🧠:[<think>Alright, let's tackle this inequality problem. So, we have these positive numbers a₁, a₂, ..., aₙ, where n is at least 2, and their sum is 1. We need to prove that the sum of aᵢ divided by (2 - aᵢ) is at least n/(2n - 1). Hmm, okay. Let me start by understanding the problem better.First, the given condition is that the sum of all aᵢ is 1. The expression we need to evaluate is the sum of each aᵢ divided by (2 - aᵢ). The inequality to prove is that this sum is greater than or equal to n divided by (2n - 1). I remember that inequalities like these sometimes can be approached using techniques like Cauchy-Schwarz, Jensen's inequality, or maybe even Lagrange multipliers. Let me think which one would be applicable here.Since the function involved here is f(x) = x/(2 - x), maybe we can check if this function is convex or concave. If it's convex, then Jensen's inequality could be useful. Let's compute the second derivative to check.First derivative of f(x): f'(x) = [1*(2 - x) - x*(-1)]/(2 - x)² = [2 - x + x]/(2 - x)² = 2/(2 - x)².Second derivative: f''(x) = [0*(2 - x)² - 2*2*(2 - x)*(-1)]/(2 - x)^4. Wait, maybe that's getting complicated. Alternatively, since the first derivative is 2/(2 - x)², then the second derivative would be 2*2/(2 - x)^3 = 4/(2 - x)^3. Since x is positive and less than 2 (because the sum of aᵢ is 1, so each aᵢ is less than 1, so 2 - aᵢ is greater than 1), so the second derivative is positive. Therefore, f(x) is convex on the interval (0,1).Jensen's inequality states that for a convex function, the average of the function at each point is at least the function at the average point. So, if we apply Jensen's inequality here, we would get:(1/n) * Σ [aᵢ/(2 - aᵢ)] ≥ [ (1/n) * Σ aᵢ ] / [2 - (1/n)*Σ aᵢ ]But Σ aᵢ is 1, so substituting that in:(1/n) * Σ [aᵢ/(2 - aᵢ)] ≥ [ (1/n) * 1 ] / [2 - (1/n)*1 ] = (1/n) / (2 - 1/n) = (1/n) / ( (2n - 1)/n ) = 1/(2n - 1)Multiply both sides by n:Σ [aᵢ/(2 - aᵢ)] ≥ n/(2n - 1)And that's exactly what we needed to prove! Wait, so using Jensen's inequality directly gives us the result. That seems straightforward. But let me verify if I applied Jensen correctly.Given that f is convex, Jensen's inequality is:f(Σ λᵢ xᵢ) ≤ Σ λᵢ f(xᵢ)But in our case, we have the sum (1/n) Σ f(aᵢ), which is the average of f(aᵢ). Since f is convex, Jensen gives:f( (1/n) Σ aᵢ ) ≤ (1/n) Σ f(aᵢ )But we need the reverse inequality. Wait, hold on. Wait, no, if f is convex, then the inequality is:(1/n) Σ f(aᵢ) ≥ f( (1/n) Σ aᵢ )Yes, that's correct. Because for convex functions, the average of the function is at least the function of the average. So since Σ aᵢ = 1, then (1/n) Σ aᵢ = 1/n. So,(1/n) Σ [aᵢ/(2 - aᵢ)] ≥ f(1/n) = (1/n)/(2 - 1/n) = (1/n)/( (2n - 1)/n ) = 1/(2n - 1)Then multiplying both sides by n gives the desired inequality. So that works out. Therefore, the inequality is proven using Jensen's inequality.But let me check if there are other ways to approach this problem, maybe for better understanding.Alternatively, perhaps using Cauchy-Schwarz. Let's consider the Cauchy-Schwarz inequality in the form:(Σ (aᵢ/(2 - aᵢ)) ) (Σ (2 - aᵢ) ) ≥ (Σ √(aᵢ/(2 - aᵢ)) * √(2 - aᵢ) )² = (Σ √aᵢ )²But I'm not sure if that helps directly. Let's compute Σ (2 - aᵢ). Since Σ aᵢ = 1, then Σ (2 - aᵢ) = 2n - 1. So, we have:Left-hand side: (Σ (aᵢ/(2 - aᵢ)) ) * (2n - 1) ≥ (Σ √aᵢ )²But then, (Σ √aᵢ )² ≤ (n Σ aᵢ ) by Cauchy-Schwarz? Wait, actually, by Cauchy-Schwarz, (Σ √aᵢ )² ≤ (Σ 1)(Σ aᵢ ) = n*1 = n. Therefore, we have:(Σ (aᵢ/(2 - aᵢ)) ) * (2n - 1) ≥ nWhich implies that Σ (aᵢ/(2 - aᵢ)) ≥ n/(2n - 1). Hey, that works too! So using Cauchy-Schwarz also gives the same result. So that's another method.Therefore, both Jensen's inequality and Cauchy-Schwarz can be applied here. Interesting. So depending on the approach, either method works. Let me verify the Cauchy-Schwarz approach again to ensure I didn't make a miscalculation.Let me write Cauchy-Schwarz as (Σ xᵢ yᵢ )² ≤ (Σ xᵢ²)(Σ yᵢ²). But in the application above, I used a different form. Let me rephrase.If I set xᵢ = √(aᵢ/(2 - aᵢ)) and yᵢ = √(2 - aᵢ), then Σ xᵢ yᵢ = Σ √(aᵢ/(2 - aᵢ)) * √(2 - aᵢ) = Σ √aᵢ. Then by Cauchy-Schwarz,(Σ √aᵢ )² ≤ (Σ xᵢ²)(Σ yᵢ²) = (Σ (aᵢ/(2 - aᵢ)) )(Σ (2 - aᵢ))Which is exactly the same as before. Therefore, (Σ √aᵢ )² ≤ (Σ (aᵢ/(2 - aᵢ)) )(2n - 1). But we also know that (Σ √aᵢ )² ≤ n, by Cauchy-Schwarz again, since (Σ √aᵢ )² ≤ (Σ 1)(Σ aᵢ ) = n*1 = n. Therefore, combining these:n ≤ (Σ (aᵢ/(2 - aᵢ)) )(2n - 1)Which leads to the desired inequality. So that works as well. Therefore, either approach is valid. Alternatively, another method could be the Titu's lemma, which is a specific case of Cauchy-Schwarz. Titu's lemma states that Σ (aᵢ² / bᵢ ) ≥ (Σ aᵢ )² / Σ bᵢ. But in our case, we have Σ (aᵢ / (2 - aᵢ)). If we consider it as Σ (aᵢ / (2 - aᵢ)), maybe we can apply Titu's lemma here. Let's see.Set aᵢ as the numerators and 2 - aᵢ as denominators. But in Titu's lemma, it's Σ (xᵢ² / yᵢ ) ≥ (Σ xᵢ )² / Σ yᵢ. If we let xᵢ = √aᵢ and yᵢ = 2 - aᵢ, then:Σ (aᵢ / (2 - aᵢ)) = Σ (xᵢ² / yᵢ ) ≥ (Σ xᵢ )² / Σ yᵢ = (Σ √aᵢ )² / (2n - 1)Again, similar to before. And since (Σ √aᵢ )² ≤ n, but here we have (Σ √aᵢ )² in the numerator, so this seems like we get the same inequality. Wait, but if we have Σ (aᵢ / (2 - aᵢ)) ≥ (Σ √aᵢ )² / (2n - 1), and we also know that (Σ √aᵢ )² ≤ n, then that would give Σ (aᵢ / (2 - aᵢ)) ≥ (something ≤ n) / (2n - 1). Wait, but that seems like the opposite direction. Hmm, maybe I made a mistake here.Wait, no. Let's clarify. If Titu's lemma gives Σ (aᵢ / (2 - aᵢ)) ≥ (Σ √aᵢ )² / (2n - 1), but we need to show Σ (aᵢ / (2 - aᵢ)) ≥ n/(2n - 1). Therefore, if we can show that (Σ √aᵢ )² ≥ n, then the inequality would follow. However, in reality, by Cauchy-Schwarz, (Σ √aᵢ )² ≤ n*Σ aᵢ = n. So we have (Σ √aᵢ )² ≤ n, which would imply that Σ (aᵢ / (2 - aᵢ)) ≥ [ (Σ √aᵢ )² ] / (2n - 1) ≤ n/(2n - 1). Wait, that seems contradictory. So maybe Titu's lemma isn't directly helpful here because it gives a lower bound in terms of (Σ √aᵢ )², which is itself bounded above by n, leading to the lower bound being at most n/(2n -1). But we need the lower bound to be exactly n/(2n -1). Therefore, this approach might not be the right one. So perhaps Cauchy-Schwarz in the way I did earlier is better.But in the previous approach, when I used Cauchy-Schwarz with xᵢ = √(aᵢ/(2 - aᵢ)) and yᵢ = √(2 - aᵢ), leading to (Σ √aᵢ )² ≤ (Σ (aᵢ/(2 - aᵢ)))(Σ (2 - aᵢ)), and then since Σ √aᵢ squared is ≤ n, we get n ≤ (Σ (aᵢ/(2 - aᵢ)))(2n -1), hence the required inequality. That works. So in this case, Titu's lemma is not directly applicable, but a different application of Cauchy-Schwarz gives the result.Alternatively, another method: Lagrange multipliers. Suppose we want to minimize Σ [aᵢ/(2 - aᵢ)] subject to Σ aᵢ = 1 and aᵢ > 0. To find the minimum, we can set up the Lagrangian.Let me denote the Lagrangian as L = Σ [aᵢ/(2 - aᵢ)] - λ(Σ aᵢ - 1). Taking partial derivatives with respect to each aᵢ:∂L/∂aᵢ = [ (1)(2 - aᵢ) + aᵢ ] / (2 - aᵢ)^2 - λ = [2 - aᵢ + aᵢ]/(2 - aᵢ)^2 - λ = 2/(2 - aᵢ)^2 - λ = 0.Therefore, for all i, 2/(2 - aᵢ)^2 = λ. This implies that all (2 - aᵢ)^2 are equal, hence all aᵢ are equal. Since Σ aᵢ = 1, each aᵢ = 1/n. Therefore, the minimal value is Σ [ (1/n) / (2 - 1/n) ] = n*(1/n)/(2 - 1/n) = 1/(2 - 1/n) = n/(2n -1). Therefore, the minimal value is achieved when all aᵢ are equal, and thus Σ [aᵢ/(2 - aᵢ)] ≥ n/(2n -1). That's another approach using optimization techniques. So by symmetry, the minimum occurs when all variables are equal, which is a common occurrence in inequalities with symmetric constraints.Therefore, multiple methods confirm the inequality. The problem seems to be a classic example where either convexity (Jensen), Cauchy-Schwarz, or Lagrangian multipliers can be applied. All approaches lead to the same conclusion.Just to double-check with an example. Let’s take n=2. Then the right-hand side is 2/(4-1)=2/3. Suppose a₁ + a₂ =1. Let's compute the left-hand side. If a₁ = a₂ =1/2, then each term is (1/2)/(2 -1/2)= (1/2)/(3/2)=1/3. Sum is 2*(1/3)=2/3, which equals the RHS. If we take a different case, say a₁ =0.9, a₂=0.1. Then the left-hand side is 0.9/(2 -0.9) +0.1/(2 -0.1)=0.9/1.1 +0.1/1.9 ≈0.818 +0.0526≈0.8706, which is greater than 2/3≈0.6667. Another case: a₁ approaching 1, a₂ approaching 0. Then the left-hand side approaches 1/(2 -1) +0/(2 -0)=1/1 +0=1, which is still greater than 2/3. So the inequality holds in these cases, and the minimal value occurs when all aᵢ are equal.Therefore, the proof seems solid with multiple methods confirming the result. The key takeaway is recognizing the convexity of the function involved, which allows the use of Jensen's inequality, or applying Cauchy-Schwarz with an appropriate choice of sequences. Alternatively, using Lagrange multipliers to find the minimum under the given constraint also leads to the same conclusion.Final Answerboxed{dfrac{n}{2n-1}}</think>Given positive numbers (a_{1}, a_{2}, ldots, a_{n}) (where (n geq 2)) satisfying (sum_{i=1}^{n} a_{i} = 1), we need to prove that:[sum_{i=1}^{n} frac{a_{i}}{2 - a_{i}} geqslant frac{n}{2n - 1}] Approach Using Jensen's Inequality1. Convexity Check: The function (f(x) = frac{x}{2 - x}) is convex on the interval ((0, 1)) because its second derivative (f''(x) = frac{4}{(2 - x)^3}) is positive for (x in (0, 1)).2. Applying Jensen's Inequality: For a convex function (f), Jensen's inequality states: [ frac{1}{n} sum_{i=1}^{n} f(a_{i}) geq fleft( frac{1}{n} sum_{i=1}^{n} a_{i} right) ] Substituting (f(x)) and using (sum_{i=1}^{n} a_{i} = 1): [ frac{1}{n} sum_{i=1}^{n} frac{a_{i}}{2 - a_{i}} geq frac{frac{1}{n}}{2 - frac{1}{n}} = frac{1}{2n - 1} ] Multiplying both sides by (n): [ sum_{i=1}^{n} frac{a_{i}}{2 - a_{i}} geq frac{n}{2n - 1} ] Approach Using Cauchy-Schwarz Inequality1. Applying Cauchy-Schwarz: Let (x_{i} = sqrt{frac{a_{i}}{2 - a_{i}}}) and (y_{i} = sqrt{2 - a_{i}}). By Cauchy-Schwarz: [ left( sum_{i=1}^{n} sqrt{a_{i}} right)^2 leq left( sum_{i=1}^{n} frac{a_{i}}{2 - a_{i}} right) left( sum_{i=1}^{n} (2 - a_{i}) right) ] Since (sum_{i=1}^{n} (2 - a_{i}) = 2n - 1) and (left( sum_{i=1}^{n} sqrt{a_{i}} right)^2 leq n) (by Cauchy-Schwarz): [ n leq left( sum_{i=1}^{n} frac{a_{i}}{2 - a_{i}} right) (2n - 1) ] Dividing both sides by (2n - 1): [ sum_{i=1}^{n} frac{a_{i}}{2 - a_{i}} geq frac{n}{2n - 1} ] Approach Using Lagrange Multipliers1. Setting Up the Lagrangian: Minimize (sum_{i=1}^{n} frac{a_{i}}{2 - a_{i}}) subject to (sum_{i=1}^{n} a_{i} = 1). The Lagrangian is: [ L = sum_{i=1}^{n} frac{a_{i}}{2 - a_{i}} - lambda left( sum_{i=1}^{n} a_{i} - 1 right) ]2. Solving for Critical Points: Taking partial derivatives and setting them to zero, we find all (a_{i}) are equal. Thus, (a_{i} = frac{1}{n}).3. Calculating the Minimum: Substituting (a_{i} = frac{1}{n}): [ sum_{i=1}^{n} frac{frac{1}{n}}{2 - frac{1}{n}} = frac{n}{2n - 1} ] ConclusionAll methods confirm the inequality holds, and the minimum value is achieved when all (a_{i}) are equal. Therefore, the final answer is:[boxed{dfrac{n}{2n-1}}],<think>Alright, so I have this inequality to prove: given positive numbers ( a_1, a_2, ldots, a_n ) where the sum of all ( a_i ) is 1, I need to show that the sum ( sum_{i=1}^n frac{a_i}{2 - a_i} ) is at least ( frac{n}{2n - 1} ). Hmm, okay. Let me think step by step.First, the problem states that all ( a_i ) are positive and their sum is 1. The inequality involves fractions where each term is ( frac{a_i}{2 - a_i} ). The right-hand side is ( frac{n}{2n - 1} ). I need to find a way to relate these terms. Maybe some inequality like Cauchy-Schwarz or Jensen's inequality could work here. Let me recall those.Jensen's inequality applies to convex or concave functions. If I can express the left-hand side as a function of ( a_i ), maybe I can check convexity. Let me consider the function ( f(x) = frac{x}{2 - x} ). Let's compute its second derivative to check for convexity.First derivative: ( f'(x) = frac{(2 - x) - (-1)x}{(2 - x)^2} = frac{2 - x + x}{(2 - x)^2} = frac{2}{(2 - x)^2} ).Second derivative: ( f''(x) = frac{2 cdot 2}{(2 - x)^3} = frac{4}{(2 - x)^3} ). Since ( a_i ) are positive and sum to 1, each ( a_i < 1 ), so ( 2 - a_i > 1 ), so the denominator is positive. Therefore, ( f''(x) > 0 ), meaning ( f(x) ) is convex on the interval ( (0, 1) ). Therefore, by Jensen's inequality for convex functions, we have:( sum_{i=1}^n frac{a_i}{2 - a_i} geq n cdot frac{frac{1}{n} sum a_i}{2 - frac{1}{n} sum a_i} ).Wait, Jensen's inequality states that for a convex function ( f ), ( frac{1}{n} sum f(a_i) geq fleft( frac{1}{n} sum a_i right) ). So multiplying both sides by ( n ), we get ( sum f(a_i) geq n fleft( frac{1}{n} sum a_i right) ).But in this case, ( sum a_i = 1 ), so ( frac{1}{n} sum a_i = frac{1}{n} ). Therefore, applying Jensen's:( sum_{i=1}^n frac{a_i}{2 - a_i} geq n cdot frac{frac{1}{n}}{2 - frac{1}{n}} = n cdot frac{1/n}{(2n - 1)/n} } = n cdot frac{1}{2n - 1} = frac{n}{2n - 1} ).Wait, that seems like exactly the inequality we need to prove. So does that mean the proof is straightforward by Jensen's inequality? Let me double-check.Given that ( f(x) = frac{x}{2 - x} ) is convex, then Jensen's gives:( frac{1}{n} sum_{i=1}^n f(a_i) geq fleft( frac{1}{n} sum a_i right) ).Multiply both sides by ( n ):( sum f(a_i) geq n fleft( frac{1}{n} right) ).Compute ( fleft( frac{1}{n} right) ):( frac{frac{1}{n}}{2 - frac{1}{n}} = frac{1}{2n - 1} ).Therefore, the sum is ( geq n cdot frac{1}{2n - 1} ), which is exactly the required inequality. So, that's it? The problem is solved by applying Jensen's inequality directly.But let me verify with a simple example. Suppose ( n = 2 ). Then the right-hand side is ( frac{2}{4 - 1} = frac{2}{3} ). Let me take ( a_1 = a_2 = 1/2 ).Compute the left-hand side: ( frac{1/2}{2 - 1/2} + frac{1/2}{2 - 1/2} = 2 cdot frac{1/2}{3/2} = 2 cdot frac{1}{3} = frac{2}{3} ). So equality holds here.Another case: suppose ( n = 2 ), but ( a_1 = 1 ), ( a_2 = 0 ). Wait, but the problem states all numbers are positive, so ( a_2 ) can't be zero. But approaching towards that limit, say ( a_1 = 1 - epsilon ), ( a_2 = epsilon ), where ( epsilon ) approaches 0. Then the left-hand side becomes ( frac{1 - epsilon}{2 - (1 - epsilon)} + frac{epsilon}{2 - epsilon} = frac{1 - epsilon}{1 + epsilon} + frac{epsilon}{2 - epsilon} ). As ( epsilon to 0 ), this approaches ( 1 + 0 = 1 ), which is greater than ( frac{2}{3} ). So the inequality holds here as well.Another test case: ( n = 3 ), each ( a_i = 1/3 ). Then left-hand side: ( 3 cdot frac{1/3}{2 - 1/3} = 3 cdot frac{1/3}{5/3} = 3 cdot frac{1}{5} = frac{3}{5} ). The right-hand side is ( frac{3}{6 - 1} = frac{3}{5} ). Equality again holds here. So when all variables are equal, equality holds. That makes sense because Jensen's inequality gives equality when all variables are equal for a convex function.Therefore, the key step was recognizing the convexity of the function ( f(x) = frac{x}{2 - x} ), applying Jensen's inequality, and verifying that the equality condition holds when all ( a_i ) are equal. Since the problem states ( n geq 2 ), and all ( a_i ) are positive summing to 1, this approach should work for all such cases.Alternatively, maybe I can approach this using the method of Lagrange multipliers or Cauchy-Schwarz, but since Jensen's gives a straightforward proof, that's probably the intended method.Just to check another approach: perhaps using the Cauchy-Schwarz inequality. Let me think. Let's consider the sum ( sum frac{a_i}{2 - a_i} ). If I rewrite each term as ( frac{a_i}{2 - a_i} = frac{a_i}{2 - a_i} ). Maybe I can write this as ( frac{a_i}{2 - a_i} = frac{a_i}{1 + (1 - a_i)} ). Not sure if that helps. Alternatively, maybe consider that ( frac{a_i}{2 - a_i} = frac{a_i}{1 + (1 - a_i)} geq frac{a_i}{2} ), but that doesn't seem helpful since summing that would give ( frac{1}{2} ), which is less than ( frac{n}{2n - 1} ).Alternatively, perhaps use the Cauchy-Schwarz inequality in the form:( sum frac{a_i}{2 - a_i} geq frac{(sum sqrt{a_i})^2}{sum (2 - a_i)} ). Let's see. By Cauchy-Schwarz, ( sum frac{x_i^2}{y_i} geq frac{(sum x_i)^2}{sum y_i} ). If I let ( x_i = sqrt{a_i} ) and ( y_i = 2 - a_i ), then:( sum frac{a_i}{2 - a_i} geq frac{(sum sqrt{a_i})^2}{sum (2 - a_i)} ).But ( sum (2 - a_i) = 2n - 1 ), since ( sum a_i = 1 ). So the denominator is ( 2n - 1 ). The numerator is ( (sum sqrt{a_i})^2 ). However, we need to compare ( (sum sqrt{a_i})^2 ) with ( n ). By Cauchy-Schwarz, ( (sum sqrt{a_i})^2 leq n sum a_i = n ). Wait, so that would give ( sum frac{a_i}{2 - a_i} geq frac{(sum sqrt{a_i})^2}{2n - 1} geq frac{n}{2n - 1} ) only if ( (sum sqrt{a_i})^2 geq n ), but actually ( (sum sqrt{a_i})^2 leq n ) because of the Cauchy-Schwarz inequality. Therefore, this approach would give ( sum frac{a_i}{2 - a_i} geq frac{(sum sqrt{a_i})^2}{2n - 1} geq frac{?}{2n -1} ). But since ( (sum sqrt{a_i})^2 leq n sum a_i = n ), this would give a lower bound of ( frac{(sum sqrt{a_i})^2}{2n -1} leq frac{n}{2n -1} ), which is the opposite direction. So this approach doesn't help.Alternatively, perhaps using the Titu's lemma (which is a form of Cauchy-Schwarz). Wait, Titu's lemma is exactly the Cauchy-Schwarz inequality applied in the form above. So maybe that's not useful here.Alternatively, consider the function ( f(x) = frac{x}{2 - x} ), which is convex. Then by Jensen, as before.Alternatively, maybe use the method of Lagrange multipliers to find the minimum of ( sum frac{a_i}{2 - a_i} ) given the constraint ( sum a_i =1 ). Let's try that.Let ( L = sum_{i=1}^n frac{a_i}{2 - a_i} - lambda left( sum_{i=1}^n a_i -1 right) ).Taking derivative with respect to ( a_i ):( frac{dL}{da_i} = frac{(2 - a_i) - a_i(-1)}{(2 - a_i)^2} - lambda = frac{2}{(2 - a_i)^2} - lambda ).Set derivative to zero: ( frac{2}{(2 - a_i)^2} = lambda ).Therefore, all ( (2 - a_i)^2 ) must be equal, so all ( a_i ) must be equal. Hence, the minimum occurs when all ( a_i = 1/n ), which gives the value ( frac{n}{2n -1} ). Therefore, this confirms that the minimum is achieved when all variables are equal, hence proving the inequality.This method also shows that the inequality is tight when all ( a_i ) are equal, which matches our test cases earlier.So whether by Jensen's inequality or Lagrange multipliers, we arrive at the conclusion that the minimal value is ( frac{n}{2n -1} ), achieved when all ( a_i = 1/n ).Therefore, the proof is straightforward with Jensen's inequality once recognizing the convexity of the function involved.Final Answerboxed{dfrac{n}{2n-1}}</think>Given positive numbers ( a_1, a_2, ldots, a_n ) such that ( sum_{i=1}^{n} a_i = 1 ), we need to prove that:[sum_{i=1}^{n} frac{a_i}{2 - a_i} geqslant frac{n}{2n - 1}]First, we recognize that the function ( f(x) = frac{x}{2 - x} ) is convex. This can be verified by computing its second derivative:[f'(x) = frac{2}{(2 - x)^2} quad text{and} quad f''(x) = frac{4}{(2 - x)^3}]Since ( f''(x) > 0 ) for ( x in (0, 1) ), the function ( f(x) ) is convex. By Jensen's inequality for convex functions, we have:[frac{1}{n} sum_{i=1}^{n} f(a_i) geq fleft( frac{1}{n} sum_{i=1}^{n} a_i right)]Given ( sum_{i=1}^{n} a_i = 1 ), we substitute ( frac{1}{n} sum_{i=1}^{n} a_i = frac{1}{n} ). Multiplying both sides of Jensen's inequality by ( n ), we get:[sum_{i=1}^{n} frac{a_i}{2 - a_i} geq n cdot fleft( frac{1}{n} right)]Calculating ( fleft( frac{1}{n} right) ):[fleft( frac{1}{n} right) = frac{frac{1}{n}}{2 - frac{1}{n}} = frac{1}{2n - 1}]Thus,[sum_{i=1}^{n} frac{a_i}{2 - a_i} geq n cdot frac{1}{2n - 1} = frac{n}{2n - 1}]Equality holds when all ( a_i ) are equal, i.e., ( a_i = frac{1}{n} ) for all ( i ). This confirms that the minimum value of the sum is indeed ( frac{n}{2n - 1} ).[boxed{dfrac{n}{2n-1}}]]

🔑:Prove thatsum_{i=1}^{n} frac{a_{i}}{2-a_{i}}=sum_{i=1}^{n}left(frac{2}{2-a_{i}}-1right)=sum_{i=1}^{n} frac{2}{2-a_{i}}-n .By the Cauchy-Schwarz inequality, we haveleft(sum_{i=1}^{n} frac{1}{2-a_{i}}right)left[sum_{i=1}^{n}left(2-a_{i}right)right] geqslant n^{2}Therefore,begin{array}{l}sum_{i=1}^{n} frac{1}{2-a_{i}} geqslant frac{n^{2}}{sum_{i=1}^{n}left(2-a_{i}right)}=frac{n^{2}}{2 n-1} sum_{i=1}^{n} frac{a_{i}}{2-a_{i}} geqslant frac{2 n^{2}}{2 n-1}-n=frac{n}{2 n-1}end{array}

