🧠:[<think>Okay, let's see. The problem is to calculate the sum (1 cdot 1! + 2 cdot 2! + 3 cdot 3! + ldots + 1580 cdot 1580!). Hmm, that's a pretty big sum. Each term is a number multiplied by its factorial. I remember factorials grow really quickly, so adding all these up manually is impossible. There must be a pattern or formula here.Let me think. Maybe there's a telescoping series approach? Telescoping series are when terms cancel each other out, leaving only a few terms. I remember something about factorial sums and telescoping. Let's try to manipulate one of the terms to see if a pattern emerges.Take the general term (k cdot k!). Let's see if we can express this in terms of another factorial. For example, we know that ((k+1)! = (k+1) cdot k!). So, expanding that, ((k+1)! = k cdot k! + k!). Wait, if we rearrange that equation, maybe we can express (k cdot k!) in terms of ((k+1)! - k!). Let's check:Starting with ((k+1)! = (k+1) cdot k!), which is (k cdot k! + k!). So, subtracting (k!) from both sides gives (k cdot k! = (k+1)! - k!). Oh! That seems useful. So each term (k cdot k!) can be rewritten as ((k+1)! - k!). Let me verify with a small example:Take (k = 1): (1 cdot 1! = 1). On the other side, ((1+1)! - 1! = 2! - 1! = 2 - 1 = 1). Okay, that works.Another one: (k = 2): (2 cdot 2! = 4). The other expression is (3! - 2! = 6 - 2 = 4). Perfect. So the transformation works.So this means that the entire sum (S = sum_{k=1}^{1580} k cdot k!) can be rewritten as (sum_{k=1}^{1580} [(k+1)! - k!]). Now, this is a telescoping series, because when we expand the terms, each negative term cancels with the positive term of the next summand.Let me write out the first few terms to visualize:When (k = 1): (2! - 1!)When (k = 2): (3! - 2!)When (k = 3): (4! - 3!)...When (k = 1580): (1581! - 1580!)So adding all these up:((2! - 1!) + (3! - 2!) + (4! - 3!) + ldots + (1581! - 1580!))Now, notice that each (-2!) cancels with the next (+2!), the (-3!) cancels with the (+3!), and so on, all the way up to (-1580!) cancels with the (+1580!). What's left is the last term of the last expression, which is (1581!), and the very first negative term, which is (-1!).Therefore, the entire sum simplifies to (1581! - 1!). Since (1! = 1), this is (1581! - 1).But wait, the problem mentions that (1580! = a). The answer is in terms of (a). Hmm. How does 1581! relate to 1580!? Well, 1581! = 1581 cdot 1580! = 1581 cdot a. So substituting back in, the sum becomes (1581 cdot a - 1).Therefore, the answer should be (1581 cdot a - 1), which can be written as (1581a - 1). Let me check again to make sure I didn't miss anything.Starting with the sum (1 cdot 1! + 2 cdot 2! + ldots + 1580 cdot 1580!), converted each term to ((k+1)! - k!), telescoped to (1581! - 1), and since (1581! = 1581 times 1580! = 1581a), substituting gives (1581a - 1). Seems solid.Just to test with a smaller number. Let's say n=2. Then the sum is 1*1! + 2*2! = 1 + 4 = 5. According to the formula, (2+1)! -1 = 6 -1 =5. Correct. Another test with n=3: 1*1! +2*2! +3*3! =1 +4 +18=23. Formula:4! -1=24-1=23. Perfect. So the formula works for small n. Therefore, the logic holds.Therefore, the answer is 1581a -1.Final Answerboxed{1581a - 1}</think>To calculate the sum (1 cdot 1! + 2 cdot 2! + 3 cdot 3! + ldots + 1580 cdot 1580!), we start by expressing each term (k cdot k!) as a telescoping series. We know that ((k+1)! = (k+1) cdot k!), which can be rearranged to (k cdot k! = (k+1)! - k!). Using this, we can rewrite each term in the sum:[k cdot k! = (k+1)! - k!]Substituting this into the sum, we get:[sum_{k=1}^{1580} k cdot k! = sum_{k=1}^{1580} [(k+1)! - k!]]Expanding the series, we observe that it is a telescoping series:[(2! - 1!) + (3! - 2!) + (4! - 3!) + ldots + (1581! - 1580!)]In this series, most terms cancel out, leaving only the first negative term and the last positive term:[1581! - 1!]Since (1! = 1), this simplifies to:[1581! - 1]Given that (1580! = a), we know (1581! = 1581 cdot 1580! = 1581a). Substituting this back, we get:[1581a - 1]Thus, the final answer is:[boxed{1581a - 1}],<think>Okay, so I need to figure out the sum 1·1! + 2·2! + 3·3! + ... + 1580·1580!. Hmm, let's see. The problem also mentions that 1580! = a, but I'm not sure how that plays into it yet. Maybe the sum can be simplified in terms of a factorial?First, I remember that factorials grow really fast. Each term in the sum is k multiplied by k!, which is the same as (k+1)! - k!. Wait, let me check that. Let's take a small example. If k = 1, then (k+1)! - k! = 2! - 1! = 2 - 1 = 1, which is equal to 1·1! = 1. If k = 2, then (3! - 2!) = 6 - 2 = 4, and 2·2! = 4. That works. For k = 3, (4! - 3!) = 24 - 6 = 18, and 3·3! = 18. Wow, that's a useful identity! So in general, k·k! = (k+1)! - k!.If that's the case, then the sum from k=1 to n of k·k! is the sum from k=1 to n of [(k+1)! - k!]. That looks like a telescoping series. Let's write out the first few terms:When k=1: 2! - 1!k=2: 3! - 2!k=3: 4! - 3!...k=n: (n+1)! - n!So when we add all these up, most of the terms should cancel out. Let's see:(2! - 1!) + (3! - 2!) + (4! - 3!) + ... + (n+1)! - n!All the intermediate terms cancel: -1! remains, and then +2! -2!, +3! -3!, etc., up until -n! cancels with the previous term's +n!. So the sum simplifies to (n+1)! - 1!.But 1! is just 1, so the total sum is (n+1)! - 1.Therefore, applying this formula to the problem where n = 1580, the sum should be (1580 + 1)! - 1 = 1581! - 1.But the problem mentions that 1580! = a. So 1581! is 1581 × 1580! = 1581 × a. Therefore, the sum is 1581a - 1.Wait, let me double-check. If 1580! = a, then 1581! = 1581 × a. So substituting back, the sum is 1581a - 1. That makes sense.But just to be absolutely sure, let's test this with a smaller number. Let's say n=3. The sum would be 1·1! + 2·2! + 3·3! = 1 + 4 + 18 = 23. According to the formula, (3+1)! - 1 = 24 - 1 = 23. Correct. Another example, n=2: 1 + 4 = 5. The formula gives 3! -1 = 6 -1 =5. Perfect. So the formula works.Therefore, applying the same logic, for n=1580, the sum is 1581! -1. Since 1581! is 1581×1580!, and 1580! is given as a, then 1581! = 1581a. So substituting, the sum is 1581a -1. Therefore, the answer should be 1581a -1.Wait, but the problem says "calculate: 1·1!+2·2!+3·3!+…+1580·1580!" and mentions that 1580! = a. So they just want the expression in terms of a? Then yes, 1581a -1. Alternatively, if they just want the factorial expression, it's 1581! -1. But since 1580! is given as a, perhaps expressing the answer in terms of a is better. Let me check the problem statement again."Given that 1580!=a, calculate: 1·1!+2·2!+3·3!+…+1580·1580! (n!=1·2·3·…·n)"So, they want the sum calculated, given that 1580! is equal to a. So likely, the answer should be expressed in terms of a. So, as 1581a -1. That's the answer.But to make sure, let me confirm the telescoping again. Each term k·k! = (k+1)! -k! So when you sum from 1 to n, you get sum_{k=1}^n [(k+1)! -k!] = (n+1)! -1! because all the middle terms cancel. So yes, sum = (n+1)! -1. So here n=1580, so sum=1581! -1. Since 1581! =1581×1580! =1581a, so sum=1581a -1. Therefore, the answer is 1581a -1. So I think that's correct.Another way to see: If you have 1·1! = 2! -1!, 2·2! =3! -2!, ..., 1580·1580! =1581! -1580!. Then adding all these, the 2! cancels with -2!, 3! cancels with -3!, ..., 1580! cancels with -1580!, leaving -1! +1581!. So total is 1581! -1. Which is 1581a -1. Perfect.Therefore, I feel confident that the answer is 1581a -1.Final Answerboxed{1581a - 1}</think>To calculate the sum (1 cdot 1! + 2 cdot 2! + 3 cdot 3! + ldots + 1580 cdot 1580!) given that (1580! = a), we use the identity (k cdot k! = (k+1)! - k!). This allows us to rewrite each term in the sum as a telescoping series:[begin{aligned}1 cdot 1! &= 2! - 1!, 2 cdot 2! &= 3! - 2!, 3 cdot 3! &= 4! - 3!, &vdots 1580 cdot 1580! &= 1581! - 1580!.end{aligned}]Adding these terms together, most intermediate terms cancel out, leaving:[(2! - 1!) + (3! - 2!) + (4! - 3!) + ldots + (1581! - 1580!) = 1581! - 1!.]Since (1! = 1), the sum simplifies to (1581! - 1). Given that (1580! = a), we can express (1581!) as (1581 cdot 1580! = 1581a). Therefore, the sum is:[1581a - 1.]Thus, the final answer is (boxed{1581a - 1}).]

🔑:Solution:1 cdot 1!+2 cdot 2!+3 cdot 3!+ldots+1580 cdot 1580!==(2-1) cdot 1!+(3-1) cdot 2!+(4-1) cdot 3!+ldots+(1581-1) cdot 1580!==2 cdot 1!-1!+3 cdot 2!-2!+4 cdot 3!-3!+ldots+1581 cdot 1580!-1580!==2!-1!+3!-2!+4!-3!+ldots+1581!-1580!==-1!+1581!=1581 cdot 1580!-1=1581 a-1Answer: 1581 a-1

🧠:[<think>Okay, let's tackle this problem step by step. First, the problem has two parts, so I need to handle them one by one. Let me start with part (1).The problem states that in triangle ABC, vectors a = (sin A, 1) and b = (cos A, √3) are parallel. Also, angle A is between 0 and π/2. For part (1), we have another condition: sin(ω - A) = 3/5, where 0 < ω < π/2, and we need to find cos ω.First, since vectors a and b are parallel, one must be a scalar multiple of the other. That means their components must be proportional. So, the ratio of their x-components should be equal to the ratio of their y-components. Let me write that down.If vector a is parallel to vector b, then there exists some scalar k such that:sin A = k * cos A 1 = k * √3From the second equation, I can solve for k. Since 1 = k√3, then k = 1/√3.Now, substitute this value of k into the first equation:sin A = (1/√3) * cos A Which simplifies to: sin A / cos A = 1/√3 So, tan A = 1/√3Hmm, tan A is 1 over √3. That corresponds to an angle of π/6, since tan(π/6) = 1/√3. Since A is between 0 and π/2, that makes sense. So, angle A is π/6 or 30 degrees. Let me note that down: A = π/6.Now, moving on to part (1): sin(ω - A) = 3/5. Since we know A is π/6, this becomes sin(ω - π/6) = 3/5. We need to find cos ω.First, let me recall that sin(θ) = 3/5 implies that θ could be in the first or second quadrant. However, the problem states that 0 < ω < π/2, so ω - A must also be within a certain range. Let's check the possible range of ω - A.Given that 0 < ω < π/2 and A = π/6, subtracting A from all parts:-π/6 < ω - A < π/2 - π/6 = π/3.So ω - A is between -π/6 and π/3. However, since sin(ω - A) = 3/5, which is positive, the angle ω - A must be in the first or second quadrant. But the range we have for ω - A is from -π/6 to π/3. The positive part of this range is 0 to π/3. Therefore, ω - A is in the first quadrant (since π/3 is approximately 60 degrees, which is still in the first quadrant). Therefore, ω - A is an angle in the first quadrant with sin(θ) = 3/5. Therefore, we can find cos(θ) using the Pythagorean identity.Let me set θ = ω - A. Then:sin θ = 3/5 cos θ = √(1 - (3/5)^2) = √(1 - 9/25) = √(16/25) = 4/5Therefore, cos θ = 4/5.But θ = ω - A, so cos(ω - A) = 4/5.We need to find cos ω. To do that, we can use the cosine of a difference formula. Let me recall that:cos ω = cos[(ω - A) + A] = cos(θ + A) = cos θ cos A - sin θ sin AWe know cos θ = 4/5, sin θ = 3/5, and we need to find cos A and sin A. Since we found earlier that A = π/6, cos A = cos(π/6) = √3/2, and sin A = sin(π/6) = 1/2.Plugging these into the formula:cos ω = (4/5)(√3/2) - (3/5)(1/2) = (4√3)/10 - 3/10 = (4√3 - 3)/10Therefore, cos ω is (4√3 - 3)/10. Let me check if that makes sense.Alternatively, since θ = ω - A is in the first quadrant, and we have sin θ = 3/5 and cos θ = 4/5, then ω = θ + A. So, cos ω can also be found using the formula for cosine of a sum. Wait, that's exactly what I did. So the answer seems correct.But let me verify once again. A = π/6, θ = ω - π/6, sin θ = 3/5, cos θ = 4/5. So:cos ω = cos(θ + π/6) = cos θ cos π/6 - sin θ sin π/6 = (4/5)(√3/2) - (3/5)(1/2) = (4√3)/10 - 3/10 = (4√3 - 3)/10Yes, that's correct. So part (1)'s answer should be (4√3 - 3)/10. Let me box this for later.Now, moving on to part (2). The problem states that BC = 2√3 and AC + AB = 4. We need to find the area of triangle ABC.First, let's recall that in triangle ABC, BC is one of the sides. Let me denote the sides as follows:- BC = a = 2√3 - AC = b - AB = c - Given that AC + AB = 4, so b + c = 4.We need to find the area of triangle ABC. Since we know angle A from part (1), which is π/6, perhaps we can use the formula for area in terms of two sides and the included angle. The area is (1/2)*b*c*sin A.But we need to find the values of b and c. We know that b + c = 4. Also, using the Law of Cosines on side BC (which is side a), we can relate sides b and c.Law of Cosines: a² = b² + c² - 2bc cos A.We know a = 2√3, A = π/6, so cos A = √3/2. Let me plug in the values:(2√3)^2 = b² + c² - 2bc*(√3/2) 12 = b² + c² - √3 bcBut we also know that b + c = 4. Let me denote S = b + c = 4, and P = bc. Then, b² + c² = S² - 2P = 16 - 2P.Substituting into the equation from the Law of Cosines:12 = (16 - 2P) - √3 P 12 = 16 - 2P - √3 P Bring 16 to the left:12 - 16 = -2P - √3 P -4 = -P(2 + √3) Multiply both sides by -1:4 = P(2 + √3) Therefore, P = 4 / (2 + √3)Rationalizing the denominator:Multiply numerator and denominator by (2 - √3):P = 4*(2 - √3) / [(2 + √3)(2 - √3)] Denominator: 4 - 3 = 1 So P = 4*(2 - √3)/1 = 8 - 4√3Therefore, bc = 8 - 4√3.Now, the area is (1/2)*b*c*sin A. Since sin A = sin(π/6) = 1/2:Area = (1/2)*(8 - 4√3)*(1/2) = (1/4)*(8 - 4√3) = 2 - √3Wait, that seems straightforward, but let me check the calculations again.Starting with area:Area = (1/2)*b*c*sin A = (1/2)*(8 - 4√3)*(1/2) = (1/4)*(8 - 4√3) = (8 - 4√3)/4 = 2 - √3Yes, that's correct. But let me verify if that's possible. The area being 2 - √3. Let me compute the numerical value to check if it's positive. Since √3 ≈ 1.732, then 2 - √3 ≈ 0.267. So the area is approximately 0.267. Let me cross-validate this with another approach.Alternatively, we can use