u is a unit vector because (x/4)^2 + (y/3)^2 = 1. Then the dot product u · v is equal to |u||v|cosθ = |v|cosθ. Since |v| = sqrt(4² + 3²) = 5. Therefore, the maximum value of the dot product is 5*1 = 5, and the minimum is -5. Therefore, z ranges from -5 to 5. That's another way to see it. So all methods lead to the same conclusion.Therefore, I think the answer is that the range of z is [-5, 5]. So the possible values of z are all real numbers between -5 and 5, inclusive.Final AnswerThe range of values for ( z = x + y ) is boxed{[-5, 5]}.</think>Given the equation (frac{x^2}{16} + frac{y^2}{9} = 1), we recognize it as the standard form of an ellipse with semi-major axis 4 and semi-minor axis 3. We need to determine the range of values for (z = x + y).First, we parameterize the ellipse using trigonometric functions. Let (x = 4 cos theta) and (y = 3 sin theta). Substituting these into (z), we get:[z = x + y = 4 cos theta + 3 sin theta]This expression is a linear combination of sine and cosine functions, which can be rewritten using the amplitude form. The amplitude (C) is calculated as:[C = sqrt{4^2 + 3^2} = sqrt{16 + 9} = sqrt{25} = 5]Thus, (z) can be expressed as (5 cos(theta - phi)), where (phi) is a phase shift. The maximum value of (z) is (5) and the minimum value is (-5).To verify, we use the method of Lagrange multipliers. We set up the Lagrangian:[L = x + y - lambda left( frac{x^2}{16} + frac{y^2}{9} - 1 right)]Taking partial derivatives and solving, we find the critical points ((16/5, 9/5)) and ((-16/5, -9/5)). Evaluating (z) at these points gives (5) and (-5) respectively.Using the Cauchy-Schwarz inequality, we consider vectors (u = (x/4, y/3)) and (v = (4, 3)). The dot product (u cdot v = x + y) is maximized and minimized by the magnitudes of the vectors, confirming the range ([-5, 5]).Thus, the range of values for (z = x + y) is (boxed{[-5, 5]}).]

🔑:Given that x and y satisfy the equation of an ellipse frac{x^2}{16} + frac{y^2}{9} = 1, we want to find the range for z = x + y. We can rewrite the provided equation as a system:begin{cases}frac{x^{2}}{16} + frac{y^{2}}{9} = 1 y = z - xend{cases}Substitute y from the second equation into the first one:frac{x^{2}}{16} + frac{(z - x)^{2}}{9} = 1Expanding the terms, we get:frac{x^{2}}{16} + frac{z^{2} - 2zx + x^{2}}{9} = 1Now, multiply through by a common denominator (144):9x^{2} + 16z^{2} - 32zx + 16x^{2} = 144Combine like terms:25x^{2} - 32zx + 16z^{2} - 144 = 0For this quadratic equation in x to have real solutions (since x is real), the discriminant must be non-negative:Delta = (-32z)^{2} - 4 cdot 25 cdot (16z^{2} - 144) geq 0Simplifying, we get:Delta = 1024z^{2} - 100 (16z^{2} - 144) = 1024z^{2} - 1600z^{2} + 14400 geq 0Further simplification gives us:-576z^{2} + 14400 geq 0Divide by -576 (reversing the inequality because of division by a negative number):z^{2} leq frac{14400}{576}Taking the square root:|z| leq frac{120}{24}Therefore:|z| leq 5The final range of possible values for z is -5 leq z leq 5, which can be written as:boxed{z in [-5, 5]}