🧠:[<think>Okay, let me try to figure out this problem. So we have 100 points in the plane colored either red or white. The condition is that each red point must be the center of a circle that passes through at least three white points. We need to find the least possible number of white points. Hmm, interesting.First, let me parse the problem again to make sure I understand. We have 100 points total, some red and some white. Each red point is the center of a circle that passes through at least three white points. The goal is to minimize the number of white points. So, we need as few white points as possible such that every red point has three white points lying on some circle centered at it. The remaining points (100 minus the number of white points) will be red.Let me rephrase the condition: For each red point, there must exist three white points that are equidistant from it. That is, those three white points lie on a circle with the red point as the center. So, each red point must have three white points at the same distance from it. But those three white points don't have to be the same for different red points. However, if multiple red points can share the same white points, that might help reduce the total number of white points needed.So, the key is to arrange white points in such a way that each red point has three white points lying on a circle centered at that red point. And we want to do this with as few white points as possible.Let me think about the relationships here. Each white point can be on multiple circles centered at different red points. So, a single white point can help satisfy the condition for multiple red points. That seems crucial. If each white point can be used by multiple red points, then perhaps we can cover all red points with a small number of white points.But each red point needs three distinct white points. So, for each red point, we need three white points such that each is at some distance (could be different for each red point) from the red point. But these white points can be shared among different red points.Wait, but the distances might complicate things. If two red points share a white point, the distance from that white point to each red point would need to be such that it lies on a circle around each red point. So, unless the two red points are concentric, which they can't be unless they are the same point, the white point would have to lie on the intersection of two circles centered at different red points. The intersection of two circles (if they intersect) can be at most two points. So, unless multiple red points lie on the perpendicular bisector of the segment joining two white points, but that seems restrictive.Wait, perhaps I need to think of configurations where multiple red points can share the same set of three white points. If three white points are co-circular, meaning they lie on a single circle, then the center of that circle would be the red point. But if we have three white points on a circle centered at a red point, but those same three white points can't be on a circle centered at another red point unless the two red points are the same. Because the center of a circle through three non-colinear points is unique.Therefore, three non-colinear white points determine a unique circle, and hence a unique center. So, if we want multiple red points to each have three white points on their circles, those sets of three white points must either be different or arranged in such a way that the same three white points lie on circles with different centers. But as per the previous thought, three non-colinear points determine a unique circle, so unless those three points lie on multiple circles with different centers, which would require them to be co-circular with respect to multiple centers. But a set of three points can lie on multiple circles only if they are colinear, because three non-colinear points define a unique circle.Wait, if three points are colinear, they lie on infinitely many circles, each circle having its center on the perpendicular bisector of the line segment joining the leftmost and rightmost points. But in that case, the centers would lie along that perpendicular bisector. So, if we have three colinear white points, then there are infinitely many red points (along the perpendicular bisector) for which those three white points lie on a circle centered at the red point. However, the problem states that each red point must have three white points on a circle centered at it. So, if we have three colinear white points, then any red point on the perpendicular bisector of the line through these three white points would have those three white points lying on its circle. Therefore, in such a case, those three white points can serve multiple red points lying on that perpendicular bisector.This seems promising. If we arrange white points in a colinear fashion, then we can have multiple red points lying on the perpendicular bisector, each of which can use those three colinear white points as points on their circle. Therefore, each such red point would have those three white points lying on their circle (which is centered at the red point). So, with three colinear white points, we can have infinitely many red points along the perpendicular bisector, each using those three white points. But since we have a finite number of red points (100 minus the number of white points), perhaps this is a way to minimize the number of white points.But let's verify. Suppose we have three colinear white points. Let me denote them as A, B, C lying on a straight line. Then, the perpendicular bisector of the line AC would be the set of centers of circles passing through A, B, and C. Since A, B, C are colinear, the perpendicular bisector of AC is the line perpendicular to AC at its midpoint. Any point on this perpendicular bisector is equidistant from A and C, but B is on the line AC, so the distance from the center to B would vary. Wait, hold on. If we take a point on the perpendicular bisector of AC, then it is equidistant to A and C, but B is somewhere on AC. So, unless the center is equidistant to all three points A, B, C, which only occurs if B is the midpoint of AC. Wait, no. If A, B, C are colinear with B not necessarily the midpoint, then the circle passing through A, B, C must have its center on the perpendicular bisector of AC. However, for such a circle, the distance from the center to A and C is the same, but the distance to B would be different unless B is equidistant from A and C as well. Wait, no. Let me think again.Suppose A, B, C are colinear points with coordinates A(-a, 0), B(0,0), C(a, 0) for some a > 0. Then, the perpendicular bisector of AC is the y-axis (since AC is along the x-axis from (-a,0) to (a,0)). The center of any circle passing through A, B, C must lie on the y-axis. Let’s pick a center at (0, k) on the y-axis. The radius of the circle would be sqrt(a^2 + k^2), since the distance from (0,k) to A(-a,0) is sqrt(a^2 + k^2). The distance from (0,k) to B(0,0) is |k|. For B to lie on the circle centered at (0,k) passing through A and C, we must have |k| = sqrt(a^2 + k^2). But this is only possible if a = 0, which would collapse A and C to the same point. Therefore, this is impossible. Wait, that suggests that three colinear points cannot lie on a circle unless they are the same point, which is not possible. Wait, that contradicts my earlier thought. So, perhaps three colinear points can't lie on a circle unless they are all the same point, which is trivial. Therefore, my initial assumption was wrong. So, three colinear points cannot lie on a circle unless the line is a diameter of the circle, but even then, three colinear points on a circle would require that they are equally spaced around the circle? Wait, no. Wait, three points on a line can't all lie on a circle unless the line is a diameter of the circle, but then three colinear points on a circle would have two points symmetric about the center and the third point coinciding with the center? Hmm, perhaps I need to clarify.Wait, three colinear points can lie on a circle only if the line is a diameter of the circle and the three points are symmetrically placed. For example, suppose we have a circle with center at (0,0), and three points on the x-axis: (-a,0), (0,0), (a,0). Then, those three points lie on the circle of radius a centered at (0,0). Wait, but then (0,0) is the center and lies on the circle only if the radius is zero, which is trivial. So, actually, if three colinear points lie on a circle, then one of them must be the center. But in our problem, the red points are centers, and the white points are on the circle. So, if a white point is the center, then the radius would be zero, which doesn't make sense. Therefore, perhaps three colinear white points cannot lie on a circle unless two are symmetric with respect to the center and the third is the center itself, but the center is a red point, not a white point. Therefore, in our problem, the three white points cannot include the red point. Therefore, three colinear white points cannot lie on a circle centered at a red point unless two of them are symmetric with respect to the red point, and the third is... but three colinear points on a circle would require that the center is on the line, right? Because the perpendicular bisector of any chord passes through the center. If three points are colinear, then all their perpendicular bisectors would be perpendicular to that line. Wait, but for three colinear points, the perpendicular bisector of each pair is the same line, perpendicular to the line of the points. Therefore, the center must lie on that perpendicular bisector. Therefore, if three colinear points lie on a circle, the center of the circle must lie on the perpendicular bisector of the line segment joining any two of them. But if the three points are colinear, their perpendicular bisector is a single line. Therefore, there's only one circle (actually infinitely many circles with centers along that perpendicular bisector) passing through those three colinear points. Wait, no, that's not correct.Wait, three colinear points can't lie on a circle unless the line is a diameter, but in that case, the center is the midpoint of the line segment. Let me take an example. Let's say we have three colinear points A, B, C on the x-axis at (-1,0), (0,0), (1,0). Then, the circle with center at (0, k) would pass through A and C if the radius is sqrt(1 + k^2). For point B (0,0) to lie on this circle, the distance from (0,k) to (0,0) must be equal to the radius sqrt(1 + k^2). So |k| = sqrt(1 + k^2). Squaring both sides, k^2 = 1 + k^2 => 0 = 1, which is impossible. Therefore, three colinear points cannot lie on a circle unless they are all the same point. So, my initial thought was wrong. Three colinear points cannot lie on a circle unless they are coincident, which is trivial. Therefore, such a configuration is impossible.Therefore, going back, each red point must have three non-colinear white points on its circle. So, three white points that form a non-degenerate triangle. Since three non-colinear points determine a unique circle, each set of three white points can only correspond to one red point (the center of their circumscribed circle). Therefore, if we have a set of three white points, they can only satisfy the condition for one red point. Therefore, to cover multiple red points, we need different sets of three white points.Therefore, this seems like a problem where each red point needs a unique set of three white points. But if we can reuse white points across different triples, then maybe we can cover all red points with fewer white points. This is similar to a combinatorial design problem where we want to cover a certain number of elements (red points) with triples (sets of three white points) such that each element is covered by at least one triple, and we want to minimize the number of white points used in all triples.In combinatorics, this is similar to the set cover problem. However, set cover is NP-hard, but perhaps we can find a lower bound or use some combinatorial design.Alternatively, think in terms of hypergraphs. Each white point is a vertex, and each red point is a hyperedge connecting three white points. We need a hypergraph with 100 hyperedges (red points) each of size 3 (three white points), and we want the minimum number of vertices (white points) such that each hyperedge is connected to three vertices. This is the problem of finding the minimal number of vertices in a 3-uniform hypergraph with 100 hyperedges. However, in our case, there's an additional geometric constraint: the three white points in each hyperedge must lie on a circle centered at the corresponding red point. So, it's not just any hypergraph, but one that can be embedded in the plane with the centers of the circles being the red points.But maybe abstracting it to a hypergraph problem can give us a lower bound. In the hypergraph setting, the minimal number of vertices required for a 3-uniform hypergraph with 100 hyperedges is such that each hyperedge is a distinct triple. However, if we allow overlapping triples, the number of vertices can be much less. The lower bound is obtained when each vertex is included in as many hyperedges as possible.The formula for the lower bound in such a case is given by the solution to the inequality:C(n, 3) >= 100,where C(n,3) is the number of combinations of n white points taken 3 at a time. Solving for n:n(n-1)(n-2)/6 >= 100Let's compute this:n(n-1)(n-2) >= 600Testing n=10: 10*9*8=720 >=600. So n=10 gives 120 triples, which is more than 100. So the minimal n is 10? But this is if all triples are allowed. However, in our geometric problem, not all triples are possible because the three white points must lie on a circle centered at the red point. Therefore, the hypergraph is constrained by the geometry, so the minimal number might be higher.But maybe the lower bound from the hypergraph is still useful. If in the abstract case, you need at least 10 white points, then in the geometric case, you might need more. However, perhaps there's a clever geometric configuration that allows many triples to share white points, so maybe the lower bound is still around 10. Wait, but 10 choose 3 is 120, so if we can arrange 10 white points such that each red point corresponds to a unique triple of white points lying on a circle centered at the red point. But the problem is that in the plane, arranging 10 white points such that every triple lies on some circle (with the center being a red point) is non-trivial. Because each circle's center is determined by the triple. So, you would need 100 different centers (red points), each corresponding to a unique triple of the 10 white points.But in reality, each triple defines a unique center (the circumcenter), so unless multiple triples share the same circumcenter, which would require that those triples are co-circular with the same center. But if you have multiple triples of white points lying on the same circle (same center), then their circumcenters would be the same. Therefore, each red point (center) can correspond to multiple triples of white points lying on its circle. However, in our problem, each red point must have at least three white points on its circle. So, if a red point has, say, k white points on its circle, then the number of triples from those k white points is C(k,3). So, each red point can account for C(k,3) triples, but in our problem, each red point only needs one triple. Wait, no. The problem states that each red point is the center of a circle passing through at least three white points. So, each red point needs at least three white points on its circle. So, if a red point has k >=3 white points on its circle, that's sufficient. However, those white points can be part of other circles (other red points) as well.Therefore, if we can have a single white point on multiple circles (i.e., multiple red points), then we can reuse white points. So, perhaps arranging white points in such a way that each white point is on many circles centered at different red points.But how can a single white point be on multiple circles? Each circle is centered at a different red point, so the distances from the white point to each red point would have to be equal to the radii of those circles. Therefore, a white point can be on multiple circles if it is equidistant to multiple red points. Wait, but the distance from the white point to each red point is fixed. So, for a white point to be on a circle centered at red point R1 with radius r1, and also on a circle centered at red point R2 with radius r2, we must have that the distance from the white point to R1 is r1, and to R2 is r2. So, unless r1 and r2 are different, but the white point is at different distances from R1 and R2.But if two red points, R1 and R2, are such that the white point lies on both the circle of radius r1 around R1 and the circle of radius r2 around R2, then the white point lies at the intersection of those two circles. The intersection of two circles can be at most two points. Therefore, if we have multiple red points, their circles (of possibly different radii) can intersect at a white point. But for each white point, it can only lie on circles (around red points) for which it is on their respective circles. So, each white point can be associated with multiple red points, but each association requires that the white point is at a specific distance from the red point.This seems complicated. Perhaps another approach. Let me think of dual incidence. Let's consider red points and white points, with an incidence relation where a red point is connected to a white point if the white point lies on the circle centered at the red point. The problem states that each red point is connected to at least three white points. We need to minimize the number of white points.This is similar to a bipartite graph where one partition is red points (100 - w in total, where w is the number of white points) and the other partition is white points (w in total). Each red point has degree at least 3. The total number of edges is at least 3*(100 - w). On the other hand, each white point can be connected to multiple red points. To minimize w, we need to maximize the number of edges each white point can contribute. However, each edge corresponds to a white point being at a certain distance from a red point.But in terms of the bipartite graph, without geometric constraints, the minimal w would satisfy 3*(100 - w) <= w*(maximum degree per white point). To minimize w, set the maximum degree as high as possible. If there is no limit, then theoretically, w could be as low as 3, but with geometric constraints, the maximum degree per white point is limited.But in reality, how many red points can a single white point be associated with? For a white point W, it can lie on circles centered at various red points R1, R2, ..., Rk. Each such circle has W on its circumference, meaning the distance from W to each Ri is the radius of the respective circle. However, in the plane, the set of centers (red points) for which W lies on their circle is the set of all points at distance r from W, where r can be any positive real. However, in our problem, the red points are fixed, and the white points are fixed. So, for a single white point W, the red points that can have W on their circle are all red points that are at some distance from W. Each red point can choose its radius such that W is on its circle. However, each red point needs two other white points on its circle as well.Therefore, a single white point can be part of multiple circles (around different red points), but each such circle must also include two other white points. Therefore, if a white point W is used by multiple red points, each of those red points must have two other white points that are also on their respective circles. Therefore, the problem is not just about how many red points a single white point can be connected to, but also about how those connections overlap with other white points.This is getting complex. Maybe we can think of an upper bound first. Suppose all red points share the same three white points. Then, those three white points would need to lie on a circle centered at each red point. But as we saw earlier, three non-colinear points define a unique circle, so all red points would have to be the same point, which is impossible since they are distinct. Therefore, we cannot have all red points share the same three white points.Alternatively, suppose we partition the red points into groups, each group sharing the same three white points. For example, if we have groups of red points, each group using the same three white points. However, as before, each group would have to lie at the same center, which is impossible unless they are the same point. Therefore, this approach doesn't work.Another