Heron's formula. For Heron's formula, we need all three sides. We know a = 2√3, and b + c = 4. But we need individual values of b and c. However, bc = 8 - 4√3. Let's check if we can find b and c.We have b + c = 4 and bc = 8 - 4√3. The quadratic equation for this would be x² - 4x + (8 - 4√3) = 0. Let's solve for x.Using the quadratic formula:x = [4 ± √(16 - 4*(8 - 4√3))]/2 = [4 ± √(16 - 32 + 16√3)]/2 = [4 ± √(-16 + 16√3)]/2But √(-16 + 16√3) is √(16(√3 - 1)) = 4√(√3 - 1). However, √(√3 - 1) is a real number because √3 ≈ 1.732, so √3 - 1 ≈ 0.732, and its square root is real. Therefore, the roots are real. But Heron's formula would require semi-perimeter, etc. However, this seems complicated, so perhaps sticking to the area formula is better.Alternatively, perhaps the answer is correct. Let me think if there's another way. Since the sides are b and c, with b + c = 4, and bc = 8 - 4√3, and angle A = 30 degrees. Then, the area is (1/2)*b*c*sin(30) = (1/2)*(8 - 4√3)*(1/2) = 2 - √3. That seems consistent.Wait, but let me check if bc = 8 - 4√3 makes sense. If b + c = 4, then the maximum possible bc is when b = c = 2, so bc = 4. But here bc = 8 - 4√3 ≈ 8 - 6.928 ≈ 1.072, which is less than 4, which is okay because 8 - 4√3 is approximately 1.072, which is possible. Therefore, the area is about 0.267, which is small but possible.Alternatively, maybe there's a miscalculation in the Law of Cosines step. Let me go back.Law of Cosines:a² = b² + c² - 2bc cos AGiven a = 2√3, so a² = 12cos A = √3/2Therefore:12 = b² + c² - 2bc*(√3/2) 12 = b² + c² - bc√3Then, since b + c = 4, b² + c² = (b + c)^2 - 2bc = 16 - 2bcSo substituting back into the equation:12 = 16 - 2bc - bc√3 12 = 16 - bc(2 + √3) Then, rearranging:bc(2 + √3) = 16 - 12 = 4 So bc = 4 / (2 + √3) Rationalizing:Multiply numerator and denominator by (2 - √3):bc = [4*(2 - √3)] / [(2 + √3)(2 - √3)] Denominator: 4 - 3 = 1 Therefore, bc = 4*(2 - √3) = 8 - 4√3That's correct. So bc is indeed 8 - 4√3, so the area is 2 - √3. Therefore, the area of triangle ABC is 2 - √3. Let me check units and dimensions. Since BC is 2√3, which is approximately 3.464, and AC + AB = 4, the sides would be such that two sides add to 4, and the third is 2√3. The area seems small but possible. Alternatively, maybe converting 2 - √3 into a different form? But 2 - √3 is already simplified, so that's the answer.Alternatively, if I think in terms of coordinates. Maybe if we place the triangle in coordinate system with point A at the origin, and angle A at the origin, but that might be more complicated. Let's see.Assume point A is at (0,0). Then, vector a = (sin A, 1) and vector b = (cos A, √3) are parallel. Wait, but these vectors might not correspond to the sides of the triangle. Wait, maybe the problem defines vectors a and b as vectors in the plane, but the triangle's sides? Wait, maybe the vectors are not necessarily the sides of the triangle. The problem states "In triangle ABC, we are given vectors a = (sin A, 1) and b = (cos A, √3), with a parallel to b". So these vectors are in the plane, but their relation to the triangle is not specified. However, since the problem is about triangle ABC, and part (2) talks about BC, AC, AB, which are sides of the triangle, but vectors a and b might be other vectors related to the triangle. However, part (1) only requires the angle A, which we found as π/6, and part (2) uses that angle A and the given side lengths.But since part (2) is independent after part (1), and we used angle A = π/6, which was determined from the vectors being parallel, which is given in the problem statement. Therefore, the previous steps are correct.Therefore, the area is 2 - √3.But let me compute 2 - √3 numerically. Since √3 ≈ 1.732, 2 - 1.732 ≈ 0.267. So the area is approximately 0.267. Let's see if that makes sense. Given sides AC + AB = 4, and BC = 2√3 ≈ 3.464. So triangle sides: two sides sum to 4, and the third is 3.464. Since the sum of the two smaller sides must be greater than the largest side. Here, 4 > 3.464, which is okay, so such a triangle exists.But 0.267 area seems small, but perhaps with angle 30 degrees, it's possible. Let me check using another method. Suppose sides AB = c, AC = b, with b + c = 4. Let's say we pick b = 1, c = 3. Then, using Law of Cosines:a² = 1² + 3² - 2*1*3*cos(30°) = 1 + 9 - 6*(√3/2) = 10 - 3√3 ≈ 10 - 5.196 ≈ 4.804 But a is given as 2√3 ≈ 3.464, so 3.464² ≈ 12. So 4.804 ≈ a² isn't equal to 12. Therefore, this is not the case. Hence, the actual values of b and c are such that when squared and subtracted by 2bc*cos(30°), we get 12.But through the previous calculation, bc = 8 - 4√3, which is approximately 1.072. Then, the area is 0.267. It seems very small, but perhaps correct. Alternatively, maybe there's a miscalculation. Wait, let's see:Area = (1/2)*b*c*sin A = (1/2)*(8 - 4√3)*(1/2) = (8 - 4√3)/4 = 2 - √3. Yes, that's correct. So if bc is approximately 1.072, then (1/2)*1.072*0.5 ≈ 0.268, which matches. Therefore, the answer is correct.Therefore, part (2)'s answer is 2 - √3.But let me check again with Heron's formula. Suppose a = 2√3, b + c = 4, and bc = 8 - 4√3. Let's suppose we compute semi-perimeter:s = (a + b + c)/2 = (2√3 + 4)/2 = 2 + √3.Then, Heron's formula:Area = √[s(s - a)(s - b)(s - c)]But we don't know individual b and c, but we can express in terms of s.Wait, since b + c = 4 and a = 2√3, so s = (2√3 + 4)/2 = 2 + √3.Then, s - a = (2 + √3) - 2√3 = 2 - √3 s - b and s - c: since b + c = 4, then s - b = (2 + √3) - b, and s - c = (2 + √3) - c. However, without individual values of b and c, this might not help. But perhaps use the formula:Area = √[s(s - a)(s - b)(s - c)]But we can express (s - b)(s - c) as (s - b)(s - c) = s² - s(b + c) + bc. Since b + c = 4, this becomes s² - 4s + bc.Therefore:Area = √[s(s - a)(s² - 4s + bc)]We know s = 2 + √3, a = 2√3, bc = 8 - 4√3.Compute s - a = 2 + √3 - 2√3 = 2 - √3 s² = (2 + √3)^2 = 4 + 4√3 + 3 = 7 + 4√3 Then s² - 4s + bc = (7 + 4√3) - 4*(2 + √3) + (8 - 4√3) = 7 + 4√3 - 8 - 4√3 + 8 - 4√3 = (7 - 8 + 8) + (4√3 - 4√3 - 4√3) = 7 - 4√3Therefore, Area = √[s(s - a)(s² - 4s + bc)] = √[(2 + √3)(2 - √3)(7 - 4√3)]First compute (2 + √3)(2 - √3) = 4 - 3 = 1 Then, Area = √[1*(7 - 4√3)] = √(7 - 4√3)Wait, but we previously found the area to be 2 - √3. Let me check if √(7 - 4√3) equals 2 - √3.Compute (2 - √3)^2 = 4 - 4√3 + 3 = 7 - 4√3. Therefore, √(7 - 4√3) = 2 - √3. Since 2 - √3 is positive (≈0.267), the square root is indeed 2 - √3. Therefore, Heron's formula confirms the area is 2 - √3. Therefore, part (2)'s answer is 2 - √3.So both answers check out. Therefore, the final answers are:(1) cos ω = (4√3 - 3)/10 (2) Area = 2 - √3Final Answer(1) boxed{dfrac{4sqrt{3} - 3}{10}} (2) boxed{2 - sqrt{3}}</think>Given triangle (ABC) with vectors (overrightarrow{a} = (sin A, 1)) and (overrightarrow{b} = (cos A, sqrt{3})) being parallel, we determine that (tan A = frac{1}{sqrt{3}}), leading to (A = frac{pi}{6}). Part (1)Given (sin(omega - A) = frac{3}{5}) with (0 < omega < frac{pi}{2}):1. Since (A = frac{pi}{6}), we have (sinleft(omega - frac{pi}{6}right) = frac{3}{5}).2. (omega - frac{pi}{6}) is in the first quadrant, so (cosleft(omega - frac{pi}{6}right) = frac{4}{5}).3. Using the angle addition formula for cosine: [ cos omega = cosleft(left(omega - frac{pi}{6}right) + frac{pi}{6}right) = cosleft(omega - frac{pi}{6}right)cosfrac{pi}{6} - sinleft(omega - frac{pi}{6}right)sinfrac{pi}{6} ] Substituting the known values: [ cos omega = frac{4}{5} cdot frac{sqrt{3}}{2} - frac{3}{5} cdot frac{1}{2} = frac{4sqrt{3}}{10} - frac{3}{10} = frac{4sqrt{3} - 3}{10} ] Thus, (cos omega = boxed{dfrac{4sqrt{3} - 3}{10}}). Part (2)Given (BC = 2sqrt{3}) and (AC + AB = 4):1. Using the Law of Cosines for side (BC): [ (2sqrt{3})^2 = AB^2 + AC^2 - 2 cdot AB cdot AC cdot cosfrac{pi}{6} ] Simplifying, we get: [ 12 = (AB + AC)^2 - 2 cdot AB cdot AC - sqrt{3} cdot AB cdot AC ] Substituting (AB + AC = 4): [ 12 = 16 - 2 cdot AB cdot AC - sqrt{3} cdot AB cdot AC ] Solving for (AB cdot AC): [ AB cdot AC = frac{4}{2 + sqrt{3}} = 8 - 4sqrt{3} ]2. The area of triangle (ABC) is: [ text{Area} = frac{1}{2} cdot AB cdot AC cdot sinfrac{pi}{6} = frac{1}{2} cdot (8 - 4sqrt{3}) cdot frac{1}{2} = 2 - sqrt{3} ] Thus, the area of triangle (ABC) is (boxed{2 - sqrt{3}}).,<think>Okay, so I need to solve this problem about triangle ABC. There are two parts: part (1) involves vectors and trigonometric identities, and part (2) is about finding the area of the triangle given some side lengths. Let me start with part (1).First, the problem says that vectors a = (sin A, 1) and b = (cos A, sqrt(3)) are parallel. If two vectors are parallel, one must be a scalar multiple of the other. That means there exists some scalar k such that:sin A = k * cos A 1 = k * sqrt(3)So from the second equation, I can solve for k. If 1 = k * sqrt(3), then k = 1 / sqrt(3). Then plug this k into the first equation:sin A = (1 / sqrt(3)) * cos A Which simplifies to sin A / cos A = 1 / sqrt(3) So tan A = 1 / sqrt(3)Hmm, tan A equals 1 over sqrt(3). The angle whose tangent is 1/sqrt(3) is 30 degrees, since tan(30°) = 1/sqrt(3). Since A is between 0 and pi/2, which is 0 to 90 degrees, A must be 30°, or pi/6 radians. So angle A is pi/6.Now, part (1) asks: If sin(omega - A) = 3/5, with 0 < omega < pi/2, find cos omega. Since we found A is pi/6, then sin(omega - pi/6) = 3/5. Let me denote theta = omega - pi/6, so sin theta = 3/5. Then theta is in the range ( -pi/6, pi/2 - pi/6 ) = (-pi/6, pi/3). But since omega is between 0 and pi/2, theta = omega - pi/6 must be between -pi/6 and pi/3. However, sin(theta) = 3/5, which is positive, so theta must be in the first quadrant. Therefore, theta is between 0 and pi/3. So theta = arcsin(3/5).Therefore, omega = pi/6 + arcsin(3/5). Then we need to find cos(omega). Let me compute cos(pi/6 + arcsin(3/5)). Using the cosine addition formula:cos(pi/6 + theta) = cos(pi/6)cos(theta) - sin(pi/6)sin(theta)We already know theta = arcsin(3/5), so sin(theta) = 3/5, and cos(theta) can be found using Pythagoras. Since sin(theta) = 3/5, then cos(theta) = sqrt(1 - (3/5)^2) = sqrt(1 - 9/25) = sqrt(16/25) = 4/5. But theta is in the first quadrant, so cos(theta) is positive. Therefore:cos(pi/6 + theta) = (sqrt(3)/2)(4/5) - (1/2)(3/5) = (4 sqrt(3)/10) - (3/10) = (4 sqrt(3) - 3)/10Therefore, cos omega = (4 sqrt(3) - 3)/10. Let me check if that makes sense. Alternatively, maybe there's another way to approach it. Let me verify.Alternatively, since sin(omega - A) = 3/5, and we can find omega - A = arcsin(3/5). Then omega = A + arcsin(3/5). Then cos omega = cos(A + arcsin(3/5)). Since A is pi/6, same as before. So this seems consistent. So yes, using the cosine addition formula gives us (4 sqrt(3) - 3)/10.Wait, let me compute this again step by step:cos(pi/6) is sqrt(3)/2, cos(theta) is 4/5. sin(pi/6) is 1/2, sin(theta) is 3/5. Therefore:sqrt(3)/2 * 4/5 = (4 sqrt(3))/10 1/2 * 3/5 = 3/10 So subtract: (4 sqrt(3) - 3)/10Yes, that's correct. So cos omega is (4 sqrt(3) - 3)/10. So that's the answer for part (1). Let me write that down.Now moving on to part (2). The problem states: If BC = 2 sqrt(3) and AC + AB = 4, find the area of triangle ABC.First, let's recall that in triangle ABC, BC is one side, and we need to relate the other sides. Since in part (1) we found angle A is pi/6, which is 30 degrees. So angle A is 30 degrees. Let me confirm: from the first part, A is pi/6. So in triangle ABC, angle at A is 30 degrees. Then BC is the side opposite angle A. Wait, in triangle notation, side BC is opposite angle A. Because in triangle ABC, side opposite angle A is BC. So BC = a, AC = b, AB = c. Wait, maybe different notation. Let's recall:In standard notation, for triangle ABC:- a is the length of BC (opposite angle A)- b is the length of AC (opposite angle B)- c is the length of AB (opposite angle C)But given that in the problem, they are using vectors a and b, but maybe that's separate. Let's be careful.Wait, in the problem, vectors a and b are given as (sin A, 1) and (cos A, sqrt(3)), but they are just vectors, not necessarily related to the sides. But in part (2), BC = 2 sqrt(3), which is the length of BC. So BC is a side, and in standard notation, BC is side a (opposite angle A). Then angle A is 30 degrees (pi/6). Then using the Law of Sines or Law of Cosines, perhaps.But given that AC + AB = 4. So AC and AB are sides adjacent to angle A. Wait, AC is side b (opposite angle B), AB is side c (opposite angle C). But AC is adjacent to angle B and angle A? Wait, perhaps it's better to visualize the triangle.Let me draw triangle ABC with angle at A being 30 degrees. So vertex A has angle 30°, side BC is opposite angle A, so BC = a = 2 sqrt(3). Then sides AB and AC are sides adjacent to angle A. The problem states that AC + AB = 4. So AB + AC = 4. So if we denote AB = c, AC = b, then b + c = 4. We need to find the area of the triangle.Given angle A, sides adjacent to A sum to 4, and the side opposite angle A is 2 sqrt(3). So how can we find the area? The area can be calculated as (1/2)*AB*AC*sin A. Since angle A is 30°, sin A is 1/2. So area = (1/2)*b*c*(1/2) = (1/4)*b*c. So if we can find the product b*c, then we can compute the area.Alternatively, using the Law of Cosines. Since we know sides b and c, and angle A, we can relate them to side a. Law of Cosines says:a² = b² + c² - 2bc cos AWe have a = 2 sqrt(3), angle A = 30°, so cos A = sqrt(3)/2. Therefore:(2 sqrt(3))² = b² + c² - 2bc*(sqrt(3)/2) 12 = b² + c² - bc sqrt(3)Also, we know that b + c = 4. Let me denote s = b + c = 4. Let me think how to express b² + c² in terms of s and bc. We know that (b + c)^2 = b² + 2bc + c², so b² + c² = s² - 2bc = 16 - 2bc.Substitute into the equation:12 = (16 - 2bc) - bc