idea: Arrange white points in such a way that each white point is used by as many red points as possible. For instance, if a white point is placed such that many red points can have circles passing through it and two other white points. The challenge is to maximize the number of red points that can share a white point.But how can a single white point be on multiple circles centered at different red points, each requiring two other distinct white points? Let's consider a white point W1. For each red point R that uses W1, R must have two other white points, say W2 and W3, such that W1, W2, W3 lie on a circle centered at R. If W1 is used by multiple red points R1, R2, ..., Rk, then each Ri must have two other white points W2i and W3i such that W1, W2i, W3i lie on a circle centered at Ri.Therefore, for each red point Ri, we need a distinct pair (W2i, W3i) such that the three points W1, W2i, W3i lie on a circle centered at Ri. However, the pairs (W2i, W3i) can overlap with other pairs for different red points. So, if we have multiple red points sharing W1 and some other white points, perhaps we can reuse those other white points.But each time we add a red point Ri that uses W1, we need two other white points. If those two white points are unique to Ri, then each red point would require two new white points, leading to a linear growth in the number of white points. But if we can reuse the same pair (W2, W3) for multiple red points, then we can have multiple red points sharing W1, W2, W3. However, as before, the three white points W1, W2, W3 can only define a unique circle, so all those red points would have to be the same center, which is not allowed. Therefore, each trio of white points can only correspond to one red point.Therefore, if we fix W1, each additional red point that uses W1 must pair it with two new white points. Therefore, the number of white points needed would be 1 + 2*(number of red points using W1). To minimize the total number of white points, we want as many red points as possible to share white points.But if we use W1 with multiple pairs (W2, W3), (W4, W5), etc., each pair requiring two new white points, then the total number of white points would be 1 + 2k, where k is the number of red points using W1. However, each of those pairs (W2, W3) can only be used once, since they define a unique red point.Alternatively, perhaps some of those pairs can be shared among multiple white points. Wait, this is getting too vague. Let me try a different approach.Suppose we have w white points. Each white point can be part of multiple triples (with other white points), each triple defining a red point. Each triple must be unique to a red point. The number of triples is at least 100 - w (since there are 100 - w red points). Each triple is a combination of three white points, so the total number of triples available is C(w, 3). Therefore, we have the inequality:C(w, 3) >= 100 - wSo,w(w - 1)(w - 2)/6 >= 100 - wMultiply both sides by 6:w(w - 1)(w - 2) >= 600 - 6wBring all terms to left:w(w - 1)(w - 2) + 6w - 600 >= 0Expand w(w - 1)(w - 2):w^3 - 3w^2 + 2w + 6w - 600 >= 0Simplify:w^3 - 3w^2 + 8w - 600 >= 0Now, solve for w:Let's test w=9:9^3 - 3*9^2 +8*9 -600 = 729 - 243 + 72 -600 = (729-243)=486; 486+72=558; 558-600= -42 <0w=10:1000 - 300 +80 -600= 1000-300=700; 700+80=780; 780-600=180 >=0So, w=10 satisfies the inequality.Therefore, according to this combinatorial bound, we need at least 10 white points. However, this assumes that any triple of white points can be used for a red point, which is not necessarily true in the geometric case. Because in the plane, not every triple of white points will have a center (red point) such that all three lie on a circle around it. Therefore, the geometric constraint might require more white points.However, maybe with a clever arrangement, we can make many triples lie on circles with different centers. For example, if all white points lie on a circle, then their circumscribed circles would have centers located at various points. Wait, but if all white points lie on a common circle, then the centers of the circles passing through three white points would be the circumcenters of those triangles. The set of all such centers would form the set of red points. But how many red points can we get this way?If we place w white points on a circle, then the number of red points (circumcenters of triples of white points) would depend on the positions of the white points. If the white points are equally spaced on a circle, then the circumcenters of triples of points would mostly be the center of the original circle. For example, if you take three consecutive points on the circle, their circumcenter is the center of the original circle. However, if you take three points that are not equally spaced, the circumcenter might be different. Wait, actually, any three points on a circle will have their circumcenter at the center of the circle. Because all points lie on the circle, so the circle is the circumcircle for any triangle formed by three points on it. Therefore, the circumcenter is always the center of the original circle. Therefore, if all white points lie on a circle, then all red points (centers) must coincide with the center of that circle. But we can only have one red point there, which is insufficient since we need 100 - w red points. Therefore, this approach would not work.Therefore, arranging white points on a single circle is not helpful. Alternatively, arrange white points in such a way that different triples have different circumcenters. For example, place white points in general position (no three on a line, no four on a circle). Then, each triple has a unique circumcenter. Therefore, if we have w white points in general position, the number of red points we can get is C(w, 3). Therefore, to get 100 - w red points, we need C(w, 3) >= 100 - w, which as before gives w >=10. Therefore, if we can place 10 white points in general position, we can get 120 red points, which is more than enough. But in our problem, we need only 100 - w red points. If w=10, then 100 -10=90 red points. But C(10,3)=120, so we can have 120 red points. Therefore, with 10 white points, we can have up to 120 red points. Since we need only 90, it's feasible. However, in reality, arranging 10 white points such that each triple has a unique circumcenter, and all those circumcenters are distinct and not overlapping with white points. But this is possible if the white points are in general position.However, in the problem, the total number of points is 100. If we have 10 white points, then we have 90 red points. But according to the above, those 90 red points would be the circumcenters of the triples of white points. However, in reality, some circumcenters might coincide or overlap with white points. But if the white points are in general position, their triples' circumcenters are all distinct and not among the white points. Therefore, this seems possible. Therefore, in theory, 10 white points might suffice.But wait, the problem states that each red point must be the center of a circle passing through at least three white points. So, each red point is precisely the circumcenter of some three white points. If we have 10 white points in general position, we can get 120 circumcenters. Therefore, we can choose 90 of them as our red points. Therefore, 10 white points would suffice. However, this seems too good, and the answer is probably higher because of some constraints I'm missing.Wait, but in reality, when you choose 10 white points in general position, the number of distinct circumcenters might be less than C(10,3)=120 because different triples can have the same circumcenter. For example, four white points forming a rectangle would have the same circumcenter for the four triples that form the diagonals. So, in that case, multiple triples share the same circumcenter. Therefore, the number of distinct red points (circumcenters) can be less than 120. Therefore, to ensure that we have 100 - w red points, we need to make sure that the number of distinct circumcenters is at least 100 - w. Therefore, arranging white points such that many triples have unique circumcenters is non-trivial.Alternatively, perhaps we can use a grid or some symmetric configuration where multiple triples share the same circumcenter, but still allowing enough distinct circumcenters. However, this complicates the matter.Alternatively, maybe a lower bound argument. If we need each red point to have three white points, and each white point can be part of multiple red points' circles, then perhaps using double counting. Let me denote:Let R be the number of red points, which is 100 - w.Let W be the number of white points, which is w.Each red point is associated with at least three white points. Let’s count the total number of incidences between red points and white points. Each incidence is a white point lying on a circle centered at a red point. Each red point has at least three incidences, so total incidences >= 3R.On the other hand, each white point can be associated with multiple red points. Let’s denote that each white point is part of k_i incidences. Then, the total number of incidences is sum_{i=1}^W k_i.Therefore, sum_{i=1}^W k_i >= 3R.But each incidence corresponds to a white point lying on a circle around a red point. For a white point to lie on a circle around a red point, it must be at a certain distance from it. In the plane, the number of red points that a single white point can be associated with is limited by how many distinct distances it can have. However, in theory, a single white point can be at any distance from multiple red points. Therefore, there is no inherent upper limit on k_i other than the number of red points. However, for each red point associated with a white point, the red point must have two other white points on its circle.Therefore, if a white point is part of k_i red points, each of those red points must have two other distinct white points. Therefore, the number of other white points needed is at least 2k_i. But since white points are shared among red points, this might not directly translate to a bound. However, if a white point is used by k_i red points, each requiring two other white points, the total number of white points needed would be at least 1 + 2k_i. But since these other white points can be shared among different red points, this isn't straightforward.Alternatively, think of it as a hypergraph where each hyperedge (red point) connects three white points. We have R = 100 - w hyperedges, each of size 3, and we want the minimal number of white points. The minimal number of vertices in a 3-uniform hypergraph with R hyperedges is such that the number of hyperedges is <= C(w, 3). Therefore, as before, C(w, 3) >= 100 - w, leading to w >=10. But this is without considering geometric constraints. If the hypergraph is realizable in the plane with the additional condition that each hyperedge's three vertices lie on a circle with the hyperedge's center (the red point), then perhaps the minimal w is higher.But maybe the answer is indeed 10, but I have to check.Wait, if we take 10 white points in general position (no three on a line, no four on a circle), then each triple of white points will have a unique circumcenter. Therefore, the number of red points would be C(10, 3) = 120. Since we need only 100 -10 = 90 red points, we can select 90 of these 120 circumcenters as our red points. The remaining 30 circumcenters would not be used as red points, but since the problem allows for any coloring where red points satisfy the condition, this is acceptable. Therefore, with 10 white points, it's possible to have 90 red points, each being the circumcenter of a unique triple of white points. Therefore, the minimal number of white points is 10.However, the problem states that there are 100 points in total. If we have 10 white points and 90 red points, that sums to 100 points. However, the red points are the circumcenters of the triples of white points. But in reality, the circumcenters (red points) might coincide with white points or other red points.Wait, if the white points are in general position, no three are colinear and no four on a circle, then the circumcenters of triples of white points will not coincide with any white points. Because if a circumcenter of three white points were one of the white points, then that white point would be equidistant to the other two, meaning the three points lie on a circle centered at the white point. But since no four points are on a circle, this is not possible unless the three points are on a circle with the fourth point as the center, which is prohibited by the general position. Therefore, the circumcenters (red points) are distinct from the white points.However, different triples of white points might result in the same circumcenter. For example, consider four white points forming a rectangle. The circumcenter of each of the four triples formed by leaving out one vertex is the center of the rectangle. Therefore, four triples share the same circumcenter. Therefore, in such a case, the number of distinct red points would be less than C(4,3)=4. However, in our case, if we arrange white points in general position without such symmetries, we can maximize the number of distinct circumcenters.Therefore, if we place 10 white points in general position (no three colinear, no four concyclic), then each triple will have a unique circumcenter, none of which coincide with white points. Therefore, we can have 120 distinct red points. Since we only need 90, we can choose any 90 of them. Therefore, 10 white points suffice.But the problem says "colored using two colors, red and white". So, the 100 points consist of some red and some white. The red points are the circumcenters, and white points are the original 10. Therefore, as long as the 90 red points (circumcenters) are distinct from the 10 white points, this is feasible.But wait, in the plane, given 10 arbitrary white points in general position, the circumcenters of their triples could potentially overlap with each other or with other white points. However, if they are in general position, as defined (no three colinear, no four concyclic), then the circumcenters won't coincide with white points, but different triples might still have the same circumcenter. For instance, two different triples might, by coincidence, have the same circumcenter even if the white points are in general position. However, in general position, this is unlikely. But to guarantee that all circumcenters are unique, we might need a stronger condition.However, maybe it's possible to carefully place the white points such that all circumcenters are distinct. For example, choosing points with irrational coordinates or some other method to avoid symmetry. This might be possible, but it's non-trivial. However, the problem doesn't require an explicit configuration, just the minimal number. So, if we can argue that 10 white points are sufficient in theory (assuming we can arrange them such that all triples have distinct circumcenters not overlapping with white points), then 10 would be the answer. However, the problem might expect a higher number due to practical geometric constraints.Alternatively, perhaps the answer is 6. Wait, let me think differently. If each white point can be used by multiple red points, and each red point needs three white points, maybe arranging white points in a grid or something where each red point uses three white points with different distances.Wait, let's think of an example. Suppose we have four white points arranged in a square. Then, the center of the square is a red point, with the four white points on its circle. But each red point needs only three white points. So, the center of the square can be a red point using any three of the four white points. However, each red point needs to be the center of a circle passing through three white points. So, the center of the square is one such red point. But if we have other red points, they would need their own circles.Alternatively, if we have three white points forming an equilateral triangle. The circumcenter is the centroid. If we place another white point such that it forms another equilateral triangle with two of the original points, but that would require symmetry.Alternatively, consider white points arranged as vertices of a regular polygon. Each three consecutive vertices define a circle, but the centers would be the same. Hmm, not helpful.Wait, going back. If we need 100 - w red points, each connected to three white points, and white points can be shared. The key is to find the minimal w such that 3*(100 - w) <= number of incidences. But each white point can participate in multiple incidences.In the purely combinatorial case, the lower bound is w >= 10. But with geometric constraints, perhaps w must be higher. For example, in geometry, each white point can be used by at most how many red points? If a white point is used by k red points, each of those red points must have two other distinct white points. Each red point is defined by three white points. Therefore, if a white point is in k triples, each triple must include two other white points. The number of pairs of other white points is C(w -1, 2). Therefore, the number of triples that a single white point can be part of is at most C(w -1, 2). However, in reality, each pair can be used by only one red point, since a pair plus the white point defines a unique red point (the circumcenter). However, if two different red points share the same pair of white points with the original white point, then their centers must be the same, which is impossible. Therefore, each pair of white points can be used with the original white point only once. Therefore, the number of red points that a single white point can be part of is at most C(w -1, 2). But since we need each red point to have three white points, the total number of red points is limited by the combinations.But this seems similar to the earlier reasoning. The total number of incidences is 3*(100 - w) <= w*C(w -1, 2). Wait, that inequality might not hold. Let me think again.Each white point can be paired with C(w -1, 2) pairs of other white points. Each such pair defines a unique red point. Therefore, each white point can be part of at most C(w -1, 2) red points. Therefore, the total number of red points is at most sum_{i=1}^w C(w -1, 2) = w*C(w -1, 2). However, each red point is counted three times in this sum, once for each white point it contains. Therefore, the actual upper bound is:Total red points <= w*C(w -1, 2) /3Therefore,100 - w <= w*C(w -1, 2)/3Compute C(w -1, 2) = (w -1)(w -2)/2Therefore,100 - w <= w*(w -1)(w -2)/6Multiply both sides by 6:600 - 6w <= w(w -1)(w -2)This is the same inequality as before:w^3 - 3w^2 + 8w - 600 >=0Which we saw holds when w >=10. Therefore, this leads us back to the same lower bound of w=10. Therefore, even with the geometric constraint, the lower bound is still 10. Therefore, the minimal number of white points is 10.But I need to verify if this is achievable. As mentioned earlier, if we place 10 white points in general position, such that no three are colinear and no four are concyclic, then each triple of white points has a unique circumcenter, distinct from all other triples and not overlapping with any white points. Therefore, we can have C(10, 3)=120 distinct red points. Since we need only 90, this is feasible. Therefore, 10 white points are sufficient.However, I have a lingering doubt. Because arranging 10 points in general position such that no four are concyclic is possible, but ensuring that all C(10,3)=120 circumcenters are distinct might be challenging. However, in theory, if we place the white points generically, this should hold. For example, choosing points with algebraically independent coordinates would ensure that no four are concyclic and no three are colinear, and all circumcenters are distinct. Therefore, in the plane, such a configuration exists, which means that 10 white points are indeed sufficient.Therefore, the minimal number of white points is 10.But wait, the problem states "the least possible number of white points". If 10 is achievable, then that's the answer. However, I recall that in some combinatorial geometry problems, the minimal number is higher due to some constraints. Let me check with smaller numbers.Suppose we try w=6. Then C(6,3)=20. So, we could have 20 red points. But we need 100 -6=94 red points. Therefore, w=6 is insufficient.w=7: C(7,3)=35. 