sqrt(3) 12 = 16 - 2bc - bc sqrt(3) Then subtract 16 from both sides:-4 = -2bc - bc sqrt(3) Multiply both sides by -1:4 = 2bc + bc sqrt(3) Factor out bc:4 = bc(2 + sqrt(3)) Therefore, bc = 4 / (2 + sqrt(3))Rationalize the denominator:Multiply numerator and denominator by (2 - sqrt(3)):bc = [4*(2 - sqrt(3))] / [(2 + sqrt(3))(2 - sqrt(3))] Denominator is 4 - 3 = 1. So bc = 4*(2 - sqrt(3)) = 8 - 4 sqrt(3)Therefore, bc = 8 - 4 sqrt(3). Then the area is (1/4)*bc = (1/4)*(8 - 4 sqrt(3)) = 2 - sqrt(3). But wait, that seems small. Let me check.Wait, area = (1/2)*b*c*sin A. Since angle A is 30°, sin A = 1/2. So area = (1/2)*b*c*(1/2) = (1/4)*b*c. So yes, if bc = 8 - 4 sqrt(3), then area is (8 - 4 sqrt(3))/4 = 2 - sqrt(3). Hmm, is that correct?But let me check the calculations again step by step.Law of Cosines:a² = b² + c² - 2bc cos AGiven:a = 2 sqrt(3) => a² = 12cos A = sqrt(3)/2Therefore:12 = b² + c² - 2bc*(sqrt(3)/2) 12 = b² + c² - bc sqrt(3)Then, since b + c = 4:(b + c)^2 = 16 = b² + 2bc + c² Therefore, b² + c² = 16 - 2bc. Substitute into previous equation:12 = (16 - 2bc) - bc sqrt(3) 12 = 16 - 2bc - bc sqrt(3) Bring 16 to left:12 - 16 = -2bc - bc sqrt(3) -4 = -bc(2 + sqrt(3)) Multiply both sides by -1:4 = bc(2 + sqrt(3)) Therefore:bc = 4 / (2 + sqrt(3)) Multiply numerator and denominator by (2 - sqrt(3)):bc = [4*(2 - sqrt(3))] / [(2 + sqrt(3))(2 - sqrt(3))] Denominator is 4 - 3 = 1 Therefore, bc = 4*(2 - sqrt(3)) = 8 - 4 sqrt(3)So bc = 8 - 4 sqrt(3). Then area is (1/4)*bc = (1/4)*(8 - 4 sqrt(3)) = 2 - sqrt(3). So 2 - sqrt(3) is approximately 2 - 1.732 = 0.267. That seems very small for a triangle with sides adding up to 4 and a side of 2 sqrt(3). Let me check if that makes sense.Alternatively, maybe I made a mistake in the area formula. Wait, area is (1/2)*ab*sin C, where a and b are two sides and C is the included angle. In this case, angle A is 30°, and the sides adjacent to angle A are AB and AC, which are of lengths c and b. Therefore, the area should be (1/2)*b*c*sin A = (1/2)*b*c*(1/2) = (1/4)*b*c. So that part is correct. So if bc is 8 - 4 sqrt(3), then area is 2 - sqrt(3). Let me compute sqrt(3) ≈ 1.732, so 2 - 1.732 ≈ 0.267. But maybe that's correct? Let me consider the triangle.Alternatively, maybe using Heron's formula. Let's check.If we have sides BC = 2 sqrt(3), AB = c, AC = b, with b + c = 4. Then perimeter semiperimeter would be (a + b + c)/2 = (2 sqrt(3) + 4)/2 = 2 + sqrt(3). Then area = sqrt[s(s - a)(s - b)(s - c)]. But without knowing individual values of b and c, this might not help. Alternatively, if bc is 8 - 4 sqrt(3), then area is 2 - sqrt(3). Let me verify with possible values of b and c.Suppose b + c = 4 and bc = 8 - 4 sqrt(3). Then we can solve for b and c. Let me set up the quadratic equation:x² - 4x + (8 - 4 sqrt(3)) = 0Solutions:x = [4 ± sqrt(16 - 4*(8 - 4 sqrt(3}))]/2 = [4 ± sqrt(16 - 32 + 16 sqrt(3))]/2 = [4 ± sqrt(-16 + 16 sqrt(3))]/2Wait, sqrt(-16 + 16 sqrt(3)) is sqrt(16(sqrt(3) - 1)) = 4 sqrt(sqrt(3) - 1). Since sqrt(3) ≈ 1.732, sqrt(3) - 1 ≈ 0.732, so sqrt(0.732) ≈ 0.855. So sqrt(-16 + 16 sqrt(3)) ≈ 4 * 0.855 ≈ 3.42. So x = [4 ± 3.42]/2. So x ≈ (4 + 3.42)/2 ≈ 7.42/2 ≈ 3.71, and x ≈ (4 - 3.42)/2 ≈ 0.58/2 ≈ 0.29. So sides would be approximately 3.71 and 0.29. Adding up to 4. Then using these sides with BC = 2 sqrt(3) ≈ 3.464. So sides would be approx 3.71, 0.29, and 3.464. Then area is approx 0.267. Hmm, seems possible. But maybe there's a mistake.Alternatively, maybe using vectors from part (1) to find something else? Wait, in part (1), we found angle A is 30°, which is used in part (2). The vectors in part (1) are given to find angle A, which is used in part (2). So the connection is angle A is 30°, which is essential for part (2). So maybe the calculation is correct. Let me think again.Alternatively, perhaps the answer is 2 - sqrt(3). But to check, let's consider a different approach. Let me use Law of Sines. In triangle ABC, with angle A = 30°, BC = a = 2 sqrt(3), and sides AB = c, AC = b. Then Law of Sines:a / sin A = b / sin B = c / sin CSo 2 sqrt(3) / sin 30° = 2 sqrt(3) / 0.5 = 4 sqrt(3). So all sides can be expressed as:b = 4 sqrt(3) * sin B c = 4 sqrt(3) * sin CSince angle A is 30°, angles B + C = 150°. Also, sides b + c = 4. So:4 sqrt(3) sin B + 4 sqrt(3) sin C = 4 Divide both sides by 4 sqrt(3):sin B + sin C = 1 / sqrt(3)But angles B + C = 150°, so let's let C = 150° - B. Then:sin B + sin(150° - B) = 1 / sqrt(3)Using sine subtraction formula:sin(150° - B) = sin 150° cos B - cos 150° sin B = (1/2) cos B - (-sqrt(3)/2) sin B = (1/2) cos B + (sqrt(3)/2) sin BTherefore:sin B + (1/2 cos B + sqrt(3)/2 sin B) = 1 / sqrt(3) Combine like terms:sin B + (sqrt(3)/2) sin B + (1/2) cos B = 1 / sqrt(3) Factor sin B:sin B (1 + sqrt(3)/2) + (1/2) cos B = 1 / sqrt(3)This seems complicated, but maybe we can write this as:Let me factor out 1/2:(1/2)[2 sin B + sqrt(3) sin B + cos B] = 1 / sqrt(3) Wait, no. Let me think of another approach.Alternatively, let me set angle B = x, so angle C = 150° - x. Then:sin x + sin(150° - x) = 1 / sqrt(3)Use the identity sin A + sin B = 2 sin[(A + B)/2] cos[(A - B)/2]So sin x + sin(150° - x) = 2 sin[(x + 150° - x)/2] cos[(x - (150° - x))/2] = 2 sin(75°) cos(2x - 150°)/2 Wait, let's compute:(A + B)/2 = (x + 150° - x)/2 = 150°/2 = 75° (A - B)/2 = (x - (150° - x))/2 = (2x - 150°)/2 = x - 75°Therefore:sin x + sin(150° - x) = 2 sin(75°) cos(x - 75°) = 1 / sqrt(3)Therefore:2 sin(75°) cos(x - 75°) = 1 / sqrt(3) Divide both sides by 2 sin(75°):cos(x - 75°) = 1 / (2 sqrt(3) sin(75°))Compute sin(75°). sin(75°) = sin(45° + 30°) = sin45 cos30 + cos45 sin30 = (sqrt(2)/2)(sqrt(3)/2) + (sqrt(2)/2)(1/2) = sqrt(6)/4 + sqrt(2)/4 = (sqrt(6) + sqrt(2))/4 ≈ 0.9659So sin(75°) ≈ 0.9659, and 2 sqrt(3) sin(75°) ≈ 2 * 1.732 * 0.9659 ≈ 3.346Thus, 1 / (2 sqrt(3) sin(75°)) ≈ 1 / 3.346 ≈ 0.298So cos(x - 75°) ≈ 0.298, which would mean x - 75° ≈ arccos(0.298) ≈ 72.7°, so x ≈ 75° + 72.7° ≈ 147.7°, but angle B can't be 147.7° since angle A is already 30°, and angles sum to 180°. Alternatively, x - 75° ≈ -72.7°, so x ≈ 75° -72.7° ≈ 2.3°, so angle B ≈ 2.3°, angle C ≈ 150° - 2.3° ≈ 147.7°. Wait, but angle C would be 147.7°, which is too big. But angles B and C sum to 150°, so if angle B is small, angle C is large. However, in a triangle, all angles must be positive. But with angle C being 147.7°, that's possible if sides are such. Then sides b and c would be:b = 4 sqrt(3) sin B ≈ 4 * 1.732 * sin 2.3° ≈ 6.928 * 0.04 ≈ 0.277 c = 4 sqrt(3) sin C ≈ 4 * 1.732 * sin 147.7° ≈ 6.928 * 0.544 ≈ 3.77Then b + c ≈ 0.277 + 3.77 ≈ 4.047, which is close to 4, but not exact due to approximation errors. But given that the exact answer gives bc = 8 - 4 sqrt(3) ≈ 8 - 6.928 ≈ 1.072, which when multiplied by 1/4 gives area ≈ 0.268, which is consistent with the Heron's formula approach. So despite seeming small, perhaps that's the correct answer.Alternatively, maybe I made a mistake in the Law of Cosines approach.Wait, let me recast the problem: given angle A = 30°, sides AB + AC = 4, side BC = 2 sqrt(3). Find the area.Another approach: use coordinates. Let me place point A at the origin (0,0), and since angle A is 30°, we can set coordinates for points B and C.But maybe better to place point A at the origin, side AB along the x-axis. Let me try that.Let me place point A at (0,0). Let me let AB = c, so point B is at (c, 0). Then point C is somewhere in the plane. Since angle at A is 30°, the coordinates of point C can be determined using angle A = 30°, AC = b. Therefore, coordinates of C are (b cos 30°, b sin 30°) = (b*(sqrt(3)/2), b*(1/2))Then BC is the distance between (c, 0) and (b*sqrt(3)/2, b/2). So BC^2 = (c - b*sqrt(3)/2)^2 + (0 - b/2)^2 = (c - (b sqrt(3)/2))^2 + (b/2)^2Given BC = 2 sqrt(3), so BC^2 = 12. Therefore:(c - (b sqrt(3)/2))^2 + (b/2)^2 = 12Also, we know that b + c = 4. Let me write c = 4 - b. Substitute into the equation:( (4 - b) - (b sqrt(3)/2) )^2 + (b/2)^2 = 12Let me simplify the first term:(4 - b - (b sqrt(3)/2)) = 4 - b(1 + sqrt(3)/2)Let me denote k = 1 + sqrt(3)/2 ≈ 1 + 0.866 ≈ 1.866Then expression becomes (4 - k*b)^2 + (b/2)^2 = 12Expand (4 - k b)^2:16 - 8 k b + k² b²Add (b² /4):16 - 8 k b + k² b² + b² /4 = 12Combine the b² terms:(k² + 1/4) b² - 8 k b + 16 = 12Subtract 12:(k² + 1/4) b² - 8 k b + 4 = 0Now plug in k = 1 + sqrt(3)/2:First compute k²:k = 1 + sqrt(3)/2 k² = (1)^2 + 2*1*(sqrt(3)/2) + (sqrt(3)/2)^2 = 1 + sqrt(3) + 3/4 = (4/4 + 4 sqrt(3)/4 + 3/4) = (7/4 + sqrt(3))Therefore, k² + 1/4 = 7/4 + sqrt(3) + 1/4 = 2 + sqrt(3)So equation becomes:(2 + sqrt(3)) b² - 8*(1 + sqrt(3)/2) b + 4 = 0Simplify the middle term:8*(1 + sqrt(3)/2) = 8 + 4 sqrt(3)So the equation is:(2 + sqrt(3)) b² - (8 + 4 sqrt(3)) b + 4 = 0Let me write this equation as:(2 + sqrt(3)) b² - (8 + 4 sqrt(3)) b + 4 = 0Let me divide all terms by (2 + sqrt(3)) to simplify:b² - [ (8 + 4 sqrt(3)) / (2 + sqrt(3)) ] b + [4 / (2 + sqrt(3))] = 0Compute the coefficients:First, [ (8 + 4 sqrt(3)) / (2 + sqrt(3)) ]Factor numerator: 4*(2 + sqrt(3))Therefore, [4*(2 + sqrt(3))] / (2 + sqrt(3)) = 4Similarly, [4 / (2 + sqrt(3))] = multiply numerator and denominator by (2 - sqrt(3)):[4*(2 - sqrt(3))]/[(2 + sqrt(3))(2 - sqrt(3))] = [8 - 4 sqrt(3)] / (4 - 3) = 8 - 4 sqrt(3)Therefore, the equation simplifies to:b² - 4 b + (8 - 4 sqrt(3)) = 0Which is the same quadratic as before: x² -4x + (8 -4 sqrt(3)) = 0, where x = b. So we have:b = [4 ± sqrt(16 - 4*(8 - 4 sqrt(3)))] / 2 = [4 ± sqrt(16 -32 +16 sqrt(3))]/2 = [4 ± sqrt(-16 +16 sqrt(3))]/2 = [4 ± 4 sqrt(sqrt(3) -1 ) ] /2 = 2 ± 2 sqrt(sqrt(3) -1 )But sqrt(3) -1 ≈ 0.732, so sqrt(0.732) ≈ 0.855, so sqrt(sqrt(3)-1) ≈ 0.855. Therefore:b ≈ 2 ± 2*0.855 ≈ 2 ± 1.71So b ≈ 3.71 or b ≈ 0.29, same as before. Therefore, sides b and c are approximately 3.71 and 0.29, which multiply to bc ≈ 3.71*0.29 ≈ 1.076, which is close to 8 -4 sqrt(3) ≈ 1.072. Hence, bc ≈ 1.072, leading to area ≈ 0.268. So this seems consistent. Therefore, the exact area is 2 - sqrt(3), which is approximately 0.2679, which matches. Therefore, the answer is 2 - sqrt(3).But wait, the problem says "find the area of triangle ABC". So the answer is 2 - sqrt(3). Let me confirm that 2 - sqrt(3) is positive. Since sqrt(3) ≈ 1.732, 2 -1.732 ≈ 0.267, which is positive, so it's valid.Therefore, part (2)'s answer is 2 - sqrt(3). Therefore, boxed answers are:(1) boxed{dfrac{4sqrt{3} - 3}{10}}(2) boxed{2 - sqrt{3}}Final Answer(1) boxed{dfrac{4sqrt{3} - 3}{10}} (2) boxed{2 - sqrt{3}}</think>Given triangle (ABC) with vectors (overrightarrow{a} = (sin A, 1)) and (overrightarrow{b} = (cos A, sqrt{3})) being parallel, we need to solve the following problems:(1) Since (overrightarrow{a} parallel overrightarrow{b}), there exists a scalar (k) such that (sin A = k cos A) and (1 = k sqrt{3}). Solving for (k), we get (k = frac{1}{sqrt{3}}). This leads to (tan A = frac{1}{sqrt{3}}), so (A = frac{pi}{6}).Given (sin(omega - A) = frac{3}{5}) with (0 < omega < frac{pi}{2}), let (theta = omega - A). Then (sin theta = frac{3}{5}) and (theta) is in the first quadrant. We find (cos theta = frac{4}{5}). Using the cosine addition formula:[cos omega = cos(A + theta) = cosleft(frac{pi}{6} + thetaright) = cosleft(frac{pi}{6}right)costheta - sinleft(frac{pi}{6}right)sintheta][= frac{sqrt{3}}{2} cdot frac{4}{5} - frac{1}{2} cdot frac{3}{5} = frac{4sqrt{3}}{10} - frac{3}{10} = frac{4sqrt{3} - 3}{10}](2) Given (BC = 2sqrt{3}) and (AC + AB = 4), we use the Law of Cosines. Let (AB = c), (AC = b), and (BC = a = 2sqrt{3}). With (A = frac{pi}{6}), we have:[a^2 = b^2 + c^2 - 2bc cos A][12 = b^2 + c^2 - 2bc cdot frac{sqrt{3}}{2}]Given (b + c = 4), we use (b^2 + c^2 = (b + c)^2 - 2bc = 16 - 2bc). Substituting, we get:[12 = 16 - 2bc - bcsqrt{3}][4 = bc(2 + sqrt{3})]Solving for (bc):[bc = frac{4}{2 + sqrt{3}} = frac{4(2 - sqrt{3})}{(2 + sqrt{3})(2 - sqrt{3})} = 8 - 4sqrt{3}]The area of triangle (ABC) is:[text{Area} = frac{1}{2}bc sin A = frac{1}{2}(8 - 4sqrt{3}) cdot frac{1}{2} = frac{8 - 4sqrt{3}}{4} = 2 - sqrt{3}]Final answers:(1) (boxed{dfrac{4sqrt{3} - 3}{10}}) (2) (boxed{2 - sqrt{3}})]

🔑:1. Since overrightarrow{a} parallel overrightarrow{b}, we have the condition cos A - sqrt{3} sin A = 0. This can be rewritten as tan A = frac{1}{sqrt{3}}. Since A in (0, frac{pi}{2}), we deduce A = frac{pi}{6}.Given sin(omega - A) = frac{3}{5}, we can express sin omega using the sine of angle addition formula: sin omega = sin (omega - A + A) = sin (omega - A) cos A + cos (omega - A) sin A, which simplifies tosin omega = frac{3}{5} cos A + sqrt{1 - left(frac{3}{5}right)^2} sin A.Substituting A = frac{pi}{6}, we get sin omega = frac{3}{5} times frac{sqrt{3}}{2} + frac{4}{5} times frac{1}{2}. Since 0 < omega < frac{pi}{2}, we use sin omega = sqrt{1 - cos^2 omega}.Hence, we get the equation frac{3sqrt{3}}{10} + frac{2}{5} = sqrt{1 - cos^2 omega}. Squaring both sides and solving for cos omega, we end up with a quadratic equation 100cos^2 omega + 60cos omega - 39 = 0. Solving this, we discard the negative root frac{-3 - 4sqrt{3}}{10} (since cos omega must be positive in this interval) and obtain boxed{cos omega = frac{4sqrt{3} - 3}{10}}.2. Since BC = 2sqrt{3} and AC + AB = 4 with A = frac{pi}{6}, we can use the Law of Cosines:BC^2 = AB^2 + AC^2 - 2 cdot AB cdot AC cdot sin A.Simplifying with the given values:12 = AB^2 + AC^2 - 2cdot AB cdot AC cdot frac{1}{2} = (AB + AC)^2 - (2 + sqrt{3}) cdot AB cdot AC = 16 - (2 + sqrt{3}) cdot AB cdot AC.Solving for AB cdot AC, we find AB cdot AC = frac{4}{2 + sqrt{3}}. Therefore, the area of triangle ABC is given by:text{Area} = frac{1}{2} cdot AB cdot AC cdot sin A = frac{1}{2} times frac{4}{2 + sqrt{3}} times frac{1}{2} = frac{2}{2 + sqrt{3}}.Rationalizing the denominator, the area becomes:boxed{text{Area} = 2 - sqrt{3}}.