100-7=93. Still insufficient.w=8: C(8,3)=56. 100-8=92. Still insufficient.w=9: C(9,3)=84. 100 -9=91. Insufficient.w=10: C(10,3)=120. 100 -10=90. Sufficient.Therefore, according to this calculation, w=10 is the minimal number.However, maybe there's a smarter way to arrange white points such that each white point is used in more triples. For instance, if white points are arranged such that multiple triples share the same two white points, thus reusing those two with a third different one. But each red point needs three white points. If two white points are used with multiple thirds, then each of those thirds would define a different red point. However, the two white points plus each third must lie on a circle centered at the respective red point.But the problem is that each red point must be the center, so the two white points would have to be equidistant from the red point, along with the third white point. Therefore, if two white points are fixed, the red points that use them must lie on the perpendicular bisector of the segment joining those two white points. Additionally, the third white point must be equidistant from the red point as the two white points.Therefore, for two white points A and B, all red points using A and B must lie on the perpendicular bisector of AB. Moreover, for each such red point R on the perpendicular bisector, there must be a third white point C such that RC = RA = RB.Therefore, for each red point R on the perpendicular bisector of AB, there must be a white point C such that C lies on the circle centered at R with radius RA. If we have multiple red points on the perpendicular bisector, each requires a distinct white point C lying on their respective circles.However, this would mean that for each red point on the perpendicular bisector, we need a distinct white point C. Therefore, this approach doesn't save white points; in fact, it requires more white points.Alternatively, if multiple red points on the perpendicular bisector can share the same third white point C, but then C must lie on multiple circles centered on the bisector, each with different radii. The intersection of such circles would be at points equidistant to A and B, so C must lie on the perpendicular bisector as well. If C is on the perpendicular bisector, then AC = BC, so all three points A, B, C are on a line, which would mean that the three points are colinear, and hence their circumcircle is undefined or a line, which isn't a valid circle. Therefore, this is not possible.Therefore, this approach doesn't help. Thus, going back, the initial calculation of w=10 seems to be the minimal number, given that C(10,3)=120 >=90.Therefore, the least possible number of white points is 10.But wait, let me confirm once more. If we have 10 white points, we can have 120 red points as their triples' circumcenters. Since we need only 90 red points, we just need 90 of those 120. The other 30 circumcenters would just not be colored red. Therefore, yes, this works. Therefore, the minimal number is 10.However, the problem says "each red point is the centre of a circle passing through at least three white points". It doesn't require the red points to be only those circumcenters. So, perhaps some red points could be centers of circles passing through more than three white points, thereby reducing the total number of required triples. But since we're trying to minimize the number of white points, having a red point with more white points on its circle doesn't help; we still need each red point to have at least three, so having more doesn't reduce the required number of white points.Alternatively, if multiple red points share the same three white points, but as established earlier, this would require the red points to coincide, which is impossible. Therefore, each red point must have a distinct set of three white points. Therefore, the number of required triples is equal to the number of red points. Therefore, the formula C(w,3) >= R, where R=100 -w, leading to w=10.Therefore, the minimal number is 10.But I recall that in some similar problems, the answer is 6. Wait, let me think of a different approach. Maybe using inversion.Suppose we use an inversion transformation. Inversion might complicate things, but perhaps we can model the problem as a dual graph.Alternatively, think of each red point having three white points. If each white point is used by t red points, then total red points R = (number of white points * t)/3. Since each red point uses three white points, and each white point is used by t red points, total incidences are 3R = w*t => R = (w*t)/3.We need R = 100 - w. So:(w*t)/3 = 100 - w => w*t = 300 - 3w => t = (300 - 3w)/w = 300/w -3.To minimize w, we need to maximize t. Since t must be an integer, we can write:t = 300/w -3.To have t >=1 (since each white point must be used by at least one red point, but actually, since each red point needs three white points, each white point must be used by at least one red point unless there are redundant white points, which we don't have in minimal configuration).Therefore, t >=1.So:300/w -3 >=1 => 300/w >=4 => w <=75.But we are to minimize w, so this doesn't help directly. However, the maximum possible t is when w is minimized. For example, if w=10, t=(300/10)-3=27. So each white point is used by 27 red points. Is this possible?But in geometry, each white point can be part of multiple circles (red points). The number of red points that a single white point can be part of is limited by the number of circles it can lie on, which is potentially infinite, but in our case, each circle is centered at a red point and must include two other white points.Therefore, for a white point W, each red point that uses W must have two other white points U and V such that W, U, V are concyclic with center at the red point. The number of such red points is equal to the number of pairs (U, V) of other white points such that W, U, V are concyclic with the center being a red point.But in the plane, given two other white points U and V, there is at most one red point (the circumcenter of W, U, V) such that W, U, V lie on a circle centered at that red point. Therefore, for each pair (U, V), there is at most one red point that can use the triple (W, U, V). Therefore, the number of red points that can use W is at most C(w -1, 2). Therefore, t <= C(w -1, 2). For w=10, t <= C(9,2)=36. Since we needed t=27, which is less than 36, it's theoretically possible. Therefore, each white point can be part of up to 36 red points, but we only need 27. Therefore, it's feasible.However, arranging this in the plane is non-trivial. Each white point W needs to be paired with 27 pairs of other white points, each pair defining a red point (circumcenter of W, U, V). To have 27 such pairs per white point without overlapping red points requires that all these circumcenters are distinct. This would require that the triples (W, U, V) for each white point W have distinct circumcenters, which is possible if the white points are in general position.Therefore, in conclusion, the minimal number of white points is 10. Therefore, the answer is boxed{10}.Final Answerboxed{10}</think>To determine the least possible number of white points, we need to ensure that each red point (center) has at least three white points on its circle. The key steps are as follows:1. Combinatorial Bound: We need to find the minimum number ( w ) of white points such that the number of combinations of these points taken three at a time is at least the number of red points ( 100 - w ). This leads to the inequality: [ binom{w}{3} geq 100 - w ] Solving this inequality, we find that ( w geq 10 ).2. Geometric Consideration: We need to verify if 10 white points can be arranged such that each red point is the circumcenter of three white points. If the white points are in general position (no three collinear, no four concyclic), each triple of white points will have a unique circumcenter. This ensures that we can have up to ( binom{10}{3} = 120 ) distinct red points, which is more than the required 90 red points (since ( 100 - 10 = 90 )).3. Feasibility: By arranging 10 white points in general position, we can ensure that each red point (circumcenter of a triple of white points) is distinct and does not coincide with any white point. This configuration satisfies the problem's requirements.Thus, the least possible number of white points is (boxed{10}).,<think>Okay, let's see. The problem is about coloring 100 points in the plane with two colors, red and white. The condition is that each red point must be the center of a circle that passes through at least three white points. We need to find the least possible number of white points. Hmm, interesting.So, we want to minimize the number of white points. That means we should maximize the number of red points. Since there are 100 points in total, the more red points we have, the fewer white points we need. But each red point needs to be the center of a circle passing through at least three white points. So each red point must have three white points that lie on a circle centered at that red point.Wait, but how does that work? If a red point is the center of a circle passing through three white points, those three white points must be equidistant from the red point. So, each red point must have three white points at the same distance from it. Hmm, but in the plane, three points lying on a circle centered at the red point means they are all at the same radius from the red point.But wait, three non-collinear points lie on a unique circle. But here, the circle is specifically centered at the red point. So, for each red point, there must be three white points lying on some circle with the red point as the center. That doesn't necessarily require the white points to be equidistant from the red point unless they're all on the same circle. Wait, no. Each white point could be on a different circle, but each red point must have at least three white points lying on a single circle centered at that red point. Wait, the problem says "a circle passing through at least three white points". So, for each red point, there exists at least one circle (could be more) centered at that red point which passes through three or more white points.So, each red point needs at least three white points lying on a circle centered at it. Therefore, each red point must have three white points lying on a common circle centered at itself.Now, to minimize the number of white points, we want each white point to be used by as many red points as possible. That is, a single white point can lie on circles centered at multiple red points. So, if a white point is on a circle around multiple red points, that would allow us to reuse that white point for multiple red points. Therefore, the strategy is to arrange white points in such a way that each white point is on the circles of as many red points as possible.So, the key idea is to share white points among multiple red points. The more red points a single white point can serve, the fewer white points we need. Therefore, we need to figure out the maximum number of red points that can share a single white point on their respective circles.But each red point needs three white points. So, if we can arrange three white points such that each of them lies on circles centered at multiple red points, then those three white points can satisfy multiple red points.Wait, but how exactly can a single white point lie on circles centered at multiple red points? For a white point to lie on a circle centered at a red point, the distance between them must be equal to the radius of that circle. So, if a white point is used by multiple red points, each of those red points must be at the same distance from the white point. Therefore, all those red points must lie on a circle centered at the white point with radius equal to that distance.Therefore, if we have a white point, we can have multiple red points lying on a circle around it, each at radius r from the white point. Then, for each such red point, the white point is on a circle (of radius r) centered at the red point. However, each red point needs three such white points. So, if we have three white points, each with a circle of red points around them, then a red point can be at the intersection of three such circles (each centered at a different white point). Wait, this is getting a bit complicated. Maybe it's better to think in terms of configurations where multiple red points share the same set of three white points.Suppose we have three white points forming a triangle. Then, the circumcircle of this triangle is centered at the circumcenter. If multiple red points are at the circumcenter, then each red point would have those three white points on their circle. Wait, but the circumcenter is a single point. If multiple red points are all at the same circumcenter, then they are all the same point, which is not possible since all points are distinct. So that's not helpful.Alternatively, maybe arrange red points such that each red point has the same three white points on its circle. But for that, the three white points must lie on a circle around each red point. So, if all red points are arranged in some way that they all share the same three white points on their respective circles. But how can three white points lie on circles centered at multiple red points? Each white point would have to be equidistant from all those red points. That would mean that all red points lie on the perpendicular bisector of the line segment joining two white points. Wait, but three white points would require the red points to lie on the intersection of multiple perpendicular bisectors.For instance, take three white points W1, W2, W3. The set of points equidistant from W1 and W2 is the perpendicular bisector of W1W2. Similarly, the set of points equidistant from W1 and W3 is the perpendicular bisector of W1W3. The intersection of these two bisectors is the circumcenter of triangle W1W2W3. So, only the circumcenter is equidistant to all three white points. Therefore, only a single red point can be equidistant to all three white points. Therefore, only one red point can have those three white points lying on a circle centered at itself. Therefore, if we want multiple red points to share the same three white points, they cannot all be equidistant to them. Therefore, each red point must have three white points lying on some circle centered at itself, but not necessarily the same circle for different red points.Wait, but if the three white points are not the same for different red points, then each red point would need its own distinct set of three white points. Which would mean that the number of white points would be at least 3 times the number of red points. But since we have 100 points total, if we have R red points and W white points, R + W = 100. If each red point needs 3 white points, then W >= 3R. But then W >= 3(100 - W), which implies 4W >= 300, so W >= 75. But is this tight? Because maybe some white points can be shared among multiple red points.But earlier reasoning suggests that each white point can be used by multiple red points only if those red points lie on a circle centered at the white point. So, if a white point is used by multiple red points, each of those red points must lie on a circle of radius r around the white point. Then, for each red point on that circle, the white point is at distance r from them, so the white point lies on a circle of radius r around each of those red points. But each red point needs two more white points on their circle.So, suppose we have a white point W1. If we place multiple red points R1, R2, ..., Rk on a circle of radius r around W1. Then, each of these red points R_i has W1 on their circle of radius r. But each R_i needs two more white points on their circle. If we can arrange two more white points W2 and W3 such that each R_i has W2 and W3 on their circle as well, then each R_i would have three white points: W1, W2, W3. But for this to happen, W2 and W3 must also lie on the circle of radius r around each R_i. Which would mean that each R_i is at distance r from W1, W2, and W3. Therefore, all the red points R_i would have to be at the intersection of three circles: the circle of radius r around W1, the circle of radius r around W2, and the circle of radius r around W3. The intersection of these three circles would be points equidistant to W1, W2, and W3. If W1, W2, W3 are not collinear, their common intersection is at most two points (the circumcenters). But if they are arranged in such a way that the three circles intersect at multiple points, but generally, three circles can intersect at up to six points, but if all three circles have the same radius, maybe more?Wait, suppose we have three white points W1, W2, W3 arranged in an equilateral triangle. Then, the circles of radius r around each of them will intersect at six points: each pair of circles intersects at two points, so three pairs give six points. But if we set the side length of the equilateral triangle appropriately relative to the radius r, perhaps those intersection points can be arranged so that each red point is at the intersection of all three circles. Wait, but in reality, three circles with centers at the vertices of an equilateral triangle and all with the same radius will intersect at two points each, but those two points are the ones above and below the triangle. So, if the radius is equal to the side length, then each pair of circles intersects at two points, but all three circles intersect at two common points. So, those two points are the ones forming the two equilateral triangles (the original and the reflected one). Therefore, in this case, there are only two points that are common to all three circles. Therefore, we can only have two red points that are equidistant to all three white points. Therefore, with three white points arranged in an equilateral triangle, we can have up to two red points that are equidistant to all three. But each red point beyond that would need to be on a different configuration.Alternatively, maybe if we have more white points, we can have more red points sharing different triplets of white points.But this seems complicated. Let's think differently. Let's try to find the minimum number of white points W such that each red point (of which there are 100 - W) has three white points lying on a circle centered at it. To minimize W, we need to maximize the number of red points each white point can be associated with.Each white point can be on multiple circles centered at different red points. Each such association corresponds to a red point that has that white point on its circle. For each red point, we need three such associations. So, if we let E be the number of such associations (i.e., the total number of times a white point is used by a red point), then E >= 3*(100 - W). Each white point can contribute to multiple Es, but how many?Each white point can be associated with multiple red points, but each association requires that the white point is at a certain distance from the red point. However, a single white point can be at that specific distance from multiple red points. For example, if multiple red points lie on a circle centered at the white point with radius r, then the white point is on each of their circles (of radius r). Therefore, the maximum number of red points that can share a single white point is unbounded, in theory, because you can have infinitely many red points on a circle around the white point. However, in our case, we have a finite number of points, but we need to arrange them in the plane without overlapping.But even so, to maximize the number of red points that can share a single white point, we can place as many red points as possible on a circle centered at the white point. The number of red points on that circle is limited by the fact that each red point must also have two other white points on their circle. So, if we have a white point W1 with k red points around it on a circle of radius r, then each of those k red points needs two more white points on their respective circles. If those two white points are shared among multiple red points, then perhaps we can minimize the total number of white points.This seems like a hypergraph problem where each red point is a hyperedge connecting three white points. We need a 3-uniform hypergraph covering all red points with as few white points as possible. The minimal number of white points would correspond to the minimal vertex cover of a 3-uniform hypergraph with 100 - W hyperedges. But hypergraph vertex cover is a dual problem; maybe it's better to think in terms of hyperedges covering vertices. Wait, no. Wait, each hyperedge is a red point, which connects three white points. So, we have 100 - W hyperedges (red points) each connecting three white points. We need to find the minimal number of white points such that each hyperedge is connected to three distinct white points, and white points can be shared among multiple hyperedges.In hypergraph terms, this is the problem of finding the minimum number of vertices such that there exists a 3-uniform hypergraph with 100 - W hyperedges. But the problem allows hyperedges (red points) to share vertices (white points). The minimal number of white points would be the minimal number of vertices in a 3-uniform hypergraph with 100 - W hyperedges. However, in hypergraph theory, the lower bound for the number of vertices given the number of hyperedges is not straightforward. But perhaps we can use a combinatorial argument.In our case, each white point can be part of multiple hyperedges (red points). Each hyperedge requires three distinct white points. To minimize the number of white points, we need to maximize the number of hyperedges each white point is part of. Let’s denote by d the average number of hyperedges (red points) each white point is part of. Then, the total number of hyperedges is (W * d)/3, since each hyperedge has three white points. Therefore, (W * d)/3 >= 100 - W. To minimize W, we need to maximize d. The maximum possible d for each white point is limited by the number of hyperedges it can be part of. However, in the plane, there might be geometric constraints that prevent a white point from being part of too many hyperedges (i.e., being on too many circles around red points). But if we ignore geometry for a moment and just consider the combinatorial problem, the minimal W would be when d is as large as possible.In the purely combinatorial case, the maximum d for each white point is 100 - W (if a white point is part of every hyperedge), but since each hyperedge needs three distinct white points, each white point can be in at most (100 - W) choose 2 hyperedges, but this is not straightforward. Alternatively, since each hyperedge is a triplet of white points, the number of hyperedges is at most C(W, 3). But we need 100 - W hyperedges. So, 100 - W <= C(W, 3). Solving 100 - W <= W(W - 1)(W - 2)/6. Let's see for W=7: 7*6*5/6=35. 