🧠:[<think>Okay, let's try to tackle this problem step by step. First, part (Ⅰ) where a = 1. Hmm. So, I need to find the intervals where the function is increasing or decreasing and also find the extreme values. Let me start by plugging a = 1 into the given function.The original function is f(x) = (1/3)x³ + [(1 - a)/2]x² - a² ln x + a² ln a. When a = 1, let's substitute that in:f(x) = (1/3)x³ + [(1 - 1)/2]x² - (1)² ln x + (1)² ln 1.Simplify each term:The second term becomes (0)/2 x² = 0. The last term is 1² ln 1 = 1 * 0 = 0. So the function simplifies to f(x) = (1/3)x³ - ln x.Alright, so now we need to find the monotonicity and extreme values of f(x) = (1/3)x³ - ln x. To do that, I should first find the derivative f’(x) because that will tell me where the function is increasing or decreasing.Calculating f’(x):f’(x) = d/dx [ (1/3)x³ - ln x ] = (1/3)*3x² - (1/x) = x² - 1/x.Simplify: f’(x) = x² - 1/x. To find critical points, set f’(x) = 0:x² - 1/x = 0. Let's solve for x. Multiply both sides by x to eliminate the denominator (since x > 0 because ln x is in the original function, so domain is x > 0):x³ - 1 = 0 => x³ = 1 => x = 1.So the critical point is at x = 1. Now, we need to test the intervals around x = 1 to see where the derivative is positive or negative.The domain is x > 0. Let's pick test points in (0,1) and (1, ∞).First interval: Let's take x = 0.5 (which is between 0 and 1):f’(0.5) = (0.5)² - 1/(0.5) = 0.25 - 2 = -1.75. Negative, so function is decreasing here.Second interval: x = 2 (which is greater than 1):f’(2) = (2)² - 1/2 = 4 - 0.5 = 3.5. Positive, so function is increasing here.Therefore, the function decreases on (0, 1) and increases on (1, ∞). Thus, there is a local minimum at x = 1. To find the extreme value, compute f(1):f(1) = (1/3)(1)^3 - ln 1 = 1/3 - 0 = 1/3.So part (Ⅰ) answer: decreasing on (0,1), increasing on (1, ∞), with a local minimum at (1, 1/3). That seems straightforward.Now part (Ⅱ): If f(x) has two distinct zeros, find the range of real number a > 0. Hmm. So we need to find a such that the equation f(x) = 0 has two distinct solutions. Let's recall the original function:f(x) = (1/3)x³ + [(1 - a)/2]x² - a² ln x + a² ln a.Wait, the function includes ln x and ln a, so domain is x > 0, and a > 0 as given.We need f(x) = 0 to have two distinct roots. To analyze this, perhaps we can consider the behavior of f(x) and see under what conditions it crosses the x-axis twice.First, let's analyze the function as x approaches 0+ and as x approaches infinity. Because the function has ln x term, as x approaches 0+, ln x approaches negative infinity, so -a² ln x approaches positive infinity. Also, the polynomial terms: (1/3)x³ and [(1 - a)/2]x² both approach 0. So as x approaches 0+, f(x) tends to positive infinity.As x approaches infinity, the leading term (1/3)x³ dominates, so f(x) tends to infinity. The other terms: [(1 - a)/2]x² is quadratic, which is dominated by cubic. The logarithmic terms become negligible. So as x approaches infinity, f(x) tends to infinity.Therefore, the function tends to infinity as x approaches 0+ and as x approaches infinity. So for the function to cross the x-axis twice, it must dip below the x-axis somewhere in between. So the function must have a minimum somewhere where f(x) < 0. If the minimum value is less than zero, then since the function goes to infinity on both ends, it will cross the axis twice. But if the minimum is exactly zero, it would touch the axis once (a double root), and if the minimum is above zero, no real roots. Wait, but the problem says two distinct zeros, so we need exactly two crossings. Therefore, the function must attain a minimum value less than zero. Therefore, the key is to find when the minimum of f(x) is less than zero.Alternatively, since the function goes to infinity on both ends, if there's a point where it's below zero, then there must be two crossings. But actually, maybe even two minima? Wait, no. The function's derivative is f’(x). Let's compute f’(x) for general a.Compute f’(x):f’(x) = derivative of (1/3)x³ + [(1 - a)/2]x² - a² ln x + a² ln a.Differentiate term by term:Derivative of (1/3)x³ is x².Derivative of [(1 - a)/2]x² is (1 - a)x.Derivative of -a² ln x is -a² / x.Derivative of a² ln a is 0 since it's a constant.Thus, f’(x) = x² + (1 - a)x - a² / x.Set f’(x) = 0 for critical points:x² + (1 - a)x - a² / x = 0.Multiply both sides by x to eliminate denominator (x > 0):x³ + (1 - a)x² - a² = 0.So, x³ + (1 - a)x² - a² = 0.Hmm, solving this cubic equation for x. Hmm. Maybe factor it?Let me try to factor this cubic equation. Let’s consider possible rational roots. By Rational Root Theorem, possible roots are factors of -a² over factors of 1. So possible roots are ±1, ±a, ±a², etc. Let me test x = a:Plug x = a into the equation:a³ + (1 - a)a² - a² = a³ + a² - a³ - a² = 0. So x = a is a root.Therefore, we can factor (x - a) out of the cubic.Using polynomial division or synthetic division:Divide x³ + (1 - a)x² - a² by (x - a):Let's use synthetic division. If root is x = a:Coefficients: 1 (x³), (1 - a) (x²), 0 (x term), -a² (constant term).Wait, cubic is x³ + (1 - a)x² + 0x - a². So synthetic division with root a:Bring down the 1. Multiply by a: 1 * a = a. Add to next coefficient: (1 - a) + a = 1. Multiply by a: 1 * a = a. Add to next coefficient: 0 + a = a. Multiply by a: a * a = a². Add to last term: -a² + a² = 0. Perfect, so the cubic factors as (x - a)(x² + x + a).Therefore, the equation factors into (x - a)(x² + x + a) = 0.So critical points are x = a and roots of x² + x + a = 0. Let's check discriminant of x² + x + a = 0: Discriminant D = 1 - 4a. Since a > 0, if 1 - 4a < 0, which is when a > 1/4, then no real roots. If a = 1/4, repeated root. If a < 1/4, two real roots. However, since x > 0 (domain), even if there are real roots, we need to check if they are positive.Suppose a < 1/4, then roots are x = [-1 ± sqrt(1 - 4a)]/2. Let's check if these are positive:sqrt(1 - 4a) is real and less than 1. So:x = [-1 + sqrt(1 - 4a)]/2 and x = [-1 - sqrt(1 - 4a)]/2.The second root is negative because sqrt(1 - 4a) < 1, so -1 - sqrt(1 - 4a) < -1 - 0 = -1, divided by 2, still negative. The first root: [-1 + sqrt(1 - 4a)]/2. Since sqrt(1 - 4a) < 1, so -1 + sqrt(...) is negative. Therefore, both roots are negative. So even when a < 1/4, the quadratic factor has no positive roots. Therefore, the only critical point in the domain x > 0 is x = a.Therefore, regardless of a > 0, the only critical point is x = a.Wait, that's interesting. So f’(x) = 0 only at x = a. Therefore, the function f(x) has only one critical point at x = a. Therefore, the function decreases from x = 0 to x = a, then increases from x = a to infinity. So x = a is the point of minimum.Thus, similar to part (Ⅰ), where a =1, the function decreases on (0, a) and increases on (a, ∞). Therefore, the minimal value is at x = a.Therefore, for f(x) to have two distinct zeros, the minimal value f(a) must be less than zero. Because as x approaches 0+ and infinity, f(x) tends to infinity, so if the minimum is below zero, the graph crosses the x-axis twice. If the minimum is zero, it's tangent to the x-axis (one zero with multiplicity two). If the minimum is positive, no zeros.Therefore, we need f(a) < 0.Compute f(a):f(a) = (1/3)a³ + [(1 - a)/2]a² - a² ln a + a² ln a.Wait, let's substitute x = a into f(x):f(a) = (1/3)a³ + [(1 - a)/2]a² - a² ln a + a² ln a.Simplify term by term:First term: (1/3)a³.Second term: [(1 - a)/2]a² = (a² - a³)/2.Third term: -a² ln a.Fourth term: +a² ln a.Notice that the third and fourth terms cancel each other: -a² ln a + a² ln a = 0.Therefore, f(a) = (1/3)a³ + (a² - a³)/2.Combine the terms:Let's compute (1/3)a³ + (a²/2 - a³/2).Convert to common denominator, which is 6:= (2/6)a³ + (3a²/6 - 3a³/6)= [2a³ + 3a² - 3a³]/6= (-a³ + 3a²)/6Factor numerator:= (-a²(a - 3))/6= [ -a²(a - 3) ] / 6So f(a) = [ -a²(a - 3) ] / 6Set f(a) < 0:[ -a²(a - 3) ] / 6 < 0.Since a > 0, a² is always positive. So the numerator is - positive * (a - 3). So:- a²(a - 3) < 0 => - (a - 3) < 0 => (a - 3) > 0 => a > 3.But wait, let's check:Original inequality: [ -a²(a - 3) ] / 6 < 0.Multiply both sides by 6 (positive, doesn't change inequality):- a²(a - 3) < 0.Multiply both sides by -1 (reverses inequality):a²(a - 3) > 0.Since a² > 0 (a > 0), then (a - 3) > 0 => a > 3.Therefore, the inequality holds when a > 3. Therefore, f(a) < 0 when a > 3.But wait, that suggests that f(a) < 0 when a > 3. But let's verify this with an example. Let's take a = 4. Compute f(4):f(4) = [ -4²(4 - 3) ] /6 = [ -16 *1 ] /6 = -16/6 = -8/3 < 0. Correct.But when a = 3, f(a) = [ -9*(0) ] /6 = 0. So at a =3, f(a) =0. Thus, for a >3, the minimal value is negative, so two zeros. For a =3, one zero. For a <3, minimal value is positive, so no zeros? Wait, hold on.Wait, but let me check when a <3. Let's take a =2. Then f(a) = [ -4*(2 -3) ] /6 = [ -4*(-1) ] /6 = 4/6 = 2/3 >0. So minimal value is positive. Therefore, the function never crosses the x-axis. When a =3, minimal value is zero, so tangent. When a >3, minimal value is negative, so two zeros.But wait, but in part (Ⅰ), when a=1, which is less than 3, we had a minimal value at x=1 of 1/3, which is positive, so function doesn't cross the x-axis. Indeed, when a=1, f(x) = (1/3)x³ - ln x. Let's check its behavior. At x approaching 0+, it goes to infinity. At x=1, f(1)=1/3. Then as x increases, the function tends to infinity. So indeed, when a=1, it never crosses the x-axis. So for a <3, the minimal value is positive, so no zeros. For a=3, one zero. For a >3, two zeros.Therefore, according to this, the range of a is a >3. But wait, the problem says "If f(x) has two distinct zeros, find the range of the real number a." So according to this, a must be greater than 3.But hold on, let's double-check because maybe there's a miscalculation.Wait, let's recalculate f(a):Original f(x) at x=a:f(a) = (1/3)a³ + [(1 - a)/2]a² - a² ln a + a² ln a.The last two terms do cancel out. Then f(a) = (1/3)a³ + (1 - a)a²/2.Which is (1/3)a³ + (a² - a³)/2.To combine these terms:Convert to common denominator 6:(2a³)/6 + (3a² - 3a³)/6 = (2a³ + 3a² - 3a³)/6 = (-a³ + 3a²)/6 = -a²(a - 3)/6. Correct.Therefore, f(a) = -a²(a - 3)/6.Therefore, f(a) <0 when -a²(a -3)/6 <0. Since a² is positive, the sign depends on -(a -3). Therefore:-(a -3) <0 => a -3 >0 => a >3. So yes, f(a) <0 when a >3.But is this the only condition? Let's think again. Because the function tends to infinity as x approaches 0 and infinity, and if the minimum is below zero, then there are exactly two zeros. So yes, the necessary and sufficient condition is that f(a) <0, which requires a >3.But wait, is there a possibility that even when a >3, there might be more than two zeros? But since the function only has one critical point (minimum at x=a), once it crosses the x-axis twice, it can't come back. So it's only two zeros. So the answer should be a >3.But let me check with a specific example. Let's take a=4. Then f(x) = (1/3)x³ + [(1 -4)/2]x² -16 ln x +16 ln4.Simplify: (1/3)x³ - (3/2)x² -16 ln x +16 ln4.Compute f(a)=f(4): (1/3)(64) - (3/2)(16) -16 ln4 +16 ln4 = 64/3 -24 +0 = 64/3 -72/3 = (-8)/3 <0. So the minimal value is negative. Then as x approaches 0+, f(x) approaches infinity, as x approaches infinity, f(x) approaches infinity. So two zeros.Another example, a=3. Then f(a)=0, so the function touches the x-axis at x=3. For a=2, f(a)= [ -4*(2-3) ] /6 = [ -4*(-1) ] /6=4/6=2/3>0. So function stays above the x-axis. Thus, conclusion: a>3.Therefore, the answer to part (Ⅱ) is a >3, so the range is (3, ∞).Wait, but let me think again. In the problem statement, when they say "two distinct zeroes," do they mean exactly two, or at least two? But in the context, since the function tends to infinity on both ends, and it's continuous, if it has a minimum below zero, there must be exactly two zeros. If the minimum is zero, one zero (double root). If the minimum is positive, no zeros. So yes, exactly two zeros when a>3.But let me check for a=4. Suppose we plot f(x)= (1/3)x³ - (3/2)x² -16 ln x +16 ln4. At x=4, the minimal value is -8/3. Let's check some points:At x approaching 0+, f(x) tends to infinity. At x=1: f(1)=1/3 - 3/2 -16 ln1 +16 ln4 ≈0.333 -1.5 -0 +16*1.386≈0.333 -1.5 +22.176≈20.009. Positive. At x=4, f(4)= -2.666. Negative. Then as x increases beyond 4, the function tends to infinity. So between x=1 and x=4, it goes from positive to negative, so a root. Then from x=4 to infinity, it goes from negative to positive, so another root. Thus, two zeros. Therefore, correct.Therefore, the range is a >3. Therefore, part (Ⅱ) answer is (3, ∞).But wait, let me check if there is any other critical point. Wait, earlier we derived that the critical point is only at x=a. So if the minimal value is below zero, there are two zeros. If the problem had multiple critical points, maybe there could be more zeros, but in this case, only one critical point. Hence, only two zeros when the minimal value is negative.Therefore, the conclusion is a must be greater than 3.Final AnswerⅠ. The function is decreasing on (0, 1) and increasing on (1, infty) with a local minimum at boxed{left(1, dfrac{1}{3}right)}.