100 -7=93>35. Not enough. W=8: 8*7*6/6=56. 100 -8=92>56. Still not enough. W=9: 9*8*7/6=84. 100 -9=91>84. W=10: 10*9*8/6=120. 100 -10=90<=120. So, if we have W=10, then combinatorially, it's possible to have up to 120 hyperedges. Since we need only 90, it's possible. But this is purely combinatorial, not considering the geometric constraints. However, in reality, arranging points in the plane might not allow such a high number of hyperedges due to overlapping circles and distinctness of points. Therefore, the combinatorial lower bound might not be achievable geometrically.But perhaps we can get close. If we can arrange white points such that each triplet of them forms a hyperedge (a red point), but we need each red point to be the center of a circle passing through three white points. Wait, but each hyperedge in the combinatorial sense is a red point connected to three white points, but geometrically, the red point must be the circumcenter of the three white points. However, three non-collinear white points determine a unique circumcenter. Therefore, each triplet of white points can correspond to at most one red point (the circumcenter). Therefore, in the geometric case, the number of red points is limited by the number of distinct circumcenters formed by the white points. But if multiple triplets of white points have the same circumcenter, then those can correspond to multiple red points at the same location, but the problem states that all points are distinct. Therefore, each red point must be a distinct point. Therefore, each red point must correspond to a unique circumcenter of some three white points. Therefore, in the geometric case, the number of red points is at most equal to the number of distinct circumcenters generated by the white points.But the number of circumcenters generated by W white points is at most C(W, 3), since each triplet defines a circumcenter, but many triplets will share the same circumcenter. For example, if multiple triplets are concyclic, they share the same circumcenter. However, if we arrange the white points in general position (no three concyclic), then each triplet would have a unique circumcenter. But arranging W points in general position such that no four are concyclic is possible, but even so, the number of circumcenters would be C(W, 3), which is large.But in our problem, we need the red points to be distinct. Therefore, if we can arrange the white points such that each red point is the circumcenter of a unique triplet of white points, and all these circumcenters are distinct, then the number of red points can be up to C(W, 3). But in practice, this is not feasible because circumcenters would start overlapping as the number of white points increases.But let's think again. To maximize the number of red points (i.e., circumcenters) with as few white points as possible, we need each white point to be part of as many triplets as possible, each producing a distinct circumcenter. However, each time a white point is part of a triplet, the circumcenter is determined by that triplet. If we can arrange white points such that different triplets produce different circumcenters, then each triplet contributes a unique red point.But how many red points can we get from W white points? If all C(W, 3) circumcenters are distinct, then we can have C(W, 3) red points. But we need only 100 - W red points. Therefore, if C(W, 3) >= 100 - W, then it's possible. Let's check:For W=7: C(7,3)=35. 100 -7=93. 35 <93. Not enough.W=8: C(8,3)=56. 100 -8=92. Still not enough.W=9: C(9,3)=84. 100 -9=91. 84 <91.W=10: C(10,3)=120. 100 -10=90. 120 >=90. So, with W=10 white points, if all C(10,3)=120 triplets have distinct circumcenters, then we can have up to 120 red points. Since we need only 90 red points (since 100 -10=90), it's possible. But arranging 10 white points such that all their triplets have distinct circumcenters is non-trivial, but perhaps possible if the white points are in general position.However, the problem doesn't state that the points have to be in general position or any specific configuration, just colored red or white with the given condition. Therefore, theoretically, if we can arrange 10 white points such that every triplet has a unique circumcenter, and place each circumcenter as a red point, then we can have 120 red points. But we need only 90. So, even if some triplets share the same circumcenter, as long as we have 90 distinct circumcenters, it's possible. Therefore, with W=10 white points, the maximum number of red points is C(10,3)=120, which is more than 90, so 10 white points might suffice. But this is purely theoretical. The problem is in the plane, so we have to consider geometric realizability.But the question is asking for the least possible number of white points. So, according to this combinatorial argument, W=10 could be possible. But maybe there's a lower bound. Let me check with W=9: C(9,3)=84. 100 -9=91. 84 <91, so insufficient. So W=10 is the minimal combinatorial number. But is this achievable geometrically?However, in reality, arranging 10 white points so that 90 of their triplets have distinct circumcenters might be challenging, but perhaps possible. But since the problem allows any configuration, as long as the conditions are met, maybe we can accept this. However, perhaps there's a smarter way.Wait, but let's think again. The problem states that each red point must be the center of a circle passing through at least three white points. It doesn't require that the three white points are distinct for each red point. Wait, no, the three white points must lie on the circle, so they have to be distinct points. So, each red point must have three distinct white points on its circle.But the same three white points can be used for multiple red points if those red points are all at the same circumradius from the three white points. Wait, but as before, multiple red points would have to be the same circumcenter, which would mean they are the same point, which is not allowed. Therefore, each red point must correspond to a unique circumcenter of some three white points. Therefore, if two different triplets of white points have the same circumcenter, then we can only have one red point there, not two. Therefore, in order to have multiple red points, they must correspond to different circumcenters. Therefore, the number of red points is equal to the number of distinct circumcenters formed by the white points.Therefore, to have R red points, we need at least R distinct circumcenters from the white points. Therefore, the number of distinct circumcenters must be at least R. Since R = 100 - W, we need the number of distinct circumcenters from W white points to be at least 100 - W.But how many distinct circumcenters can W white points generate? In the worst case, if every three white points determine a unique circumcenter, then it's C(W, 3). But in reality, some triplets might have the same circumcenter. For example, if four white points lie on a circle, then each triplet of them has the same circumcenter (the center of the circle). Therefore, by arranging white points on concentric circles or other configurations, we can have multiple triplets sharing the same circumcenter.However, to maximize the number of distinct circumcenters, we should arrange the white points such that no four are concyclic, so that each triplet has a unique circumcenter. If we can do that, then the number of circumcenters is C(W, 3). Therefore, if we have W white points in general position (no four concyclic), then the number of red points can be C(W, 3). So, to get 100 - W red points, we need C(W, 3) >= 100 - W. As before, with W=10, C(10, 3)=120 >=90=100 -10. So, theoretically, 10 white points can give up to 120 red points. Therefore, 10 white points would be sufficient. But the problem is asking for the minimal number. Since for W=9, C(9,3)=84 <91=100 -9. Therefore, W=10 is the minimal number according to this reasoning.But is this achievable? Can we place 10 white points in the plane such that no four are concyclic and all their circumcenters are distinct? I think that in general position, this is possible. For example, if we place the 10 white points such that no four lie on a circle, and no three are collinear, then each triplet will have a unique circumcenter. Since three non-collinear points define a unique circle, and with no four concyclic, each circle defined by three points won't contain a fourth. Therefore, all circumcenters will be distinct. Therefore, arranging 10 white points in such a general position would allow us to have C(10,3)=120 distinct circumcenters, which can serve as red points. Since we need only 90 red points, we can choose any 90 of these 120 circumcenters. Therefore, with 10 white points, we can have 90 red points, totaling 100 points. Therefore, the minimal number of white points is 10.But let me verify this logic. Each red point is the circumcenter of three white points. Each triplet of white points gives a unique red point (the circumcenter). Since we need 90 red points, we need 90 distinct triplets of white points. With 10 white points, there are 120 triplets. So, selecting 90 of them, each giving a distinct circumcenter, is possible. Therefore, 10 white points suffice. So, the minimal number is 10.However, another thought: each white point can be part of multiple triplets. But each triplet corresponds to a unique red point. However, the red points must be distinct. So, if we arrange the white points so that each triplet has a unique circumcenter, then all red points will be distinct. Therefore, as long as the white points are in general position (no four concyclic), then this is possible. Therefore, 10 white points should be sufficient.But wait, the problem states "each red point is the center of a circle passing through at least three white points". It doesn't require that the circle doesn't pass through more than three white points. So, if a red point is the center of a circle passing through four white points, that's also acceptable. However, in our previous reasoning, if four white points lie on a circle, then all triplets from those four white points will have the same circumcenter (the center of the circle). Therefore, we can have multiple red points at the same center only if they are the same point, which is not allowed. Therefore, to have distinct red points, we need each red point to be the circumcenter of a unique set of three white points. Therefore, even if four white points lie on a circle, the circumcenter is the same, so we can only have one red point there, not four. Therefore, to maximize the number of red points, it's better to have no four white points concyclic, so that each triplet gives a unique red point.Therefore, arranging the white points in general position (no three collinear, no four concyclic) allows each triplet to have a unique circumcenter, leading to the maximum number of red points. Therefore, with 10 white points, we can have 120 red points, but we only need 90. Therefore, 10 white points suffice.But let me check if W=9 is possible. For W=9, C(9,3)=84. So, 84 red points. But we need 100 -9=91 red points. 84 <91. Therefore, even if all triplets of 9 white points have unique circumcenters, we can only get 84 red points. But we need 91. Therefore, W=9 is insufficient. Therefore, the minimal W is 10.But wait, perhaps there's another way. If some white points are used in multiple triplets but with different radii. For example, a white point can be on circles of different radii around different red points. So, a single white point can be part of multiple circles around different red points, each with a different radius. In this case, the red points don't have to be the circumcenters of triplets including that white point. Wait, but the circle must be centered at the red point and pass through the white point. So, if a white point is on a circle around a red point, then the distance between them is the radius. So, different red points can have the same white point on their circles if they are at different distances from the white point.Therefore, perhaps a white point can be part of multiple circles around different red points with different radii. Therefore, a single white point can be used by multiple red points, as long as each red point has two other white points on their circle. So, in this case, the number of white points can be lower.Wait, let's clarify. Suppose we have a white point W1. Red points can be placed at various distances from W1. Each red point that includes W1 on its circle must have two other white points on the same circle. If those two other white points are different for different red points, then W1 can be shared among multiple red points. For example, red point R1 has W1, W2, W3 on its circle. Red point R2 has W1, W4, W5 on its circle. Red point R3 has W1, W6, W7 on its circle, etc. In this way, W1 is shared among multiple red points, each time paired with two new white points. Therefore, each time we use W1, we consume two additional white points. Therefore, the number of white points needed would be 1 + 2*(number of red points sharing W1). But if multiple red points share W1 with the same pair of white points, then those red points must be equidistant from W1, W2, W3, which would force them to be the same point, which is not allowed. Therefore, each red point that shares W1 must do so with a unique pair of other white points.Therefore, if a white point is used by k red points, it must be paired with 2k distinct white points. Therefore, the total number of white points needed would be at least 1 + 2k. However, these 2k white points can themselves be shared among other red points.This seems similar to a design problem where we need to cover red points with triples of white points, allowing overlaps. The minimal number of white points would be the solution to the set cover problem, but with the constraint that each red point is covered by a unique triple. However, set cover is NP-hard, but perhaps we can find a lower bound.Alternatively, consider that each white point can be part of multiple triples, but each triple requires three white points. Let’s model this as a hypergraph where each hyperedge connects three vertices (white points) and we need to cover 100 - W hyperedges (red points) with as few vertices as possible. The minimal number of vertices W is such that the number of hyperedges C(W, 3) is at least 100 - W. As before, this gives W=10.But perhaps in the geometric case, we can do better? Because in the hypergraph model, we ignore the geometric constraints, but in reality, the triples must correspond to circles centered at red points.Alternatively, think of each red point as requiring three white points such that the red point is equidistant to those three. So, each red point is the circumcenter of three white points. To have multiple red points, those must be circumcenters of different triplets. Therefore, the number of red points is limited by the number of distinct circumcenters, which, as before, is C(W, 3) if no four white points are concyclic.Therefore, the combinatorial lower bound is W=10, and this is achievable geometrically by placing the white points in general position. Therefore, the minimal number of white points is 10.But wait, let's verify with an example. Suppose we have W=10 white points in general position (no three collinear, no four concyclic). Then, each triplet of white points defines a unique circumcenter. These circumcenters are all distinct because no four white points are concyclic, so no two different triplets share the same circumcenter. Therefore, we can have C(10,3)=120 distinct circumcenters (red points). Since we need only 90 red points (since 100 -10=90), we can select 90 of these 120 circumcenters. Therefore, it's possible. Hence, 10 white points suffice.Alternatively, maybe we can use fewer white points by allowing some overlap in the circumcenters. But if we allow overlapping circumcenters, then those would correspond to the same red point, which is not allowed as all points must be distinct. Therefore, we cannot have overlapping circumcenters. Hence, each red point must be a distinct circumcenter of a unique triplet. Therefore, the combinatorial lower bound of W=10 is indeed the minimal number.Therefore, the least possible number of white points is 10.Final Answerboxed{10}</think>To determine the least possible number of white points, we need to ensure that each red point is the center of a circle passing through at least three white points. The key idea is to minimize the number of white points while maximizing the number of red points, given that each red point must be the circumcenter of three white points.1. Combinatorial Argument: Each red point requires three white points. To minimize the number of white points, we need to maximize the number of red points each white point can serve. This is analogous to a hypergraph problem where each red point (hyperedge) connects three white points (vertices).2. Geometric Realization: For each red point to be the circumcenter of three white points, the number of red points is limited by the number of distinct circumcenters formed by the white points. If white points are arranged in general position (no three collinear, no four concyclic), each triplet of white points will have a unique circumcenter.3. Calculation: We need to find the smallest ( W ) such that the number of distinct circumcenters (red points) is at least ( 100 - W ). This translates to finding ( W ) such that ( binom{W}{3} geq 100 - W ).4. Verification: - For ( W = 9 ), ( binom{9}{3} = 84 ), which is less than ( 100 - 9 = 91 ). - For ( W = 10 ), ( binom{10}{3} = 120 ), which is greater than ( 100 - 10 = 90 ).Thus, with ( W = 10 ) white points arranged in general position, we can form 120 distinct circumcenters (red points), which is more than the required 90 red points. Therefore, the least possible number of white points is (boxed{10}).]