Ⅱ. The range of a is boxed{(3, infty)}.</think>Given the function ( f(x) = frac{1}{3}x^3 + frac{1-a}{2}x^2 - a^2 ln x + a^2 ln a ) where ( a > 0 ):Ⅰ. When ( a = 1 ), the function simplifies to ( f(x) = frac{1}{3}x^3 - ln x ). To find the intervals of monotonicity and extreme values, we first compute the derivative:[ f'(x) = x^2 - frac{1}{x}. ]Setting ( f'(x) = 0 ) gives the critical point ( x = 1 ). Testing intervals around ( x = 1 ):- For ( 0 < x < 1 ), ( f'(x) < 0 ), so ( f(x) ) is decreasing.- For ( x > 1 ), ( f'(x) > 0 ), so ( f(x) ) is increasing.The function has a local minimum at ( x = 1 ). Evaluating ( f(1) ):[ f(1) = frac{1}{3}(1)^3 - ln 1 = frac{1}{3}. ]Ⅱ. To determine the range of ( a ) such that ( f(x) ) has two distinct zeroes, we analyze the critical points and the minimum value. The derivative for general ( a ) is:[ f'(x) = x^2 + (1 - a)x - frac{a^2}{x}. ]Setting ( f'(x) = 0 ) and solving, we find the critical point ( x = a ). Evaluating ( f(x) ) at ( x = a ):[ f(a) = frac{1}{3}a^3 + frac{(1 - a)a^2}{2} - a^2 ln a + a^2 ln a = frac{-a^2(a - 3)}{6}. ]For ( f(x) ) to have two distinct zeroes, the minimum value ( f(a) ) must be negative:[ frac{-a^2(a - 3)}{6} < 0 implies a > 3. ] Final AnswerⅠ. The function is decreasing on ((0, 1)) and increasing on ((1, infty)) with a local minimum at (boxed{left(1, dfrac{1}{3}right)}).Ⅱ. The range of ( a ) is (boxed{(3, infty)}).,<think>Okay, so I need to solve this problem with two parts. Let me start with part (I) where a = 1. The function becomes f(x) = (1/3)x³ + [(1 - 1)/2]x² - (1)² ln x + (1)² ln 1. Simplifying that, since (1 - a)/2 when a=1 is 0, so the second term disappears. Then, -a² ln x becomes -ln x, and a² ln a is 1*ln 1 which is 0. So the function simplifies to f(x) = (1/3)x³ - ln x. Cool, so now I need to find the intervals of monotonicity and the extreme values.First, to find monotonicity, I need the first derivative. Let's compute f'(x). The derivative of (1/3)x³ is x², and the derivative of -ln x is -1/x. So f'(x) = x² - 1/x. To find critical points, set f'(x) = 0. So x² - 1/x = 0. Multiply both sides by x to eliminate the denominator (assuming x > 0 because of the ln x term). Then x³ - 1 = 0 → x³ = 1 → x = 1. So the critical point is at x=1.Now, to determine intervals of increase and decrease, we can test values around x=1. Since the domain of f(x) is x > 0. Let's pick x = 0.5: f'(0.5) = (0.5)² - 1/(0.5) = 0.25 - 2 = -1.75 < 0. Then x=2: f'(2) = 4 - 1/2 = 3.5 > 0. So the function is decreasing on (0,1) and increasing on (1, ∞). Therefore, there's a minimum at x=1. To find the extreme value, compute f(1): (1/3)(1)³ - ln 1 = 1/3 - 0 = 1/3. So the function has a minimum at (1, 1/3). That should answer part (I).Now part (II): If f(x) has two distinct zeroes, find the range of a. Hmm, the original function is f(x) = (1/3)x³ + (1 - a)/2 x² - a² ln x + a² ln a. So we need to find a > 0 such that f(x) crosses the x-axis twice. Since it's a continuous function on x > 0, and involves polynomials and logarithms, maybe we can analyze its behavior as x approaches 0+ and infinity, then use calculus to ensure there's a certain number of zeros.First, let's analyze the limits. As x approaches 0+, ln x approaches -infinity, so the term -a² ln x becomes +infinity. The polynomial terms: (1/3)x³ and (1 - a)/2 x² both approach 0. So f(x) approaches +infinity as x→0+. As x approaches infinity, the dominant term is (1/3)x³, which approaches infinity. So f(x) approaches infinity as x→∞. Therefore, the function tends to +infinity at both ends. For there to be two distinct zeros, the function must dip below the x-axis somewhere in between. That would require that the function has a minimum somewhere below zero. Therefore, we need f(x) to have a critical point where the function value is negative. So maybe we can find the critical points, check the function's value there, and set it less than zero.But first, let's find the critical points for general a. Compute f'(x):f'(x) = derivative of (1/3)x³ is x², derivative of [(1 - a)/2]x² is (1 - a)x, derivative of -a² ln x is -a²/x, and the derivative of a² ln a is 0. So f'(x) = x² + (1 - a)x - a²/x.Set f'(x) = 0: x² + (1 - a)x - a²/x = 0. Multiply both sides by x to eliminate the denominator (x > 0):x³ + (1 - a)x² - a² = 0.So the equation is x³ + (1 - a)x² - a² = 0.We need to solve this cubic equation for x. Hmm, solving cubic equations can be tricky. Maybe we can factor it? Let's try to see if x = a is a root:Plug x = a: a³ + (1 - a)a² - a² = a³ + a² - a³ - a² = 0. So yes! x = a is a root. Therefore, we can factor the cubic as (x - a)(x² + px + q). Let's perform polynomial division or use the factor theorem.Divide x³ + (1 - a)x² - a² by (x - a). Using synthetic division:Coefficients: 1 (x³), (1 - a) (x²), 0 (x), -a² (constant).Wait, synthetic division might be messy here. Alternatively, since we know x = a is a root, write the cubic as (x - a)(x² + bx + c). Expanding:(x - a)(x² + bx + c) = x³ + bx² + cx - a x² - abx - ac = x³ + (b - a)x² + (c - ab)x - ac.Compare with original cubic: x³ + (1 - a)x² + 0x - a². Therefore:1. Coefficient of x³: 1, matches.2. Coefficient of x²: (b - a) = (1 - a) → b - a = 1 - a → b = 1.3. Coefficient of x: (c - ab) = 0. Since b=1, then c - a*1 = 0 → c = a.4. Constant term: -ac = -a² → -a*c = -a² → c = a. Which matches previous result.Thus, the cubic factors as (x - a)(x² + x + a). Therefore, the critical points are x = a and roots of x² + x + a = 0.But discriminant of x² + x + a is 1 - 4a. Since a > 0, discriminant is 1 - 4a. So if 1 - 4a ≥ 0, then there are real roots. So for a ≤ 1/4, the quadratic has real roots, so the cubic has three real roots: x = a and the two from the quadratic. But since x > 0, we need to check if those roots are positive. The quadratic x² + x + a = 0, if discriminant is non-negative, then roots are [-1 ± sqrt(1 - 4a)]/2. Since a > 0, 1 - 4a must be ≥ 0, so a ≤ 1/4. Then sqrt(1 - 4a) is real. The roots are [-1 + sqrt(1 - 4a)]/2 and [-1 - sqrt(1 - 4a)]/2. The second root is negative because sqrt(1 - 4a) < 1 (since a > 0), so -1 - sqrt(1 - 4a) is less than -1 - 0 = -1, divided by 2, still negative. The first root is [-1 + sqrt(1 - 4a)]/2. Let's see if that is positive. Since sqrt(1 - 4a) < 1, so -1 + sqrt(1 - 4a) < 0. Therefore, even the first root is negative. Therefore, for a ≤ 1/4, quadratic has two negative roots. Therefore, the only critical point in x > 0 is x = a. If a > 1/4, the quadratic has no real roots, so the only critical point is x = a. Wait, that's interesting. So regardless of a, the only critical point in x > 0 is x = a. Because when a ≤ 1/4, the quadratic roots are negative, and when a > 1/4, no real roots. So in all cases, critical point is x = a. Hmm, but that contradicts the factoring. Wait, let me check again.Wait, if x² + x + a = 0 has roots only when discriminant ≥ 0. But even when discriminant is ≥ 0, the roots are negative. So in x > 0, the only critical point is x = a. Therefore, regardless of the value of a > 0, the only critical point is x = a. Wait, but when we factored the cubic, we had (x - a)(x² + x + a). So if a is such that x² + x + a = 0 has real roots, but they are negative, so x = a is the only critical point in x > 0. So, in all cases, x = a is the only critical point. Therefore, the function f(x) has only one critical point at x = a for all a > 0. Therefore, to find if the function has two distinct zeros, since f(x) tends to +infinity as x approaches 0 and infinity, the function must dip below the x-axis at x = a. Therefore, the minimum value at x = a must be negative.Therefore, we need f(a) < 0. Let's compute f(a):f(a) = (1/3)a³ + [(1 - a)/2]a² - a² ln a + a² ln a.Wait, the last two terms: -a² ln a + a² ln a cancel each other. So f(a) simplifies to (1/3)a³ + [(1 - a)/2]a².Simplify that: (1/3)a³ + (1 - a)(a²)/2 = (1/3)a³ + (a² - a³)/2.Let's combine the terms:Multiply all terms by 6 to eliminate denominators:6*(1/3)a³ = 2a³,6*(a²/2) = 3a²,6*(-a³/2) = -3a³.So total is 2a³ + 3a² - 3a³ = -a³ + 3a² = a²(3 - a).Therefore, f(a) = [ -a³ + 3a² ] / 6? Wait, wait, let me check again.Wait, original expression:(1/3)a³ + (a² - a³)/2 = (1/3 - 1/2)a³ + (1/2)a².Compute 1/3 - 1/2 = (2 - 3)/6 = -1/6.So f(a) = (-1/6)a³ + (1/2)a² = ( -a³ + 3a² ) / 6 = a²(3 - a)/6.So f(a) = [a²(3 - a)] / 6.Therefore, f(a) = (a²(3 - a))/6.So to have f(a) < 0, we need (a²(3 - a))/6 < 0. Since a > 0, a² is positive, and 6 is positive. Therefore, the sign is determined by (3 - a). So (3 - a) < 0 → a > 3. Therefore, f(a) < 0 when a > 3. Wait, but if a > 3, then f(a) is negative. But then, the function tends to +infinity as x approaches 0 and infinity, and has a single critical point at x = a where it's a minimum. Therefore, if f(a) < 0, the function crosses the x-axis twice. Therefore, the range of a is a > 3.But wait, let me verify. Let's take a = 4. Then f(a) = (16*(3 - 4))/6 = (16*(-1))/6 = -16/6 < 0. So the function has a minimum below zero, hence two zeros. If a = 3, then f(a) = (9*(0))/6 = 0. So at a = 3, the minimum is zero, so the function touches the x-axis, hence one zero (double root). If a < 3, then f(a) > 0, so the minimum is above zero, so the function never crosses the x-axis, right? But wait, when a approaches 0, let's see. For example, take a approaching 0+. Then f(x) = (1/3)x³ + (1 - a)/2 x² - a² ln x + a² ln a.As a approaches 0+, the term a² ln a tends to 0 (since a² approaches 0 and ln a approaches -infty, but a² ln a tends to 0). The term -a² ln x is negligible for x not approaching 0. So f(x) approximates (1/3)x³ + (1/2)x². Which is a function that tends to infinity as x approaches infinity and as x approaches 0. But the function (1/3)x³ + (1/2)x² is always positive, right? For x > 0, all terms are positive, so f(x) is positive everywhere. Therefore, when a is very small, f(x) is positive everywhere. So when a < 3, the minimum at x = a is positive (since f(a) = (a²(3 - a))/6 > 0 for a < 3). Therefore, the entire function is above the x-axis, so no zeros. When a = 3, f(a) = 0, so the function touches the x-axis at x = a. When a > 3, the minimum is negative, so the function crosses the x-axis twice. Therefore, the answer for part (II) is a > 3. So the range of a is (3, ∞).But wait, wait. Wait, when a = 1 (from part I), the function had a minimum at x = 1 with value 1/3, which is positive, so f(x) never crosses zero. But when a = 4, as we saw, it's negative. So this seems correct. Therefore, the answer is a must be greater than 3.But wait, hold on. Let me check for a = 2. If a = 2, then f(a) = (4*(3 - 2))/6 = 4*1/6 = 2/3 > 0. Therefore, the minimum is positive, so function is always positive. For a = 4, f(a) is negative. Therefore, only when a > 3, f(a) is negative, so two zeros. Therefore, the range is a > 3.But let's confirm with another approach. Suppose we set f(x) = 0:(1/3)x³ + [(1 - a)/2]x² - a² ln x + a² ln a = 0.This seems complicated. Alternatively, maybe we can use the intermediate value theorem. Since as x approaches 0+, f(x) approaches +infty, and as x approaches infinity, f(x) approaches +infty. So for two zeros, there must be two points where f(x) crosses from positive to negative and back. But since there is only one critical point (minimum at x = a), the function decreases from x=0 to x=a, then increases from x=a to infinity. So if the minimum at x=a is below zero, then the function must cross the x-axis twice. If it's exactly zero, one zero (tangent), and if it's above zero, no zeros. Therefore, the condition is f(a) < 0, which gives a > 3. So yes, the range is a > 3.Therefore, part (II) answer is a > 3.Final Answer(I) The function is decreasing on (0, 1) and increasing on (1, infty) with a minimum at x=1 of boxed{dfrac{1}{3}}.(II) The range of a is boxed{(3, infty)}.</think>Given the function ( f(x) = frac{1}{3}x^3 + frac{1-a}{2}x^2 - a^2 ln x + a^2 ln a ) where ( a > 0 ):(Ⅰ) When ( a = 1 ), the function simplifies to ( f(x) = frac{1}{3}x^3 - ln x ). 1. Compute the first derivative: ( f'(x) = x^2 - frac{1}{x} ).2. Find critical points by setting ( f'(x) = 0 ): ( x^2 - frac{1}{x} = 0 ) which simplifies to ( x^3 = 1 ), giving ( x = 1 ).3. Analyze intervals around ( x = 1 ): - For ( 0 < x < 1 ), ( f'(x) < 0 ) (decreasing). - For ( x > 1 ), ( f'(x) > 0 ) (increasing).4. The function has a minimum at ( x = 1 ). Evaluating ( f(1) ): ( f(1) = frac{1}{3} - 0 = frac{1}{3} ).Thus, the function is decreasing on ( (0, 1) ) and increasing on ( (1, infty) ) with a minimum value of ( boxed{frac{1}{3}} ).(Ⅱ) To determine the range of ( a ) such that ( f(x) ) has two distinct zeroes:1. Compute the first derivative: ( f'(x) = x^2 + (1 - a)x - frac{a^2}{x} ).2. Find critical points by solving ( x^3 + (1 - a)x^2 - a^2 = 0 ). Factoring gives ( (x - a)(x^2 + x + a) = 0 ), yielding ( x = a ) as the only positive critical point.3. Evaluate ( f(a) ): [ f(a) = frac{1}{3}a^3 + frac{(1 - a)}{2}a^2 - a^2 ln a + a^2 ln a = frac{a^2(3 - a)}{6} ]4. For ( f(x) ) to have two distinct zeroes, ( f(a) < 0 ). This occurs when ( 3 - a < 0 ), i.e., ( a > 3 ).Thus, the range of ( a ) is ( boxed{(3, infty)} ).]

🔑:(Ⅰ) When a=1, the function becomes f(x)= frac{1}{3}x^{3}-ln x, where x > 0.To find the intervals of monotonicity, we determine the first derivative of f(x):f'(x)=x^{2}- frac{1}{x}= frac{x^{3}-1}{x}, quad x > 0From critical points, we find that- When 0 1, f'(x) > 0, indicating that the function is increasing.Therefore, the function f(x) is decreasing on the interval (0,1) and increasing on (1,+infty). The function has a minimum value of f(1)= frac{1}{3} and no maximum value. Hence, the extremal value is:boxed{text{Minimum at } x = 1: f(1) = frac{1}{3}}(Ⅱ) Since f'(x)=x^{2}+(1-a)x- frac{a^{2}}{x}= frac{x^{3}+(1-a)x^{2}-a^{2}}{x},we can factor out common terms to simplify the derivative:f'(x)= frac{(x-a)(x^{2}+x+a)}{x}Since x > 0 and a > 0, f'(x) changes sign at x = a. Specifically, f(x) is decreasing for 0 a. The minimum value occurs at x = a:f_{text{min}}(x) = f(a)= frac{1}{3}a^{3}+ frac{1-a}{2}a^{2}= frac{1}{6}a^{2}(3-a)For f(x) to have two zeros, its minimum value must be negative:f(a) = frac{1}{6}a^{2}(3-a) 3Additionally, we evaluate the function at x = 2a and x = 1 to confirm that it is indeed positive at these points:f(2a) = frac{1}{3}(2a)^{3}+ frac{1-a}{2}(2a)^{2}-a^{2}ln (2) = frac{2}{3}a^{3}+(2-ln 2)a^{2} > 0f(1) = frac{1}{3}cdot 1^{3}+ frac{1-a}{2}cdot 1^{2}-a^{2}ln (1/a) = frac{1}{3}+ frac{1}{2}- frac{a}{2}+a^{2}ln a > frac{1}{3}+ frac{1}{2}- frac{a}{2}+a^{2} geq (a- frac{1}{4})^{2}+ frac{37}{48} > 0In conclusion, f(x) has two distinct zeroes when a > 3, so the range of a that satisfies the condition is (3,+infty). Therefore, the range of possible a values is:boxed{a in (3, +infty)}

🧠:[<think>Okay, so I need to find the vertex coordinates of the function y = 2(x - 4)^2 + 5. Hmm, let me think. I remember that quadratic functions are often written in vertex form, which is y = a(x - h)^2 + k. In this form, the vertex is at the point (h, k). Let me check if that's correct. Wait, yes, vertex form is y = a(x - h)^2 + k, where (h, k) is the vertex. So comparing that to the given equation, y = 2(x - 4)^2 + 5, the value of h should be 4 and k should be 5. So then the vertex is at (4, 5). Is that right?Let me verify. If I expand the equation, maybe I can see the standard form and confirm the vertex. Let's see. Expanding 2(x - 4)^2 + 5: first, expand (x - 4)^2. That's x^2 - 8x + 16. Then multiply by 2: 2x^2 - 16x + 32. Then add 5: 2x^2 - 16x + 37. So the standard form is y = 2x^2 - 16x + 37. To find the vertex from standard form, I can use the formula h = -b/(2a). Here, a = 2, b = -16. So h = -(-16)/(2*2) = 16/4 = 4. Then plug h back into the equation to find k. Let's calculate k = y(4). Substitute x = 4 into the standard form: 2*(4)^2 - 16*(4) + 37. Compute 2*16 = 32, 16*4 = 64. So 32 - 64 + 37 = (32 + 37) - 64 = 69 - 64 = 5. So y = 5 when x = 4. That confirms that the vertex is indeed at (4, 5). Alternatively, since the original equation was already in vertex form, there was no need to convert it to standard form. Just comparing directly: (x - h)^2 implies h = 4 and k = 5. The coefficient a = 2 just affects the width and direction of the parabola but not the vertex. So regardless of the a value, the vertex remains (h, k). Therefore, even if I change the coefficient a, as long as the equation is in vertex form, the vertex is straightforward to read off. So in this case, h is 4, k is 5. So the vertex coordinates are (4, 5). Wait, maybe I should double-check once more. Let me pick another point on the parabola to see if the vertex makes sense. For example, when x = 4, y = 2*(0)^2 + 5 = 5. So (4,5) is a point. If x = 3, then y = 2*(3 - 4)^2 +5 = 2*(1) +5 = 7. Similarly, x = 5, same thing: 2*(1)^2 +5 = 7. So the parabola is symmetric around x = 4, which is the vertex. So points at x = 3 and x =5 both have y =7. Then if I go to x =2: y = 2*(2-4)^2 +5 = 2*(4) +5 = 13. And x=6: same, 2*(2)^2 +5=13. So yes, the vertex at (4,5) is the lowest point because the coefficient a=2 is positive, so the parabola opens upwards. Therefore, the vertex is a minimum.So all the checks confirm that the vertex is (4,5). Therefore, I can be confident that the answer is correct.Final AnswerThe vertex coordinates are boxed{(4, 5)}.</think>To find the vertex coordinates of the function ( y = 2(x - 4)^2 + 5 ), we start by recognizing that the function is given in vertex form ( y = a(x - h)^2 + k ), where the vertex is at ( (h, k) ).Comparing the given function ( y = 2(x - 4)^2 + 5 ) with the vertex form, we identify:- ( h = 4 )- ( k = 5 )Thus, the vertex coordinates are ( (4, 5) ).To verify, we convert the function to standard form by expanding and simplifying:[begin{align*}y &= 2(x - 4)^2 + 5 &= 2(x^2 - 8x + 16) + 5 &= 2x^2 - 16x + 32 + 5 &= 2x^2 - 16x + 37end{align*}]Using the standard form ( y = ax^2 + bx + c ), we calculate the vertex ( x )-coordinate ( h ) using ( h = -frac{b}{2a} ):[h = -frac{-16}{2 cdot 2} = frac{16}{4} = 4]Substituting ( x = 4 ) back into the original function to find ( k ):[y = 2(4 - 4)^2 + 5 = 2 cdot 0 + 5 = 5]Both methods confirm the vertex coordinates are ( (4, 5) ).[boxed{(4, 5)}],<think>Okay, let's see. I have this problem here: "The vertex coordinates of the function y = 2(x - 4)^2 + 5 are..." Hmm, vertex coordinates. Right, quadratic functions. I remember that quadratic functions can be written in vertex form, which is useful because it directly gives the vertex. So, vertex form is like y = a(x - h)^2 + k, right? And in this form, the vertex is at (h, k). Let me make sure. If the equation is in that form, then yeah, h and k are the x and y coordinates of the vertex. But I need to be careful with the signs. Because it's (x - h), so if it's (x - 4), then h is 4, not -4. That makes sense. Looking at the given function: y = 2(x - 4)^2 + 5. Comparing this to the standard vertex form, a is 2, which affects the width and direction of the parabola, but not the vertex. Then, h is 4, and k is 5. So the vertex should be at (4, 5). Wait, is that all? Let me double-check. Another way to find the vertex is using the formula for the vertex of a parabola in standard form. The standard form is y = ax^2 + bx + c. The x-coordinate of the vertex is at -b/(2a). But this function is already in vertex form, so converting it might be a waste of time, but maybe it's a good idea to verify.Let me expand the given equation to standard form. So starting with y = 2(x - 4)^2 + 5. First, expand (x - 4)^2: that's x^2 - 8x + 16. Then multiply by 2: 2x^2 - 16x + 32. Then add 5: 2x^2 - 16x + 37. So in standard form, it's y = 2x^2 - 16x + 37. Now, using the formula -b/(2a) for the x-coordinate. Here, a = 2, b = -16. So x = -(-16)/(2*2) = 16/4 = 4. Then plug x = 4 back into the equation to find y. Calculating y: y = 2(4)^2 -16(4) + 37. Let's compute step by step. 4 squared is 16, times 2 is 32. Then 16 times 4 is 64, so -16 times 4 is -64. Then 32 - 64 + 37. 32 -64 is -32, plus 37 is 5. So y = 5. Therefore, the vertex is at (4, 5). Wait, that's the same result as before. So both methods give me (4, 5). That seems solid. Maybe I can think of another way. Like, the vertex is the maximum or minimum point of the parabola. Since the coefficient a is 2, which is positive, the parabola opens upwards, so the vertex is a minimum. The minimum occurs at x = 4, and plugging that in, y = 5. Is there any chance I could have messed up the signs? Let me check when expanding (x - 4)^2. That's (x - 4)(x - 4) = x^2 - 4x -4x +16 = x^2 -8x +16. Yep, that's correct. Then multiplying by 2 gives 2x^2 -16x +32. Adding 5 gives 2x^2 -16x +37. Correct. Then -b/(2a) with b = -16. So -(-16) is 16, divided by 2*2 is 4. So x = 4. Correct. Then plugging back in, y is 5. Alternatively, if I use the original vertex form equation. The term (x - 4) indicates that h is 4, and then k is 5. So (h, k) is (4,5). So that's straightforward. Another thought: sometimes if the equation is written as y = a(x + h)^2 + k, people might get confused. But in this case, it's (x - 4), so h is positive 4. If it was (x + 4), then h would be -4. But here it's minus 4, so h is 4. So, after all these checks, I think the vertex coordinates are (4, 5). There's no step I can see where I made an error. Both expanding to standard form and using the vertex form directly give the same answer, so I must be right.Final AnswerThe vertex coordinates are boxed{(4, 5)}.</think>The vertex coordinates of the function ( y = 2(x - 4)^2 + 5 ) can be determined by recognizing that the function is given in vertex form. The vertex form of a quadratic function is ( y = a(x - h)^2 + k ), where ((h, k)) is the vertex. Comparing the given function ( y = 2(x - 4)^2 + 5 ) to the vertex form, we identify ( h = 4 ) and ( k = 5 ). Therefore, the vertex coordinates are ((4, 5)).To verify, we can convert the function to standard form. Expanding ( 2(x - 4)^2 + 5 ):[begin{align*}2(x - 4)^2 + 5 &= 2(x^2 - 8x + 16) + 5 &= 2x^2 - 16x + 32 + 5 &= 2x^2 - 16x + 37.end{align*}]The standard form is ( y = 2x^2 - 16x + 37 ). Using the vertex formula ( x = -frac{b}{2a} ):[x = -frac{-16}{2 cdot 2} = frac{16}{4} = 4.]Substituting ( x = 4 ) back into the original function to find ( y ):[y = 2(4 - 4)^2 + 5 = 2(0)^2 + 5 = 5.]Both methods confirm the vertex coordinates are ((4, 5)).[boxed{(4, 5)}]]

🔑:The vertex coordinates of the quadratic function y = 2(x - 4)^2 + 5 are (4, 5).Therefore, the answer is boxed{(4, 5)}.

🧠:[<think>Alright, let's tackle this problem. So, the question is asking which of the given pairs does not represent quantities with opposite meanings. The options are A, B, C, D. Hmm, okay. Let me go through each one step by step.Starting with option A: Moving forward 5 meters and moving backward 5 meters. Well, forward and backward are opposite directions. So, moving 5 meters in one direction and then 5 meters in the opposite direction would typically be considered opposites. For example, if you move forward 5 meters and then backward 5 meters, you end up where you started. So, these seem like opposites. So, A probably represents opposite quantities.Next, option B: Income of 30 yuan and expenditure of 10 yuan. Hmm. Income is money coming in, expenditure is money going out. So, they are opposite in terms of cash flow. However, the amounts are different: 30 versus 10. Does the difference in amount matter here? The question is about whether they have opposite meanings, not necessarily equal magnitudes. So, even though 30 and 10 are different numbers, income and expenditure themselves are opposites. But wait, maybe the problem is that they aren't equal in magnitude? But the question is about opposite meanings, not necessarily cancelling each other out. Let me think. If income is considered positive and expenditure negative, then income 30 is +30 and expenditure 10 is -10. So they are opposite in sign but different in magnitude. So, perhaps B is still considered opposite in meaning? Maybe. I need to check the other options to compare.Option C: Walking 10 meters east and walking 10 meters north. Directions here are east and north. These are perpendicular directions, not opposite. Opposite directions would be east and west, or north and south. So, east and north are different directions, but not opposites. Therefore, walking east and walking north are movements in different directions, but not opposite. So, this pair does not have opposite meanings. So, C might be the correct answer here.Option D: Exceeding 5 grams and falling short of 2 grams. Exceeding means being over a certain amount, and falling short means being under. So, these are opposites in terms of being over vs. under. However, the amounts are different: 5 grams over and 2 grams under. Similar to option B, the magnitudes are different, but the concepts are opposites. So, even though the numbers aren't the same, the ideas of exceeding and falling short are opposites. So, D might still be considered as having opposite meanings.Going back to B: Income and expenditure are opposites regardless of the amounts. So even if you gain 30 and spend 10, they are still opposite actions. Similarly, D is about exceeding and falling short, which are opposite concepts. Therefore, the problem is with option C, because east and north are not opposite directions. Therefore, the pair that does not represent quantities with opposite meanings is C.Wait, let me confirm once more. The question is about quantities with opposite meanings. For example, in A, moving forward and backward are opposite directions. The quantities (5 meters) are the same, but direction is opposite. In B, income and expenditure are opposite actions, but the amounts are different. The question is whether the pair represents quantities with opposite meanings. So, maybe "quantities" here refers to numbers that are opposites. For example, +5 and -5 are opposites. But in B, it's +30 and -10. Are they considered opposites? Hmm. If "quantities" refers to the numerical values being opposites (like additive inverses), then B would not be opposites since 30 and 10 aren't additive inverses. Similarly, D is 5 and 2, which are different. However, the problem says "quantities with opposite meanings," not necessarily numerically opposite. So, maybe "income" and "expenditure" are considered opposite in meaning, even if the amounts differ. Similarly, "exceeding" and "falling short" are opposites. So, in that case, the only pair that isn't opposite in meaning is C, since east and north are not opposite directions. So, the answer is C.Yes, that seems right. So, C doesn't have opposite meanings, while the others do, even if their magnitudes differ. So, the correct answer is C.Final Answerboxed{C}</think>To determine which pair does not represent quantities with opposite meanings, we analyze each option:- Option A: Moving forward 5 meters and moving backward 5 meters. These are opposite directions (forward vs. backward) with the same magnitude (5 meters). They represent opposite quantities. - Option B: Income of 30 yuan and expenditure of 10 yuan. Income and expenditure are opposite actions (money coming in vs. money going out). Although the amounts differ (30 vs. 10), the concepts are opposite in meaning. - Option C: Walking 10 meters east and walking 10 meters north. East and north are perpendicular directions, not opposite. Opposite directions would be east vs. west or north vs. south. Therefore, these do not represent opposite quantities. - Option D: Exceeding 5 grams and falling short of 2 grams. Exceeding (over) and falling short (under) are opposite concepts. Although the amounts differ (5 vs. 2), the meanings are opposite.The pair that does not represent quantities with opposite meanings is Option C because east and north are not opposite directions.boxed{C},<think>Okay, let's see. The question is asking which pair does NOT represent quantities with opposite meanings. The options are A, B, C, D. I need to figure out which one of these pairs isn't opposites.Starting with option A: Moving forward 5 meters and moving backward 5 meters. Hmm, forward and backward are opposite directions, right? So if you move forward 5 and then backward 5, they cancel each other out in terms of position. So that seems like opposite meanings. So A is probably not the answer.Then option B: Income of 30 yuan and expenditure of 10 yuan. Income is money coming in, expenditure is money going out. But the amounts are different here—30 vs. 10. If they were both the same amount, like income 30 and expenditure 30, that would be opposite. But since they are different, does that mean they aren't exact opposites? Maybe. But I need to think. If the question is about opposite meanings in terms of direction or effect, income adds to your total, expenditure subtracts. So even if the amounts are different, they are still opposite actions. But maybe the key here is that they aren't the same magnitude. So maybe B is the one that's not opposite? Not sure yet. Let me check the others.Option C: Walking 10 meters east and walking 10 meters north. East and north are different directions, perpendicular to each other. They aren't opposite directions. Opposite would be east and west, or north and south. So moving east and moving north are just different directions, not opposites. So this pair, C, seems like it does not have opposite meanings. So maybe C is the answer?Option D: Exceeding 5 grams and falling short of 2 grams. Exceeding is going over, falling short is going under. But the amounts are 5 and 2. So if you exceed by 5 and fall short by 2, they are both deviations from a target but in opposite directions. But the numbers are different. So like, is the opposite meaning about direction regardless of amount? So even if they exceed by 5 and fall short by 5, that would be exact opposites. But here it's 5 and 2. However, the question is about the pair not representing quantities with opposite meanings. Since exceeding and falling short are opposite in direction, even if the amounts differ, maybe D still counts as opposite meanings. Hmm, not sure. Let's think again.The key is "quantities with opposite meanings". So A is opposite directions with same distance, so opposite. B is income and expenditure but different amounts. So income and expenditure are opposite actions, but the amounts aren't negatives of each other. But does "quantities" refer to the numerical values or the actions? If it's about the numerical values as quantities, then 30 and -10 are not opposites, whereas if they were 30 and -30, they would be. But if the question is more about the nature of the quantities (income vs. expenditure), then they are opposites regardless of the amount. The problem is a bit ambiguous. Similarly, D has exceeding and falling short, which are opposite in direction, but different amounts. So maybe B and D are similar in that they have opposite actions but different amounts, whereas A and C are different. Wait, but C is different directions but same amount, but not opposite directions. So C is different.The question is which does NOT represent quantities with opposite meanings. So if opposite meanings can be direction or type (income/expenditure), even with different magnitudes, then which one is not opposite. C is different directions (east and north) which are not opposites, so that's not opposite. So C is the answer. B and D are opposite in type but different in magnitude, but maybe still considered opposites. So the answer is C.But let me double-check. Let's parse each option again:A: Forward and backward. Definitely opposite directions. So they are opposites.B: Income and expenditure. These are opposite actions (gaining vs. losing money), but the amounts are different. So even though they are opposites in action, the quantities are not exact negatives. But the question says "quantities with opposite meanings". If "quantities" here refers to the numbers themselves, then 30 and 10 are not opposites, but if it refers to the nature (income vs. expenditure), then they are. But the problem is in Chinese, so maybe the translation is precise. The original problem might have a specific term. Given that, maybe B is not opposites because the numbers are different. But I need to think about standard math problems. In math, opposite quantities are typically numbers that are negatives of each other, like +5 and -5. So for example, moving forward 5 and backward 5 would be +5 and -5. Similarly, exceeding 5 grams is +5, falling short 2 grams is -2. So those are not exact opposites in magnitude, but opposite in sign. But the pair in D is exceeding 5 and falling short 2. So they aren't exact opposites. But the problem is asking for pairs that do NOT represent quantities with opposite meanings. So if the key is that they have to be exact negatives (same magnitude, opposite sign), then B (30 vs. -10) and D (5 vs. -2) would not be exact opposites, but the question is which pair does NOT represent quantities with opposite meanings. So perhaps all except A and C? Wait no, C is different. Let me clarify.Wait, if we consider "quantities with opposite meanings" as numbers that are negatives (i.e., additive inverses), then:A: +5 and -5 (opposite)B: +30 and -10 (not additive inverses)C: +10 east and +10 north (not additive inverses; directions aren't represented as negatives)D: +5 and -2 (not additive inverses)But in that case, only A is a pair of opposite quantities, and the rest are not. But the question is asking which does NOT represent quantities with opposite meanings, so the answer would be B, C, D. But since the options are single answers, perhaps the problem is intended differently.Alternatively, perhaps "quantities with opposite meanings" refers to concepts that are opposites, not necessarily numerical inverses. For example, income and expenditure are opposite concepts, even if the amounts differ. Similarly, exceeding and falling short are opposite concepts. Moving forward/backward are opposites, east/north are not. So in that case, C is the pair that does not have opposite meanings. So answer is C.Alternatively, in D, exceeding and falling short are opposites, even if the amounts are different. So D still is considered opposite in meaning. So C is the only one where the directions are not opposites. Therefore, the answer is C.Final Answerboxed{C}</think>To determine which pair does not represent quantities with opposite meanings, we analyze each option:- Option A: Moving forward 5 meters and moving backward 5 meters. These are opposite directions (forward/backward) with the same magnitude (5 meters). They are opposites. - Option B: Income of 30 yuan and expenditure of 10 yuan. Income and expenditure are opposite actions (gaining vs. losing money), but the amounts (30 and 10) are different. However, the key is the nature of the actions, not the amounts. They are considered opposites. - Option C: Walking 10 meters east and walking 10 meters north. East and north are perpendicular directions, not opposites. Opposite directions would be east/west or north/south. This pair does not represent opposite meanings. - Option D: Exceeding 5 grams and falling short of 2 grams. Exceeding and falling short are opposite actions (over vs. under), even though the amounts (5 and 2) differ. They are considered opposites.Thus, the pair that does not represent quantities with opposite meanings is C.boxed{C}]

🔑:To determine which pair does not represent quantities with opposite meanings, let's analyze each option:- Option A: Moving forward 5 meters and moving backward 5 meters. - Moving forward and moving backward are opposite actions. If you move forward by a certain distance, moving backward by the same distance brings you back to your starting point. - Therefore, these quantities have opposite meanings.- Option B: Income of 30 yuan and expenditure of 10 yuan. - Income increases your total amount of money, while expenditure decreases it. - These are opposite financial transactions affecting your net worth in opposite ways. - Thus, these quantities have opposite meanings.- Option C: Walking 10 meters east and walking 10 meters north. - Walking east and walking north are perpendicular directions, not opposite. They do not cancel each other out but rather contribute to movement in two different dimensions. - Therefore, these quantities do not have opposite meanings.- Option D: Exceeding 5 grams and falling short of 2 grams. - To exceed a measure means to go beyond it, while to fall short means not reaching a specified measure. - These are opposite outcomes in terms of measurement accuracy or expectations. - Hence, these quantities have opposite meanings.Given the analysis, the pair that does not represent quantities with opposite meanings is found in Option C: Walking 10 meters east and walking 10 meters north.Therefore, the correct answer is boxed{C}.