🔑:1. Understanding the problem: We need to find the minimum number of white points such that each red point is the center of a circle passing through at least three white points. We are given a total of 100 points in the plane, colored either red or white.2. Choosing white points: Let's denote the number of white points as ( n ). We need to ensure that each red point is the center of a circle passing through at least three white points. 3. Counting red points: The number of ways to choose 3 points out of ( n ) white points is given by the binomial coefficient: [ binom{n}{3} = frac{n(n-1)(n-2)}{6} ] This represents the maximum number of red points, as each red point can be the center of a circle passing through a unique combination of three white points.4. Ensuring total points: We need the total number of points (red + white) to be at least 100. Therefore, we have: [ n + binom{n}{3} geq 100 ]5. Testing values of ( n ): - For ( n = 9 ): [ binom{9}{3} = frac{9 cdot 8 cdot 7}{6} = 84 ] Total points: [ 9 + 84 = 93 < 100 ] This is not sufficient. - For ( n = 10 ): [ binom{10}{3} = frac{10 cdot 9 cdot 8}{6} = 120 ] Total points: [ 10 + 120 = 130 > 100 ] This is sufficient.6. Conclusion: Since ( n = 10 ) white points is sufficient and ( n = 9 ) is not, the least possible number of white points is ( 10 ).[boxed{10}]

🧠:[<think>Okay, so I need to simplify the sum of cos squared terms from cos²α up to cos²nα, and similarly for the sine squared terms. Hmm, let me start by recalling some trigonometric identities that might help here. First, I remember that cos²θ can be written as (1 + cos2θ)/2, right? Similarly, sin²θ is (1 - cos2θ)/2. Maybe if I apply these identities to each term in the sum, I can simplify the expression. Let me try that for the cosine sum first.So, for each term cos²kα, where k goes from 1 to n, substituting the identity gives me (1 + cos2kα)/2. Therefore, the entire sum becomes the sum from k=1 to n of [1/2 + (cos2kα)/2]. This can be split into two separate sums: (1/2) times the sum from 1 to n of 1, plus (1/2) times the sum from 1 to n of cos2kα.The first part is straightforward. The sum of 1 added n times is just n, so that part becomes (1/2)*n = n/2. The second part is (1/2) times the sum of cos2kα from k=1 to n. So now, I need to compute the sum of cos2kα for k=1 to n. Hmm, how do I sum a series of cosines with arguments in arithmetic progression?I recall there's a formula for the sum of cosines of angles in arithmetic progression. Let me try to remember it. The formula is:Sum_{k=0}^{n-1} cos(a + kd) = [sin(n*d/2)/sin(d/2)] * cos(a + (n-1)d/2)Similarly for sine. But in our case, the sum starts at k=1 to n, and the angle is 2kα. So let's adjust variables accordingly. Let me set a = 2α (since when k=1, the term is cos2α), and the common difference d = 2α. So the sum Sum_{k=1}^{n} cos(2kα) can be rewritten as Sum_{k=1}^{n} cos(a + (k-1)d), where a = 2α and d = 2α. Wait, actually, if k starts at 1, then when k=1, we have cos(2α), which is cos(a + 0*d), so a = 2α, and each subsequent term increases by d = 2α. Therefore, the number of terms is n, starting from k=0 to k=n-1 in the standard formula. But here, our index starts at k=1, but in terms of the formula, that's equivalent to starting at k=0 with an initial term a = 2α and then adding 2α each time for n terms. Wait, maybe I need to adjust the formula correctly.Alternatively, maybe think of Sum_{k=1}^n cos(2kα) = Re[Sum_{k=1}^n e^{i2kα}]. Since cosine is the real part of the complex exponential. Then, the sum of exponentials is a geometric series. Let's try this approach.The sum Sum_{k=1}^n e^{i2kα} is a geometric series with first term e^{i2α} and ratio e^{i2α}. So, the sum is e^{i2α}*(1 - e^{i2nα}) / (1 - e^{i2α}), provided that e^{i2α} ≠ 1, which is true unless α is a multiple of π. Let's assume α is not a multiple of π for generality.Simplifying the numerator and denominator:Numerator: e^{i2α} - e^{i2(n+1)α}Denominator: 1 - e^{i2α}Multiply numerator and denominator by e^{-iα} to make it symmetric:Numerator: e^{iα} - e^{i(2n+1)α}Denominator: e^{-iα} - e^{iα} = -2i sinαWait, maybe a better approach is to factor out terms. Let's write the sum S = Sum_{k=1}^n e^{i2kα} = e^{i2α}(1 - e^{i2nα}) / (1 - e^{i2α})Then, to compute the real part, which is the sum of cosines. Let's compute this expression.First, let's write 1 - e^{i2α} = -2i sinα e^{iα}, using the identity 1 - e^{iθ} = -2i sin(θ/2) e^{iθ/2}. Wait, let's verify that.We know that 1 - e^{iθ} = e^{iθ/2}(e^{-iθ/2} - e^{iθ/2}) = e^{iθ/2}(-2i sin(θ/2)). Therefore, 1 - e^{iθ} = -2i sin(θ/2) e^{iθ/2}Similarly, the numerator: e^{i2α} - e^{i2(n+1)α} = e^{i(n+1)α} [e^{-i(n)α} - e^{i(n)α}] = e^{i(n+1)α} (-2i sin(nα))Wait, maybe not. Let's compute numerator and denominator step by step.Numerator: e^{i2α} - e^{i2(n+1)α} = e^{i2α}(1 - e^{i2nα})Denominator: 1 - e^{i2α} = -2i sinα e^{iα} (using θ = 2α, so sin(θ/2) = sinα, and e^{iθ/2} = e^{iα})Therefore, the sum S is [e^{i2α}(1 - e^{i2nα})] / (-2i sinα e^{iα}) = [e^{iα}(1 - e^{i2nα})] / (-2i sinα )Then, 1 - e^{i2nα} = -2i sin(nα) e^{i nα}, again using the same identity. So substituting that in:S = [e^{iα}*(-2i sin(nα) e^{i nα})] / (-2i sinα ) = [ -2i sin(nα) e^{i(n+1)α} ] / (-2i sinα )Simplify numerator and denominator:The -2i cancels out, leaving [ sin(nα) e^{i(n+1)α} ] / sinαTherefore, S = [ sin(nα) / sinα ] e^{i(n+1)α }Taking the real part of S gives the sum of cosines:Re(S) = [ sin(nα) / sinα ] cos((n+1)α )Therefore, Sum_{k=1}^n cos(2kα) = [ sin(nα) / sinα ] cos((n+1)α )Wait, let me check this formula. If I set n=1, then the left side is cos2α, and the right side is [sinα / sinα] cos(2α) = cos2α. That works. For n=2, left side is cos2α + cos4α. Let's compute the right side: [sin2α / sinα] cos3α = [2 cosα] cos3α. Using sin2α = 2 sinα cosα. Then, [2 cosα] cos3α. Is this equal to cos2α + cos4α?Using the identity cosA + cosB = 2 cos[(A+B)/2] cos[(A-B)/2]. Let's apply this to cos2α + cos4α. Then, A=2α, B=4α, so 2 cos3α cos(-α) = 2 cos3α cosα. Which is equal to 2 cosα cos3α. So yes, the right side gives 2 cosα cos3α, which equals cos2α + cos4α. So the formula holds for n=2. Therefore, the formula seems correct.Therefore, Sum_{k=1}^n cos2kα = [sin(nα)/sinα] cos(n+1)αTherefore, going back to the original sum of cos² terms:Sum_{k=1}^n cos²kα = n/2 + (1/2)*[sin(nα)/sinα] cos(n+1)αSimilarly, for the sine squared sum, we can do the same approach.Sum_{k=1}^n sin²kα = Sum_{k=1}^n [1 - cos2kα]/2 = n/2 - (1/2)*Sum_{k=1}^n cos2kα = n/2 - (1/2)*[sin(nα)/sinα] cos(n+1)αTherefore, both sums can be expressed in terms of n, α, and the trigonometric functions sin(nα)/sinα and cos(n+1)α.So, putting it all together, the simplified forms are:For the cosine squared sum:n/2 + [sin(nα) cos(n+1)α]/(2 sinα)And for the sine squared sum:n/2 - [sin(nα) cos(n+1)α]/(2 sinα)Alternatively, these can be written using trigonometric identities to simplify further. Let me check if the term [sin(nα) cos(n+1)α] can be simplified.Using the identity sinA cosB = [sin(A+B) + sin(A-B)]/2So, sin(nα) cos(n+1)α = [sin(nα + (n+1)α) + sin(nα - (n+1)α)]/2 = [sin((2n+1)α) + sin(-α)]/2 = [sin((2n+1)α) - sinα]/2Therefore, substituting back into the expressions:For the cosine sum:n/2 + [sin((2n+1)α) - sinα]/(4 sinα) = n/2 + [sin((2n+1)α)/(4 sinα) - sinα/(4 sinα)] = n/2 + sin((2n+1)α)/(4 sinα) - 1/4Similarly, combining constants:n/2 - 1/4 + sin((2n+1)α)/(4 sinα) = (2n - 1)/4 + sin((2n+1)α)/(4 sinα)But this may not necessarily be simpler. Alternatively, perhaps leaving it in terms of sin(nα) cos(n+1)α is better.Alternatively, let's check if there's another identity. Let's consider the denominator sinα and the numerator sin(nα) cos(n+1)α. Maybe expressing sin(nα) as sin((n+1)α - α). Let's try:sin(nα) = sin((n+1)α - α) = sin((n+1)α) cosα - cos((n+1)α) sinαTherefore, sin(nα) cos(n+1)α = [sin((n+1)α) cosα - cos((n+1)α) sinα] cos(n+1)α= sin((n+1)α) cosα cos(n+1)α - cos²(n+1)α sinαHmm, not sure if that helps. Maybe not necessary. Let's see if there's another approach.Alternatively, perhaps use the identity for sum_{k=1}^n cos2kα and then relate it to the original problem. Since we already derived that sum_{k=1}^n cos2kα = [sin(nα)/sinα] cos(n+1)α, which is already a simplified form. Therefore, substituting back into the expressions for the sums:Sum cos²kα = n/2 + (1/(2 sinα)) sin(nα) cos(n+1)αSum sin²kα = n/2 - (1/(2 sinα)) sin(nα) cos(n+1)αAlternatively, factoring out 1/(2 sinα), we can write:Sum cos²kα = [n sinα + sin(nα) cos(n+1)α]/(2 sinα)Similarly for sine squared.But perhaps we can find a more compact form. Let me check if the term sin(nα) cos(n+1)α can be combined with n sinα in some identity.Alternatively, let's recall that when we have expressions like sinA cosB, sometimes it's useful to express them as sums. As we did earlier, sin(nα) cos(n+1)α = [sin( (2n+1)α ) - sinα ] / 2. Therefore:Sum cos²kα = n/2 + [ sin((2n+1)α ) - sinα ] / (4 sinα )Similarly, simplifying the second term:[ sin((2n+1)α ) - sinα ] / (4 sinα ) = [ sin((2n+1)α ) / (4 sinα ) ] - [ sinα / (4 sinα ) ] = sin((2n+1)α )/(4 sinα ) - 1/4Therefore, the entire sum becomes:n/2 - 1/4 + sin((2n+1)α )/(4 sinα )Which can be written as (2n - 1)/4 + sin((2n+1)α )/(4 sinα )Similarly for the sine squared sum:Sum sin²kα = n/2 - [ sin((2n+1)α ) - sinα ] / (4 sinα ) = n/2 - sin((2n+1)α )/(4 sinα ) + 1/4 = (2n + 1)/4 - sin((2n+1)α )/(4 sinα )But I'm not sure if this is a more useful form. Perhaps the initial expression is better.Alternatively, another approach: notice that cos²kα + sin²kα = 1, so the sum of both series from k=1 to n is n. So, if we denote S_c = Sum cos²kα and S_s = Sum sin²kα, then S_c + S_s = n. Which is consistent with our earlier results, since S_c = n/2 + C and S_s = n/2 - C, so their sum is n.But the problem asks to simplify each sum individually, so we need to present both expressions.Another check: let's test with a small n. Let's take n=1. Then Sum cos²α = cos²α. According to the formula: n/2 + [sin(nα)/sinα] cos(n+1)α /2. So n=1: 1/2 + [sinα/sinα] cos2α /2 = 1/2 + (1 * cos2α)/2 = (1 + cos2α)/2 = cos²α, which is correct.Similarly, for the sine sum: n/2 - [sin(nα)/sinα] cos(n+1)α /2. For n=1: 1/2 - [sinα/sinα] cos2α /2 = (1 - cos2α)/2 = sin²α, correct.Another test with n=2. Let α be some angle, say α=30°, so π/6 radians.Compute Sum cos²kα for k=1,2:cos²(30°) + cos²(60°). cos(30°)=√3/2, so squared is 3/4. cos(60°)=1/2, squared is 1/4. Sum is 3/4 + 1/4 = 1. According to the formula:n=2, α=π/6.Formula: 2/2 + [sin(2*π/6)/sin(π/6)] * cos(3*π/6) /2 = 1 + [sin(π/3)/sin(π/6)] * cos(π/2)/2.sin(π/3)=√3/2, sin(π/6)=1/2, so their ratio is √3. cos(π/2)=0. Therefore, the second term is 0. So total sum is 1 + 0 = 1. Which matches the manual calculation. Good.For the sine sum: n=2, Sum sin²kα = sin²30° + sin²60° = (1/2)^2 + (√3/2)^2 = 1/4 + 3/4 = 1. According to the formula: 2/2 - [sin(2α)/sinα] cos3α /2 = 1 - [sin(π/3)/sin(π/6)]*cos(π/2)/2 = 1 - [√3/ (1/2)]*0/2 = 1 - 0 =1. Correct.Another test with n=3, α=π/4.Compute Sum cos²kπ/4 for k=1,2,3:cos²(π/4) + cos²(2π/4) + cos²(3π/4) = (√2/2)^2 + (0)^2 + (-√2/2)^2 = 0.5 + 0 + 0.5 = 1.Formula: n/2 + [sin(nα)/sinα] cos(n+1)α /2. Here n=3, α=π/4.n/2 = 3/2. sin(nα)=sin(3π/4)=√2/2. sinα=sin(π/4)=√2/2. cos(n+1)α=cos(4π/4)=cosπ= -1. Therefore:3/2 + [ (√2/2)/(√2/2) ] * (-1)/2 = 3/2 + [1 * (-1)/2] = 3/2 - 1/2 = 1. Correct.Similarly, the sine sum: Sum sin²kπ/4 for k=1,2,3:sin²(π/4) + sin²(2π/4) + sin²(3π/4) = 0.5 + 1 + 0.5 = 2.Formula: n/2 - [sin(nα)/sinα] cos(n+1)α /2 = 3/2 - [√2/2 / √2/2] * (-1)/2 = 3/2 - [1*(-1)/2] = 3/2 + 1/2 = 2. Correct.So the formula seems to hold up in these test cases. Therefore, the simplified expressions are:For the sum of cos²kα from k=1 to n:n/2 + [sin(nα) cos(n+1)α]/(2 sinα)And for the sum of sin²kα from k=1 to n:n/2 - [sin(nα) cos(n+1)α]/(2 sinα)Alternatively, since these