🧠:[<think>Okay, so I need to prove that the sum of the coefficients from a_{n+1} to a_{2n} in the expansion of (x + 2x² + ... + nxⁿ)² is equal to (1/24)n(n+1)(5n² + 5n + 2). Let me try to break this down step by step.First, let me understand the problem. We have a polynomial on the left side, which is the square of another polynomial. The original polynomial is x + 2x² + 3x³ + ... + nxⁿ. When we square this polynomial, the resulting polynomial will have terms up to x²ⁿ, and the coefficients a_k for each xᵏ will be determined by the products of the coefficients from the original polynomial. The problem is asking for the sum of the coefficients from xⁿ⁺¹ up to x²ⁿ, which is the sum from k = n+1 to 2n of a_k. And we need to show that this sum equals that specific formula involving n.Hmm, okay. So perhaps I can start by expanding the square (x + 2x² + ... + nxⁿ)² and then figure out how to compute the sum of the coefficients from degree n+1 to 2n. Let me recall that when you multiply two polynomials, the coefficient of xᵏ in the product is the sum of all products of coefficients of xⁱ in the first polynomial and xʲ in the second polynomial where i + j = k.So, in our case, the original polynomial is P(x) = x + 2x² + 3x³ + ... + nxⁿ. Then, P(x)² = (x + 2x² + ... + nxⁿ)². The coefficient a_k of xᵏ in P(x)² is the sum over all i and j such that i + j = k of (coefficient of xⁱ in P(x)) * (coefficient of xʲ in P(x)).But the coefficients of P(x) are given by: for xⁱ, the coefficient is i if 1 ≤ i ≤ n, and 0 otherwise. So, the coefficient of xⁱ is i if i is between 1 and n, inclusive, and 0 otherwise.Therefore, a_k = Σ_{i=1 to n} i * (k - i) if k - i is also between 1 and n. Wait, but since we're squaring P(x), each a_k is the sum over i=1 to n of (i) * (j), where j = k - i, but j must also be between 1 and n. So for each k, the coefficient a_k is the sum of i*(k - i) for all i such that 1 ≤ i ≤ k - 1 and 1 ≤ k - i ≤ n. Which simplifies to i ranging from max(1, k - n) to min(n, k - 1). Because if k - i needs to be at least 1, so i ≤ k - 1, and also k - i ≤ n, so i ≥ k - n. But since i must be at least 1, the lower limit is max(1, k - n + 1?) Wait, maybe I need to check.Wait, if k - i ≥ 1, then i ≤ k - 1. Also, k - i ≤ n, so i ≥ k - n. But since i must be at least 1, then the lower bound is max(1, k - n). Similarly, the upper bound is min(n, k - 1). Therefore, for each k, the coefficient a_k is the sum over i = max(1, k - n) to min(n, k - 1) of i*(k - i).Therefore, the sum we need is Σ_{k = n+1}^{2n} a_k = Σ_{k = n+1}^{2n} [Σ_{i = max(1, k - n)}^{min(n, k - 1)} i*(k - i)].Hmm, that seems a bit complicated, but maybe we can switch the order of summation. Let's try to switch the order of summations. Instead of summing over k first and then i, perhaps sum over i first and then k. But we need to adjust the limits accordingly.So, sum_{k = n+1}^{2n} a_k = sum_{k = n+1}^{2n} sum_{i = max(1, k - n)}^{min(n, k - 1)} i*(k - i).Let me see if I can reverse the order. For each i, what values of k will include i in their sum? For a given i, k must be such that max(1, k - n) ≤ i ≤ min(n, k - 1). Let's solve for k.Since k is from n+1 to 2n, let's consider k in that range. Then for each i, k must satisfy:max(1, k - n) ≤ i ≤ min(n, k - 1).But since k ≥ n + 1, then k - n ≥ 1. Therefore, max(1, k - n) = k - n. So for k ≥ n + 1, the lower limit is k - n. Similarly, the upper limit is min(n, k - 1). Since k ranges from n + 1 to 2n, then k - 1 ranges from n to 2n - 1. Therefore, min(n, k - 1) = n when k - 1 ≤ n, which would be when k ≤ n + 1. But since k starts at n + 1, when k = n + 1, min(n, n) = n. For k > n + 1, k - 1 > n, so min(n, k - 1) = n. Wait, no. Wait, when k is greater than n + 1, k - 1 is greater than n, so min(n, k - 1) is n. Wait, but when k is between n + 1 and 2n, then k - 1 ranges from n to 2n - 1. So min(n, k - 1) would be n when k - 1 ≥ n, which is always true here because k ≥ n + 1. Therefore, the upper limit is n. Wait, but when k is n + 1, k - 1 = n, so min(n, n) = n. For k = n + 2, k - 1 = n + 1, so min(n, n + 1) = n. Similarly, up to k = 2n, k - 1 = 2n - 1, min(n, 2n -1) = n. So actually, for all k from n + 1 to 2n, the upper limit is n. Therefore, for each k in [n + 1, 2n], the limits on i are from (k - n) to n. Therefore, the inner sum is over i = (k - n) to n. So, the coefficient a_k is sum_{i = k - n}^n i*(k - i).Therefore, the total sum we need is sum_{k = n + 1}^{2n} sum_{i = k - n}^n i*(k - i).Now, let's switch the order of summation. Let me let j = k - n. Then, when k ranges from n + 1 to 2n, j ranges from 1 to n. Wait, k = n + j, where j ranges from 1 to n. Therefore, we can write the sum as sum_{j = 1}^n sum_{i = j}^n i*( (n + j) - i ). Because substituting k = n + j, then i starts at k - n = j, and the upper limit is n. Then, k - i = (n + j) - i. Therefore, the sum becomes sum_{j=1}^n sum_{i = j}^n [i*(n + j - i)].So now, the double sum is over j from 1 to n and i from j to n of i*(n + j - i). Let me check if this substitution is correct. When k = n + j, j from 1 to n, then k ranges from n + 1 to 2n. Then, i starts at k - n = j and goes up to n. So, yes, this substitution seems valid.Therefore, the total sum is equal to Σ_{j=1}^n Σ_{i=j}^n i(n + j - i). Maybe we can switch the order of summation here. Let's fix i and sum over j. For each i, j ranges from 1 to i. Because when i is fixed, j goes from 1 up to i (since in the original inner sum, i starts at j, so j ≤ i). Wait, no. Wait, in the current expression, for each j, i ranges from j to n. Therefore, if we switch the order, for each i, j ranges from 1 to i. Wait, let me check:Original summation: j from 1 to n, and for each j, i from j to n. So if we switch the order, i will go from 1 to n, and for each i, j goes from 1 to i. Therefore, the sum becomes Σ_{i=1}^n Σ_{j=1}^i [i(n + j - i)]. But hold on, let's verify. For example, when i = 1, j can only be 1. When i = 2, j can be 1 and 2. Etc., up to i = n, where j can be 1 to n. So yes, the switched sum is Σ_{i=1}^n Σ_{j=1}^i [i(n + j - i)].But perhaps this is a better way to compute the sum. Let's proceed.So, Σ_{i=1}^n Σ_{j=1}^i [i(n + j - i)].First, note that the term inside is i(n + j - i) = i(n - i) + i*j.Therefore, we can split the sum into two parts:Σ_{i=1}^n Σ_{j=1}^i [i(n - i) + i*j] = Σ_{i=1}^n [Σ_{j=1}^i i(n - i) + Σ_{j=1}^i i*j].The first inner sum is Σ_{j=1}^i i(n - i) = i(n - i) * Σ_{j=1}^i 1 = i(n - i) * i = i²(n - i).The second inner sum is Σ_{j=1}^i i*j = i * Σ_{j=1}^i j = i * [i(i + 1)/2] = i²(i + 1)/2.Therefore, putting it all together, the total sum is Σ_{i=1}^n [i²(n - i) + i²(i + 1)/2] = Σ_{i=1}^n i²(n - i) + (1/2) Σ_{i=1}^n i²(i + 1).So, we need to compute these two sums separately and then add them.Let me compute the first sum: S1 = Σ_{i=1}^n i²(n - i).And the second sum: S2 = (1/2) Σ_{i=1}^n i²(i + 1).Let's compute S1 first.S1 = Σ_{i=1}^n i²(n - i) = n Σ_{i=1}^n i² - Σ_{i=1}^n i³.We know that Σ_{i=1}^n i² = n(n + 1)(2n + 1)/6,and Σ_{i=1}^n i³ = [n(n + 1)/2]^2.Therefore, S1 = n * [n(n + 1)(2n + 1)/6] - [n(n + 1)/2]^2.Let me compute this:First term: n * [n(n + 1)(2n + 1)/6] = n²(n + 1)(2n + 1)/6.Second term: [n(n + 1)/2]^2 = n²(n + 1)² / 4.Therefore, S1 = [n²(n + 1)(2n + 1)/6] - [n²(n + 1)² / 4].Let me factor out n²(n + 1)/12:S1 = n²(n + 1)/12 [2(2n + 1) - 3(n + 1)].Compute the expression inside the brackets:2(2n + 1) = 4n + 2,3(n + 1) = 3n + 3,So 4n + 2 - 3n - 3 = n - 1.Therefore, S1 = n²(n + 1)/12 * (n - 1) = n²(n + 1)(n - 1)/12.Which simplifies to n²(n² - 1)/12.Okay, so S1 = n²(n² - 1)/12.Now, compute S2 = (1/2) Σ_{i=1}^n i²(i + 1).First, expand i²(i + 1) = i³ + i².Therefore, S2 = (1/2)(Σ_{i=1}^n i³ + Σ_{i=1}^n i²).We already know these sums:Σ_{i=1}^n i³ = [n(n + 1)/2]^2,Σ_{i=1}^n i² = n(n + 1)(2n + 1)/6.Therefore,S2 = (1/2)[ (n²(n + 1)² / 4) + (n(n + 1)(2n + 1)/6) ].Let's factor out common terms. Let's factor out n(n + 1)/12:S2 = (1/2) * n(n + 1)/12 [ 3n(n + 1) + 2(2n + 1) ].Wait, let me check the factoring:First term: n²(n + 1)² / 4 = [n(n + 1)/12] * 3n(n + 1)Second term: n(n + 1)(2n + 1)/6 = [n(n + 1)/12] * 2(2n + 1)Therefore,S2 = (1/2) * [n(n + 1)/12 (3n(n + 1) + 2(2n + 1)) ]Compute inside the brackets:3n(n + 1) + 2(2n + 1) = 3n² + 3n + 4n + 2 = 3n² + 7n + 2.Therefore,S2 = (1/2) * [n(n + 1)/12 (3n² + 7n + 2)] = n(n + 1)(3n² + 7n + 2)/24.So now, combining S1 and S2:Total sum = S1 + S2 = [n²(n² - 1)/12] + [n(n + 1)(3n² + 7n + 2)/24].To add these together, let's express them with a common denominator of 24:First term: n²(n² - 1)/12 = 2n²(n² - 1)/24,Second term: n(n + 1)(3n² + 7n + 2)/24.Therefore,Total sum = [2n²(n² - 1) + n(n + 1)(3n² + 7n + 2)] / 24.Let's factor out n from both terms in the numerator:= n[2n(n² - 1) + (n + 1)(3n² + 7n + 2)] / 24.Now compute the expression inside the brackets:First part: 2n(n² - 1) = 2n³ - 2n.Second part: (n + 1)(3n² + 7n + 2). Let's multiply this out.Multiply (n + 1)(3n² + 7n + 2):= n*(3n² + 7n + 2) + 1*(3n² + 7n + 2)= 3n³ + 7n² + 2n + 3n² + 7n + 2Combine like terms:3n³ + (7n² + 3n²) + (2n + 7n) + 2= 3n³ + 10n² + 9n + 2.So, the entire numerator inside the brackets becomes:2n³ - 2n + 3n³ + 10n² + 9n + 2Combine like terms:(2n³ + 3n³) + 10n² + (-2n + 9n) + 2= 5n³ + 10n² + 7n + 2.Therefore, the total sum is:n(5n³ + 10n² + 7n + 2) / 24.Now, let's factor the numerator if possible. Let's see:5n³ + 10n² + 7n + 2.Perhaps factor by grouping.Group the first two terms and the last two terms:(5n³ + 10n²) + (7n + 2) = 5n²(n + 2) + 1*(7n + 2). Doesn't seem to factor nicely.Alternatively, try rational root theorem. Possible roots are ±1, ±2, ±1/5, etc.Test n = -1: 5(-1)^3 + 10(-1)^2 + 7(-1) + 2 = -5 + 10 -7 + 2 = 0. So n = -1 is a root. Therefore, (n + 1) is a factor.Let's perform polynomial division or use synthetic division to factor out (n + 1):Divide 5n³ + 10n² + 7n + 2 by (n + 1):Using synthetic division:-1 | 5   10   7   2       -5   -5   -2       ——————       5    5   2   0So, the quotient is 5n² + 5n + 2. Therefore,5n³ + 10n² + 7n + 2 = (n + 1)(5n² + 5n + 2).Therefore, the total sum becomes:n(n + 1)(5n² + 5n + 2) / 24.Which matches the formula given in the problem: (1/24)n(n + 1)(5n² + 5n + 2).Therefore, we've proved that the sum from k = n + 1 to 2n of a_k is indeed equal to this expression. So, the proof is complete.Final Answerboxed{dfrac{1}{24}n(n+1)(5n^2 +5n +2)}</think>To prove the given identity, we start by considering the polynomial ( P(x) = x + 2x^2 + 3x^3 + cdots + nx^n ). When squared, the resulting polynomial ( P(x)^2 ) has coefficients ( a_k ) for each term ( x^k ). We need to find the sum of the coefficients from ( a_{n+1} ) to ( a_{2n} ).The coefficient ( a_k ) in ( P(x)^2 ) is given by the sum of products of coefficients from ( P(x) ) such that the degrees add up to ( k ). Specifically, ( a_k = sum_{i=1}^{n} i cdot (k - i) ) where ( k - i ) is also between 1 and ( n ). For ( k ) in the range ( n+1 ) to ( 2n ), the valid range for ( i ) is from ( k - n ) to ( n ).We switch the order of summation to simplify the calculation. By substituting ( k = n + j ) and reindexing, we transform the problem into summing over ( i ) and ( j ). This leads