expressions can be written as:Sum cos²kα = [n sinα + sin(nα) cos(n+1)α]/(2 sinα)Sum sin²kα = [n sinα - sin(nα) cos(n+1)α]/(2 sinα)But perhaps we can combine terms further. Let's see:Let’s consider the numerator for the cosine sum: n sinα + sin(nα) cos(n+1)α.Using the identity sinA cosB = [sin(A+B) + sin(A-B)]/2, so:sin(nα) cos(n+1)α = [sin(2n+1)α + sin(-α)]/2 = [sin((2n+1)α) - sinα]/2Therefore, the numerator becomes:n sinα + [sin((2n+1)α) - sinα]/2 = (2n sinα)/2 + [sin((2n+1)α) - sinα]/2 = [2n sinα + sin((2n+1)α) - sinα]/2 = [ (2n -1) sinα + sin((2n+1)α) ] / 2Thus, the sum of cos²kα is:[ (2n -1) sinα + sin((2n+1)α) ] / (4 sinα )Which can be split into:(2n -1)/4 + sin((2n+1)α)/(4 sinα )But I don't think this is necessarily simpler. Therefore, the original form is probably the most concise expression.Therefore, the final simplified forms are:For the cosine sum:frac{n}{2} + frac{sin(n alpha) cos((n + 1)alpha)}{2 sin alpha}And for the sine sum:frac{n}{2} - frac{sin(n alpha) cos((n + 1)alpha)}{2 sin alpha}Alternatively, since these expressions are related, but I think this is as simplified as they can get unless there's a further identity that can collapse the second term into something else, but I don't recall one. Alternatively, using product-to-sum formulas might lead to a different expression, but it's likely equivalent in complexity.Another approach: maybe express the sum in terms of sin((2n+1)α) or something similar. Let's explore that.We have:Sum_{k=1}^n cos2kα = [sin(nα) cos(n+1)α]/sinαBut earlier, we also found that this sum can be written as [sin((2n+1)α) - sinα]/(2 sinα)Therefore, substituting this into the cosine squared sum:Sum cos²kα = n/2 + [sin((2n+1)α) - sinα]/(4 sinα )Which can be written as:Sum cos²kα = (2n sinα + sin((2n+1)α) - sinα)/(4 sinα )= ( (2n -1) sinα + sin((2n+1)α ) ) / (4 sinα )But again, not sure if this is simpler. It might be useful depending on the context. Alternatively, this can be expressed using telescoping series or another method, but I think the initial expression is the most straightforward.Therefore, I think the answer is as derived above. Let me just write them in LaTeX.For the sum of cos²kα:cos^{2} alpha + cos^{2} 2alpha + ldots + cos^{2} nalpha = frac{n}{2} + frac{sin(n alpha) cos((n + 1)alpha)}{2 sin alpha}And for the sum of sin²kα:sin^{2} alpha + sin^{2} 2alpha + ldots + sin^{2} nalpha = frac{n}{2} - frac{sin(n alpha) cos((n + 1)alpha)}{2 sin alpha}Therefore, these are the simplified forms of the given sums.Final AnswerThe simplified forms are:For the cosine sum:boxed{frac{n}{2} + frac{sin(n alpha) cos((n + 1)alpha)}{2 sin alpha}}For the sine sum:boxed{frac{n}{2} - frac{sin(n alpha) cos((n + 1)alpha)}{2 sin alpha}}</think>To simplify the sums (cos^{2} alpha + cos^{2} 2alpha + ldots + cos^{2} nalpha) and (sin^{2} alpha + sin^{2} 2alpha + ldots + sin^{2} nalpha), we start by using the trigonometric identities (cos^{2} theta = frac{1 + cos 2theta}{2}) and (sin^{2} theta = frac{1 - cos 2theta}{2}).For the cosine sum:[sum_{k=1}^{n} cos^{2} kalpha = sum_{k=1}^{n} frac{1 + cos 2kalpha}{2} = frac{n}{2} + frac{1}{2} sum_{k=1}^{n} cos 2kalpha]For the sine sum:[sum_{k=1}^{n} sin^{2} kalpha = sum_{k=1}^{n} frac{1 - cos 2kalpha}{2} = frac{n}{2} - frac{1}{2} sum_{k=1}^{n} cos 2kalpha]Next, we need to evaluate the sum (sum_{k=1}^{n} cos 2kalpha). Using the formula for the sum of cosines in arithmetic progression, we find:[sum_{k=1}^{n} cos 2kalpha = frac{sin(nalpha) cos((n+1)alpha)}{sin alpha}]Substituting this result back into the expressions for the sums of cosine squared and sine squared terms, we get:For the cosine sum:[sum_{k=1}^{n} cos^{2} kalpha = frac{n}{2} + frac{1}{2} left( frac{sin(nalpha) cos((n+1)alpha)}{sin alpha} right) = frac{n}{2} + frac{sin(n alpha) cos((n + 1)alpha)}{2 sin alpha}]For the sine sum:[sum_{k=1}^{n} sin^{2} kalpha = frac{n}{2} - frac{1}{2} left( frac{sin(nalpha) cos((n+1)alpha)}{sin alpha} right) = frac{n}{2} - frac{sin(n alpha) cos((n + 1)alpha)}{2 sin alpha}] Final AnswerThe simplified forms are:For the cosine sum:[boxed{frac{n}{2} + frac{sin(n alpha) cos((n + 1)alpha)}{2 sin alpha}}]For the sine sum:[boxed{frac{n}{2} - frac{sin(n alpha) cos((n + 1)alpha)}{2 sin alpha}}],<think>Okay, so I need to find a way to simplify the sum of cos squared terms from α up to nα, and similarly for sin squared terms. Let me start by recalling some trigonometric identities that might help here. I remember that there's an identity for cos²θ and sin²θ that can convert them into expressions involving cos2θ. Specifically, cos²θ = (1 + cos2θ)/2 and sin²θ = (1 - cos2θ)/2. Maybe using these identities can simplify the summation because it would turn each squared term into a constant plus a cosine of a multiple angle. That might make the sum easier to handle.So, let's try applying this identity to both sums. Starting with the cosine squared sum:Sum_{k=1}^n cos²(kα) = Sum_{k=1}^n [ (1 + cos(2kα) ) / 2 ]Similarly, for the sine squared sum:Sum_{k=1}^n sin²(kα) = Sum_{k=1}^n [ (1 - cos(2kα) ) / 2 ]So, breaking each sum into two parts, the first part is just summing 1/2 n times, which would give n/2. The second part is (1/2) times the sum of cos(2kα) for cosine squared, and minus (1/2) times the sum of cos(2kα) for sine squared. Therefore, both sums can be written in terms of n/2 plus or minus half of the sum of cos(2kα) from k=1 to n.So, the key now is to compute Sum_{k=1}^n cos(2kα). If I can find a closed-form expression for this sum, then I can plug it back into the expressions for the cosine squared and sine squared sums.I remember that there's a formula for the sum of cosines of an arithmetic sequence of angles. The general formula for Sum_{k=0}^{n-1} cos(a + kd) is [sin(n d / 2) / sin(d / 2)] * cos(a + (n - 1)d / 2). Similarly for sine. But in our case, the sum starts at k=1 and goes to n, so it's Sum_{k=1}^n cos(2kα). Let me adjust the formula accordingly.Let me let a = 2α (since the first term when k=1 is cos(2α*1)) and the common difference d = 2α. Wait, but in the standard formula, the sum is from k=0 to n-1 of cos(a + kd). Comparing that to our sum, which is from k=1 to n of cos(2kα). Let me make the substitution: let m = k - 1. Then when k=1, m=0, and when k=n, m = n-1. So the sum becomes Sum_{m=0}^{n-1} cos(2(m + 1)α) = Sum_{m=0}^{n-1} cos(2α + 2mα). Therefore, this is of the form Sum_{m=0}^{n-1} cos(a + m d), where a = 2α and d = 2α.Therefore, applying the formula:Sum = [sin(n * d / 2) / sin(d / 2)] * cos(a + (n - 1)d / 2)Substituting a = 2α and d = 2α:Sum = [sin(n * 2α / 2) / sin(2α / 2)] * cos(2α + (n - 1) * 2α / 2)Simplify:Sum = [sin(nα) / sin(α)] * cos(2α + (n - 1)α)Simplify the argument of the cosine:2α + (n - 1)α = (2 + n - 1)α = (n + 1)αTherefore, Sum = [sin(nα) / sinα] * cos((n + 1)α)Therefore, the sum Sum_{k=1}^n cos(2kα) = [sin(nα) / sinα] * cos((n + 1)α)Wait, but let's check if this formula is correct. Let me verify with n=1. If n=1, then the sum should be cos(2α). Plugging into the formula:[sin(1*α) / sinα] * cos(2α) = [sinα / sinα] * cos2α = cos2α. Correct. For n=2, the sum is cos2α + cos4α. Let's compute the formula:[sin(2α)/sinα] * cos(3α) = [2sinα cosα / sinα] * cos3α = 2cosα * cos3α. Now, using the identity cosA + cosB = 2cos((A+B)/2)cos((A-B)/2). Let's compute cos2α + cos4α = 2cos3α cos(-α) = 2cos3α cosα. Which matches the formula. So yes, the formula works for n=2. Therefore, it seems the formula is correct.So, going back, Sum_{k=1}^n cos(2kα) = [sin(nα) / sinα] * cos((n + 1)α)Therefore, substituting back into the original expressions:Sum cos²(kα) from k=1 to n = n/2 + (1/2) * [sin(nα) / sinα] * cos((n + 1)α)Similarly, Sum sin²(kα) from k=1 to n = n/2 - (1/2) * [sin(nα) / sinα] * cos((n + 1)α)Therefore, both sums can be written as n/2 plus or minus (1/(2 sinα)) sin(nα) cos((n + 1)α)Alternatively, perhaps we can further simplify the term sin(nα) cos((n + 1)α). Using the identity sinA cosB = [sin(A + B) + sin(A - B)] / 2.Let me compute sin(nα) cos((n + 1)α):= [sin(nα + (n + 1)α) + sin(nα - (n + 1)α)] / 2= [sin((2n + 1)α) + sin(-α)] / 2= [sin((2n + 1)α) - sinα] / 2Therefore, substituting back into the expression:Sum cos²(kα) = n/2 + (1/(2 sinα)) * [sin((2n + 1)α) - sinα] / 2= n/2 + [sin((2n + 1)α) - sinα] / (4 sinα)Similarly, for the sine sum:Sum sin²(kα) = n/2 - [sin((2n + 1)α) - sinα] / (4 sinα)Simplifying the terms:Let me compute [sin((2n + 1)α) - sinα] / (4 sinα):First, factor numerator using sine subtraction formula:sin((2n + 1)α) - sinα = 2 cos( ((2n + 1)α + α)/2 ) sin( ((2n + 1)α - α)/2 )= 2 cos( (2n + 2)α / 2 ) sin(2nα / 2 )= 2 cos((n + 1)α) sin(nα)Therefore, [sin((2n + 1)α) - sinα] = 2 cos((n + 1)α) sin(nα)Therefore, substituting back into the expression:[sin((2n + 1)α) - sinα] / (4 sinα) = [2 cos((n + 1)α) sin(nα)] / (4 sinα) = [cos((n + 1)α) sin(nα)] / (2 sinα)But wait, this brings us back to the original term we had earlier. So perhaps the alternate route didn't lead to a simpler form. Therefore, the expression for the sum of cos²(kα) is n/2 + [sin(nα) cos((n + 1)α)] / (2 sinα), and similarly for the sine sum.Alternatively, maybe we can express this in terms of cotangent or other trigonometric functions. Let me see.Alternatively, note that sin(nα) / sinα is a common term. Let's denote that as a separate term. Let me write:Sum cos²(kα) = n/2 + [sin(nα) cos((n + 1)α)] / (2 sinα)Similarly, for the sine sum:Sum sin²(kα) = n/2 - [sin(nα) cos((n + 1)α)] / (2 sinα)Is there a way to combine these terms further? Perhaps using product-to-sum formulas again? Let's see.Alternatively, we can use the identity for sin(nα) cos((n + 1)α). Let me recall that sinA cosB = [sin(A + B) + sin(A - B)] / 2, which we already used earlier. So:sin(nα) cos((n + 1)α) = [sin(nα + (n + 1)α) + sin(nα - (n + 1)α)] / 2= [sin((2n + 1)α) + sin(-α)] / 2= [sin((2n + 1)α) - sinα] / 2Therefore, substituting back into the sum expression:Sum cos²(kα) = n/2 + [sin((2n + 1)α) - sinα] / (4 sinα)Similarly for the sine sum.But perhaps expressing it in terms of sin((2n + 1)α) might not lead to a more simplified form unless there's a telescoping or another identity. Alternatively, maybe we can use the identity for the sum of cos(2kα). Wait, but we already derived that. Let me check if there's another approach.Alternatively, another way to compute the sum of cos²(kα) is to use complex exponentials. Remember that cosθ = (e^{iθ} + e^{-iθ}) / 2, so cos²θ = (e^{i2θ} + 2 + e^{-i2θ}) / 4. Then, summing over k from 1 to n:Sum cos²(kα) = Sum [ (e^{i2kα} + 2 + e^{-i2kα} ) / 4 ] = (1/4) Sum e^{i2kα} + (1/4) Sum e^{-i2kα} + (1/4) * 2nWhich simplifies to (1/4)(Sum e^{i2kα} + Sum e^{-i2kα}) + n/2But Sum e^{i2kα} from k=1 to n is a geometric series with ratio e^{i2α}, starting from e^{i2α} to e^{i2nα}. Similarly for the other sum.The sum of a geometric series is (r^{n} - r)/(r - 1). So:Sum_{k=1}^n e^{i2kα} = e^{i2α} * (1 - e^{i2nα}) / (1 - e^{i2α})Similarly, Sum_{k=1}^n e^{-i2kα} = e^{-i2α} * (1 - e^{-i2nα}) / (1 - e^{-i2α})Therefore, substituting back:Sum cos²(kα) = (1/4)[ e^{i2α}(1 - e^{i2nα}) / (1 - e^{i2α}) + e^{-i2α}(1 - e^{-i2nα}) / (1 - e^{-i2α}) ] + n/2This might be more complicated, but perhaps simplifying these terms can lead to the same result as before. Let me attempt it.First, compute the two sums:First sum: e^{i2α}(1 - e^{i2nα}) / (1 - e^{i2α}) = [e^{i2α} - e^{i2(n + 1)α}] / (1 - e^{i2α})Second sum: e^{-i2α}(1 - e^{-i2nα}) / (1 - e^{-i2α}) = [e^{-i2α} - e^{-i2(n + 1)α}] / (1 - e^{-i2α})Note that 1 - e^{i2α} = -e^{iα} * 2i sinα (using Euler's formula: 1 - e^{iθ} = -2i e^{iθ/2} sin(θ/2))Similarly, 1 - e^{-i2α} = -e^{-iα} * 2i sinαTherefore, let's rewrite the denominators:First sum denominator: 1 - e^{i2α} = -2i e^{iα} sinαSecond sum denominator: 1 - e^{-i2α} = -2i e^{-iα} sinαTherefore, the first sum becomes:[e^{i2α} - e^{i2(n + 1)α}] / (-2i e^{iα} sinα) = [e^{i2α} - e^{i2(n + 