us to consider the sum:[sum_{j=1}^{n} sum_{i=j}^{n} i(n + j - i)]We split this into two sums and simplify each part. After switching the order of summation and simplifying, we find:1. The first sum ( S_1 = sum_{i=1}^{n} i^2(n - i) )2. The second sum ( S_2 = frac{1}{2} sum_{i=1}^{n} i^2(i + 1) )Using known formulas for sums of powers of integers, we compute:[S_1 = frac{n^2(n^2 - 1)}{12}][S_2 = frac{n(n + 1)(3n^2 + 7n + 2)}{24}]Combining these results and factoring, we obtain:[sum_{k=n+1}^{2n} a_k = frac{n(n + 1)(5n^2 + 5n + 2)}{24}]Thus, the final answer is:[boxed{dfrac{1}{24}n(n+1)(5n^2 +5n +2)}],<think>Okay, so I need to prove that the sum of the coefficients from ( a_{n+1} ) to ( a_{2n} ) in the polynomial ( (x + 2x^2 + cdots + nx^n)^2 ) is equal to ( frac{1}{24} n(n+1)(5n^2 + 5n + 2) ). Let me try to break this down step by step.First, let me understand the problem. We have a polynomial ( P(x) = x + 2x^2 + 3x^3 + cdots + nx^n ). Then we square this polynomial, resulting in another polynomial ( Q(x) = P(x)^2 ), which will have terms up to ( x^{2n} ). The coefficients of ( Q(x) ) are the ( a_k )s, so ( Q(x) = a_0 + a_1x + a_2x^2 + cdots + a_{2n}x^{2n} ). The problem asks for the sum of the coefficients from ( a_{n+1} ) to ( a_{2n} ), which is ( sum_{k=n+1}^{2n} a_k ).I need to find this sum and show that it's equal to the given expression involving ( n ). Hmm. Let me recall that the sum of coefficients of a polynomial can be found by evaluating the polynomial at ( x = 1 ). But here, we need the sum of coefficients from degree ( n+1 ) to ( 2n ). So maybe I can compute ( Q(1) ), which is the sum of all coefficients ( a_0 + a_1 + cdots + a_{2n} ), and then subtract the sum of coefficients from ( a_0 ) to ( a_n ). That would give me the desired sum ( sum_{k=n+1}^{2n} a_k ).Yes, that makes sense. So first, let me compute ( Q(1) ). Since ( Q(x) = P(x)^2 ), then ( Q(1) = [P(1)]^2 ). Let's compute ( P(1) ):( P(1) = 1 + 2 cdot 1 + 3 cdot 1 + cdots + n cdot 1 = 1 + 2 + 3 + cdots + n ).The sum of the first ( n ) positive integers is ( frac{n(n+1)}{2} ). Therefore, ( Q(1) = left( frac{n(n+1)}{2} right)^2 ).So the total sum of all coefficients ( a_0 + a_1 + cdots + a_{2n} ) is ( left( frac{n(n+1)}{2} right)^2 ).Next, I need to find the sum ( S = sum_{k=0}^n a_k ). Then the desired sum ( sum_{k=n+1}^{2n} a_k = Q(1) - S ).Therefore, the key is to compute ( S = sum_{k=0}^n a_k ). How can I compute this?Alternatively, since ( S ) is the sum of coefficients of ( Q(x) ) from ( x^0 ) to ( x^n ), perhaps I can use generating functions or some combinatorial interpretation. Let me think.Another approach: Since ( Q(x) = (x + 2x^2 + cdots + nx^n)^2 ), expanding this product would result in terms where each term is ( i x^i cdot j x^j = ij x^{i+j} ). Therefore, the coefficient ( a_k ) in ( Q(x) ) is the sum of all products ( ij ) such that ( i + j = k ), where ( 1 leq i, j leq n ).So, more formally, ( a_k = sum_{i=1}^n sum_{j=1}^n ij cdot delta_{i+j,k} ), where ( delta_{i+j,k} ) is 1 if ( i + j = k ), 0 otherwise. Therefore, the sum ( sum_{k=0}^n a_k ) is the sum over all ( i, j ) from 1 to ( n ) of ( ij ) such that ( i + j leq n ).Hence, ( S = sum_{i=1}^n sum_{j=1}^{n - i} ij ), since for each ( i ), ( j ) can go up to ( n - i ) to keep ( i + j leq n ).Therefore, we can compute ( S ) as:( S = sum_{i=1}^{n-1} sum_{j=1}^{n - i} ij )Note that when ( i = n ), ( n - i = 0 ), so there are no terms. So the upper limit for ( i ) is ( n - 1 ).This seems a bit complicated, but maybe there's a way to compute this double sum. Let me try to compute the inner sum first. For fixed ( i ), the inner sum is ( i sum_{j=1}^{n - i} j ). The sum of the first ( m ) integers is ( frac{m(m + 1)}{2} ), so here, ( m = n - i ), so:Inner sum = ( i cdot frac{(n - i)(n - i + 1)}{2} )Therefore, ( S = sum_{i=1}^{n - 1} frac{i(n - i)(n - i + 1)}{2} )Hmm, this seems manageable. Let's write that out:( S = frac{1}{2} sum_{i=1}^{n - 1} i(n - i)(n - i + 1) )Let me make a substitution to simplify the expression. Let ( k = n - i ). When ( i = 1 ), ( k = n - 1 ); when ( i = n - 1 ), ( k = 1 ). So reversing the order of summation:( S = frac{1}{2} sum_{k=1}^{n - 1} (n - k) cdot k cdot (k + 1) )So,( S = frac{1}{2} sum_{k=1}^{n - 1} k(k + 1)(n - k) )Let me expand the term inside the sum:( k(k + 1)(n - k) = k(k + 1)n - k^2(k + 1) )So,( S = frac{n}{2} sum_{k=1}^{n - 1} k(k + 1) - frac{1}{2} sum_{k=1}^{n - 1} k^2(k + 1) )Let's compute each sum separately.First sum: ( sum_{k=1}^{n - 1} k(k + 1) )This is ( sum_{k=1}^{n - 1} (k^2 + k) = sum_{k=1}^{n - 1} k^2 + sum_{k=1}^{n - 1} k )We know that ( sum_{k=1}^m k = frac{m(m + 1)}{2} ), so ( sum_{k=1}^{n - 1} k = frac{(n - 1)n}{2} )Similarly, ( sum_{k=1}^m k^2 = frac{m(m + 1)(2m + 1)}{6} ), so ( sum_{k=1}^{n - 1} k^2 = frac{(n - 1)n(2n - 1)}{6} )Therefore, first sum becomes:( frac{(n - 1)n(2n - 1)}{6} + frac{(n - 1)n}{2} = frac{(n - 1)n}{6} [2n - 1 + 3] = frac{(n - 1)n}{6}(2n + 2) = frac{(n - 1)n cdot 2(n + 1)}{6} = frac{(n - 1)n(n + 1)}{3} )Second sum: ( sum_{k=1}^{n - 1} k^2(k + 1) )Let me expand this:( sum_{k=1}^{n - 1} (k^3 + k^2) = sum_{k=1}^{n - 1} k^3 + sum_{k=1}^{n - 1} k^2 )Again, using known formulas:( sum_{k=1}^m k^3 = left( frac{m(m + 1)}{2} right)^2 ), so ( sum_{k=1}^{n - 1} k^3 = left( frac{(n - 1)n}{2} right)^2 )And ( sum_{k=1}^{n - 1} k^2 = frac{(n - 1)n(2n - 1)}{6} ), as before.Therefore, second sum is:( left( frac{(n - 1)n}{2} right)^2 + frac{(n - 1)n(2n - 1)}{6} )So putting it all together:( S = frac{n}{2} cdot frac{(n - 1)n(n + 1)}{3} - frac{1}{2} left[ left( frac{(n - 1)n}{2} right)^2 + frac{(n - 1)n(2n - 1)}{6} right] )Let's compute each term step by step.First term: ( frac{n}{2} cdot frac{(n - 1)n(n + 1)}{3} = frac{n^2(n - 1)(n + 1)}{6} = frac{n^2(n^2 - 1)}{6} )Second term: ( frac{1}{2} left[ frac{(n - 1)^2n^2}{4} + frac{(n - 1)n(2n - 1)}{6} right] )Let me compute the inside first:( frac{(n - 1)^2n^2}{4} + frac{(n - 1)n(2n - 1)}{6} )Factor out ( frac{(n - 1)n}{12} ):= ( frac{(n - 1)n}{12} [3(n - 1)n + 2(2n - 1)] )Compute inside the brackets:3(n - 1)n + 2(2n - 1) = 3n(n - 1) + 4n - 2 = 3n^2 - 3n + 4n - 2 = 3n^2 + n - 2Therefore, the second term becomes:( frac{1}{2} cdot frac{(n - 1)n(3n^2 + n - 2)}{12} = frac{(n - 1)n(3n^2 + n - 2)}{24} )Therefore, putting both terms together:( S = frac{n^2(n^2 - 1)}{6} - frac{(n - 1)n(3n^2 + n - 2)}{24} )Let me factor out ( frac{(n - 1)n}{24} ) from both terms:First term: ( frac{n^2(n^2 - 1)}{6} = frac{n^2(n - 1)(n + 1)}{6} = frac{(n - 1)n cdot n(n + 1)}{6} )Multiply numerator and denominator by 4 to get denominator 24:= ( frac{(n - 1)n cdot 4n(n + 1)}{24} )Second term is already over 24.Therefore:( S = frac{(n - 1)n}{24} [4n(n + 1) - (3n^2 + n - 2)] )Compute the expression inside the brackets:4n(n + 1) = 4n^2 + 4nSubtract (3n^2 + n - 2):4n^2 + 4n - 3n^2 - n + 2 = (4n^2 - 3n^2) + (4n - n) + 2 = n^2 + 3n + 2Therefore:( S = frac{(n - 1)n(n^2 + 3n + 2)}{24} )Factor ( n^2 + 3n + 2 = (n + 1)(n + 2) )So:( S = frac{(n - 1)n(n + 1)(n + 2)}{24} )Wait, but let's check if that's correct. Wait, if n^2 + 3n + 2 factors into (n + 1)(n + 2), which it does. So then:( S = frac{n(n - 1)(n + 1)(n + 2)}{24} )Hmm, but let's verify with n=2. Let me test n=2 to check.If n=2, then original polynomial is (x + 2x^2)^2 = x^2 + 4x^3 + 4x^4. The coefficients from x^0 to x^2: a_0=0, a_1=0, a_2=1. So sum S=0+0+1=1.Using the formula for S: [2(2 -1)(2 +1)(2 +2)] /24 = [2*1*3*4]/24 = 24/24=1. Correct.Another test: n=3. The polynomial is (x + 2x² +3x³)^2. Let's compute coefficients up to x^3.The square would be:x*(x) = x²x*(2x²) = 2x³x*(3x³) = 3x⁴2x²*x = 2x³2x²*2x² =4x⁴2x²*3x³=6x^53x³*x=3x⁴3x³*2x²=6x^53x³*3x³=9x^6So collecting terms:x², (2x³ +2x³)=4x³, (3x⁴ +4x⁴ +3x⁴)=10x⁴, etc. But S is sum from a0 to a3. a0=0, a1=0, a2=1, a3=4. So sum S=0+0+1+4=5.Using the formula: [3*2*4*5]/24 = (3*2*4*5)/24 = 120/24=5. Correct. So formula works.Thus, general formula for S is ( frac{n(n - 1)(n + 1)(n + 2)}{24} ). Alternatively, simplifying:( frac{n(n^2 - 1)(n + 2)}{24} = frac{n(n + 2)(n^2 - 1)}{24} ). But maybe we can write it as ( frac{n(n + 1)(n - 1)(n + 2)}{24} ). However, let's see if we can express it differently.Alternatively, note that ( n(n + 1)(n - 1)(n + 2) = n(n + 1)(n^2 + n - 2) ). Hmm, not sure if that helps.But moving on. Remember that the total sum Q(1) is ( left( frac{n(n + 1)}{2} right)^2 ), which is ( frac{n^2(n + 1)^2}{4} ).Therefore, the desired sum is ( Q(1) - S = frac{n^2(n + 1)^2}{4} - frac{n(n - 1)(n + 1)(n + 2)}{24} )Let me factor out common terms. Both terms have ( n(n + 1) ). Let's factor that out:= ( frac{n(n + 1)}{24} [6n(n + 1) - (n - 1)(n + 2)] )Compute the expression inside the brackets:First term: 6n(n + 1) = 6n² + 6nSecond term: (n - 1)(n + 2) = n² + 2n - n - 2 = n² + n - 2Therefore, the expression becomes:6n² + 6n - (n² + n - 2) = 6n² + 6n - n² - n + 2 = 5n² +5n +2Therefore, the desired sum is:( frac{n(n + 1)}{24} (5n² +5n +2) )Which is exactly the expression we needed to prove: ( frac{1}{24}n(n + 1)(5n² +5n +2) ).Therefore, the proof is complete.Final Answerboxed{dfrac{1}{24}n(n+1)(5n^2 + 5n + 2)}</think>To prove that the sum of the coefficients from ( a_{n+1} ) to ( a_{2n} ) in the polynomial ( (x + 2x^2 + cdots + nx^n)^2 ) is equal to ( frac{1}{24} n(n+1)(5n^2 + 5n + 2) ), we start by considering the polynomial ( P(x) = x + 2x^2 + 3x^3 + cdots + nx^n ). Squaring this polynomial gives ( Q(x) = P(x)^2 ), which has coefficients ( a_k ).First, we compute the total sum of all coefficients of ( Q(x) ) by evaluating ( Q(1) ):[Q(1) = [P(1)]^2 = left( sum_{k=1}^n k right)^2 = left( frac{n(n+1)}{2} right)^2 = frac{n^2(n+1)^2}{4}]Next, we need the sum of coefficients from ( a_0 ) to ( a_n ). This sum, denoted ( S ), is found by considering the terms ( ij ) such that ( i + j leq n ):[S = sum_{i=1}^{n-1} sum_{j=1}^{n-i} ij]Rewriting the inner sum and substituting ( k = n - i ), we get:[S = frac{1}{2} sum_{k=1}^{n-1} k(k+1)(n-k)]Expanding and simplifying the sum, we find:[S = frac{n(n-1)(n+1)(n+2)}{24}]Finally, the desired sum is the total sum minus ( S ):[sum_{k=n+1}^{2n} a_k = frac{n^2(n+1)^2}{4} - frac{n(n-1)(n+1)(n+2)}{24}]Factoring out common terms and simplifying, we obtain:[sum_{k=n+1}^{2n} a_k = frac{n(n+1)}{24} (5n^2 + 5n + 2)]Thus, the result is:[boxed{dfrac{1}{24}n(n+1)(5n^2 + 5n + 2)}]]

🔑:[Proof] Clearly, a_{0}=a_{1}=0, a_{2}=1.For 3 leqslant k leqslant n,begin{aligned}a_{k} & =1 cdot(k-1)+2 cdot(k-2)+cdots+(k-1) cdot 1 & =k(1+2+cdots+k-1)-left[1^{2}+2^{2}+cdots+(k-1)^{2}right] & =frac{1}{2} k^{2}(k-1)-frac{1}{6}(k-1) k(2 k-1) & =frac{1}{6}(k-1) k(k+1) & =C_{k+1}^{3} .end{aligned}Thus, sum_{k=0}^{n} a_{k}=sum_{k=2}^{n} C_{k+1}^{3}.By the property of binomial coefficients C_{n+1}^{m}=C_{n}^{m}+C_{n}^{m-1}, we havesum_{k=r}^{n} C_{k}^{r}=C_{n+1}^{r+1} text {. }Therefore, sum_{k=0}^{n} a_{k}=C_{n+2}^{4}.Let f(x)=a_{0}+a_{1} x+a_{2} x^{2}+cdots+a_{2 n} x^{2 n}=left(x+2 x^{2}+cdots+n x^{n}right)^{2} text {, }thenbegin{aligned}sum_{k=n+1}^{2 n} a_{k} & =f(1)-sum_{k=0}^{n} a_{k} & =left[n(n+1) cdot frac{1}{2}right]^{2}-frac{1}{24}(n+2)(n+1) n(n-1) & =frac{1}{24} n(n+1)left(5 n^{2}+5 n+2right)end{aligned}