1)α}] / (-2i e^{iα} sinα)Similarly, second sum:[e^{-i2α} - e^{-i2(n + 1)α}] / (-2i e^{-iα} sinα )Let me factor numerator and denominator for the first sum:Numerator: e^{i2α} - e^{i2(n + 1)α} = e^{i2α}(1 - e^{i2nα})Denominator: -2i e^{iα} sinαTherefore, first sum:[e^{i2α}(1 - e^{i2nα})] / (-2i e^{iα} sinα) = [e^{iα}(1 - e^{i2nα})] / (-2i sinα )Similarly, for the second sum:Numerator: e^{-i2α} - e^{-i2(n + 1)α} = e^{-i2α}(1 - e^{-i2nα})Denominator: -2i e^{-iα} sinαTherefore, second sum:[e^{-i2α}(1 - e^{-i2nα})] / (-2i e^{-iα} sinα ) = [e^{-iα}(1 - e^{-i2nα})] / (-2i sinα )Therefore, combining both sums:First sum + Second sum = [e^{iα}(1 - e^{i2nα}) + e^{-iα}(1 - e^{-i2nα})] / (-2i sinα )Let me compute the numerator:e^{iα}(1 - e^{i2nα}) + e^{-iα}(1 - e^{-i2nα})= e^{iα} - e^{i(2n + 1)α} + e^{-iα} - e^{-i(2n + 1)α}= (e^{iα} + e^{-iα}) - (e^{i(2n + 1)α} + e^{-i(2n + 1)α})= 2cosα - 2cos((2n + 1)α)Therefore, numerator = 2[cosα - cos((2n + 1)α)]Therefore, first sum + second sum = 2[cosα - cos((2n + 1)α)] / (-2i sinα ) = [cos((2n + 1)α) - cosα] / (i sinα )Therefore, putting back into the original expression:Sum cos²(kα) = (1/4) * [cos((2n + 1)α) - cosα] / (i sinα ) + n/2Wait, but this seems problematic because we have an imaginary unit in the denominator. But cosine terms are real, so there must be a mistake here. Wait, perhaps when converting to exponentials, there was an error. Let me check.Wait, the sum we calculated was (1/4)(Sum e^{i2kα} + Sum e^{-i2kα}) + n/2. Then we expressed the sums in terms of exponentials and ended up with a term involving i. But the original sum is real, so the imaginary parts should cancel out, but here we have a term with i in the denominator, which suggests that perhaps there was a miscalculation.Wait, let me check the steps again. When we computed the first sum + second sum, we had:First sum + Second sum = [e^{iα}(1 - e^{i2nα}) + e^{-iα}(1 - e^{-i2nα})] / (-2i sinα )Then expanding the numerator:e^{iα} - e^{i(2n + 1)α} + e^{-iα} - e^{-i(2n + 1)α}= (e^{iα} + e^{-iα}) - (e^{i(2n + 1)α} + e^{-i(2n + 1)α})= 2cosα - 2cos((2n + 1)α)Therefore, numerator is 2[cosα - cos((2n + 1)α)]Therefore, numerator / (-2i sinα ) = 2[cosα - cos((2n + 1)α)] / (-2i sinα ) = [cos((2n + 1)α) - cosα] / (i sinα )But since the entire term is multiplied by (1/4), we have:Sum cos²(kα) = (1/4) * [cos((2n + 1)α) - cosα] / (i sinα ) + n/2However, this introduces an imaginary unit, which shouldn't be present in a sum of real numbers. Therefore, this suggests that there is an error in the calculation. Wait, but complex exponentials can be used to represent real functions, so perhaps we need to take the real part?Wait, but the original sum of cos²(kα) is real, and when we expressed it in terms of complex exponentials, the sum should also be real. However, in the process, the intermediate steps introduced an imaginary unit. Therefore, this suggests that there was a miscalculation in the sign or somewhere else.Wait, let me check the denominators again. The denominator for the first sum was 1 - e^{i2α} = -2i e^{iα} sinα. Let me verify this:1 - e^{i2α} = e^{iα}(e^{-iα} - e^{iα}) = e^{iα}(-2i sinα). Therefore, yes, 1 - e^{i2α} = -2i e^{iα} sinα. Similarly, 1 - e^{-i2α} = -2i e^{-iα} sin(-α) = 2i e^{-iα} sinα. Wait, hold on. Wait, 1 - e^{-i2α} = 1 - cos2α + i sin2α. Alternatively, using the same formula:1 - e^{-iθ} = -e^{-iθ/2}(e^{iθ/2} - e^{-iθ/2}) = -e^{-iθ/2}(2i sin(θ/2))Therefore, 1 - e^{-i2α} = -e^{-iα}(2i sinα). Therefore, the denominator for the second sum is -2i e^{-iα} sinα. Therefore, the denominators are both -2i e^{±iα} sinα.Therefore, when we took the first sum:[e^{i2α} - e^{i2(n + 1)α}] / (-2i e^{iα} sinα ) = [e^{iα} - e^{i(2n + 1)α}] / (-2i sinα )Similarly, the second sum:[e^{-i2α} - e^{-i2(n + 1)α}] / (-2i e^{-iα} sinα ) = [e^{-iα} - e^{-i(2n + 1)α}] / (-2i sinα )Therefore, adding both sums:[e^{iα} - e^{i(2n + 1)α} + e^{-iα} - e^{-i(2n + 1)α}] / (-2i sinα )= [ (e^{iα} + e^{-iα}) - (e^{i(2n + 1)α} + e^{-i(2n + 1)α}) ] / (-2i sinα )= [2cosα - 2cos( (2n + 1)α ) ] / (-2i sinα )= [cosα - cos( (2n + 1)α ) ] / (-i sinα )Multiply numerator and denominator by i:= i [cosα - cos( (2n + 1)α ) ] / sinαTherefore, the entire term (1/4)(Sum + Sum) is (1/4) * i [cosα - cos( (2n + 1)α ) ] / sinαBut this is imaginary, which contradicts the fact that the original sum is real. Therefore, this suggests that there's a mistake in the approach.Wait, perhaps when expressing cos²θ in terms of complex exponentials, I made a mistake. Let's re-examine that step.cos²θ = (e^{iθ} + e^{-iθ})² / 4 = (e^{i2θ} + 2 + e^{-i2θ}) / 4. Therefore, cos²θ = (e^{i2θ} + e^{-i2θ}) / 4 + 2/4 = (cos2θ)/2 + 1/2. Which matches the identity we used earlier: cos²θ = (1 + cos2θ)/2. Therefore, that part is correct.Then, Sum cos²(kα) = Sum [ (1 + cos2kα ) / 2 ] = n/2 + (1/2) Sum cos2kα. Therefore, substituting the sum of cos2kα which we already calculated as [sin(nα)/sinα] cos(n + 1)α. Therefore, the sum is n/2 + (1/2)( [sin(nα)/sinα] cos(n + 1)α )Which is real, as expected. Therefore, the mistake must be in the complex exponential approach, perhaps in the calculation steps. Since we already have a valid expression using trigonometric identities, maybe it's better to stick with that result rather than the complex exponential method which seems to have led to an inconsistency.Therefore, going back to the earlier result:Sum cos²(kα) = n/2 + [sin(nα) cos(n + 1)α ] / (2 sinα )Similarly, Sum sin²(kα) = n/2 - [sin(nα) cos(n + 1)α ] / (2 sinα )Alternatively, we can write this as:Sum cos²(kα) = [n/2] + [sin(nα) cos((n + 1)α)] / (2 sinα )Is there a way to combine these terms into a more compact form? Let me think.Alternatively, perhaps using the identity for sin(nα) cos((n + 1)α). Let me recall that:sinA cosB = [sin(A + B) + sin(A - B)] / 2So, sin(nα) cos((n + 1)α) = [sin(nα + (n + 1)α) + sin(nα - (n + 1)α)] / 2= [sin((2n + 1)α) + sin(-α)] / 2= [sin((2n + 1)α) - sinα] / 2Therefore, substituting back into the sum expression:Sum cos²(kα) = n/2 + [sin((2n + 1)α) - sinα] / (4 sinα )Similarly, Sum sin²(kα) = n/2 - [sin((2n + 1)α) - sinα] / (4 sinα )This can be written as:Sum cos²(kα) = n/2 + [sin((2n + 1)α) - sinα] / (4 sinα )= [2n sinα + sin((2n + 1)α) - sinα] / (4 sinα )= [ (2n - 1) sinα + sin((2n + 1)α) ] / (4 sinα )Hmm, perhaps there's a telescoping identity here. Let me recall that sin((2n + 1)α) can be related to a telescoping sum. Alternatively, maybe using the identity for the sum of sines.Alternatively, note that the expression [sin((2n + 1)α) - sinα] can be expressed as 2 cos((n + 1)α) sin(nα), which we saw earlier. Therefore:Sum cos²(kα) = n/2 + [2 cos((n + 1)α) sin(nα)] / (4 sinα )= n/2 + [cos((n + 1)α) sin(nα)] / (2 sinα )= n/2 + [cos((n + 1)α) / (2 sinα ) ] sin(nα )But sin(nα) / sinα is a known term. In fact, this term appears in the formula for the sum of cosines. For example, Sum_{k=1}^n cos(kα) = [sin(nα/2) / sin(α/2)] cos((n + 1)α/2 ). But perhaps not directly helpful here.Alternatively, since we have [sin(nα) / sinα] * cos((n + 1)α), maybe this can be written in terms of a cotangent or something.Alternatively, if we write:[sin(nα) cos((n + 1)α) ] / sinα = [sin(nα) / sinα] cos((n + 1)α )Note that sin(nα)/sinα is the Dirichlet kernel, which is related to sum of cosines. But unless there's a specific identity, I think the expression we have is already as simplified as it can get.Alternatively, perhaps we can use the identity:sin(nα) cos((n + 1)α) = [sin(nα + (n + 1)α) + sin(nα - (n + 1)α)] / 2 = [sin((2n + 1)α) + sin(-α)] / 2 = [sin((2n + 1)α) - sinα]/2But this just takes us back to the previous expression.Alternatively, consider writing the entire expression as:Sum cos²(kα) = [n sinα + sin(nα) cos((n + 1)α) ] / (2 sinα )But I don't think that's simpler.Alternatively, factor out 1/(2 sinα):Sum cos²(kα) = 1/(2 sinα) [ n sinα + sin(nα) cos((n + 1)α) ]But again, not sure if this is helpful.Alternatively, perhaps check for specific values of n to see if the formula holds.Let me test n=1:Sum cos²(α) = [1/2] + [sin(α) cos(2α) ] / (2 sinα )= 1/2 + [cos2α]/2= (1 + cos2α)/2 = cos²α, which is correct.Similarly, n=2:Sum cos²α + cos²2α = [2/2] + [sin2α cos3α]/(2 sinα )= 1 + [2 sinα cosα cos3α ]/(2 sinα )= 1 + cosα cos3αBut cosα cos3α = [cos4α + cos2α]/2Therefore, Sum = 1 + [cos4α + cos2α]/2But according to direct calculation:Sum cos²α + cos²2α = (1 + cos2α)/2 + (1 + cos4α)/2 = 1 + (cos2α + cos4α)/2Which matches. Therefore, the formula works.Alternatively, if we take n=2, Sum cos²kα = 1 + [cos2α + cos4α]/2. Which is the same as 1 + (sum of cos2kα)/2, which is the same as applying the original formula.Therefore, the formula is correct.Given that, perhaps this is as simplified as it can get. So, the answer for the sum of cos squared terms is n/2 plus sin(nα)cos((n + 1)α)/(2 sinα), and the sum of sin squared terms is n/2 minus the same term. Therefore, unless there's a further identity to apply, this is the simplified form.Alternatively, is there a way to express this in terms of sin(2nα) or something else? Let me think.Alternatively, consider that:sin(nα) cos((n + 1)α) = [sin(nα + (n + 1)α) + sin(nα - (n + 1)α)] / 2= [sin((2n + 1)α) + sin(-α)] / 2= [sin((2n + 1)α) - sinα] / 2Therefore, substituting back:Sum cos²(kα) = n/2 + [sin((2n + 1)α) - sinα]/(4 sinα )Similarly, for the sine sum.Alternatively, we can write this as:Sum cos²(kα) = (n + 1)/2 + [sin((2n + 1)α) - sinα - 2n sinα ]/(4 sinα )Wait, no, that seems more complicated.Alternatively, using telescoping sums. Let me consider the sum of cos2kα from k=1 to n. We have:Sum_{k=1}^n cos2kα = [sin(nα) cos(n + 1)α ] / sinαTherefore, the sum of cos squares is:n/2 + [sin(nα) cos(n + 1)α ] / (2 sinα )Similarly, for sine squares.Alternatively, if we denote S = Sum_{k=1}^n cos2kα, then the sum of cos² is n/2 + S/2, and sum of sin² is n/2 - S/2.Given that S is expressed as [sin(nα) cos(n + 1)α ] / sinα, then yes, this is the simplified form.Alternatively, consider writing S as real part of a complex exponential series. But as before, that led to complications.Given that the expression is already in terms of sine and cosine functions, and further attempts to simplify lead back to the same expression or introduce complexities, I think this is the simplest form.Therefore, the final simplified expressions are:Sum_{k=1}^n cos²(kα) = n/2 + [sin(nα) cos((n + 1)α)] / (2 sinα )Sum_{k=1}^n sin²(kα) = n/2 - [sin(nα) cos((n + 1)α)] / (2 sinα )Alternatively, these can be written as:Sum_{k=1}^n cos²(kα) = frac{n}{2} + frac{sin(nalpha) cos((n + 1)alpha)}{2 sinalpha}Sum_{k=1}^n sin²(kα) = frac{n}{2} - frac{sin(nalpha) cos((n + 1)alpha)}{2 sinalpha}These expressions are as simplified as possible using standard trigonometric identities. Therefore, these are the simplified forms of the given sums.Final AnswerThe simplified expressions are:cos^{2} alpha + cos^{2} 2alpha + ldots + cos^{2} nalpha = boxed{frac{n}{2} + frac{sin(nalpha) cos((n + 1)alpha)}{2 sin alpha}}andsin^{2} alpha + sin^{2} 2alpha + ldots + sin^{2} nalpha = boxed{frac{n}{2} - frac{sin(nalpha) cos((n + 1)alpha)}{2 sin alpha}}</think>To simplify the sums (cos^{2} alpha + cos^{2} 2alpha + ldots + cos^{2} nalpha) and (sin^{2} alpha + sin^{2} 2alpha + ldots + sin^{2} nalpha), we start by using the trigonometric identities (cos^{2} theta = frac{1 + cos 2theta}{2}) and (sin^{2} theta = frac{1 - cos 2theta}{2}).For the cosine squared sum:[sum_{k=1}^{n} cos^{2}(kalpha) = sum_{k=1}^{n} frac{1 + cos(2kalpha)}{2} = frac{n}{2} + frac{1}{2} sum_{k=1}^{n} cos(2kalpha)]For the sine squared sum:[sum_{k=1}^{n} sin^{2}(kalpha) = sum_{k=1}^{n} frac{1 - cos(2kalpha)}{2} = frac{n}{2} - frac{1}{2} sum_{k=1}^{n} cos(2kalpha)]Next, we need to evaluate the sum (sum_{k=1}^{n} cos(2kalpha)). Using the formula for the sum of cosines of an arithmetic sequence, we find:[sum_{k=1}^{n} cos(2kalpha) = frac{sin(nalpha)}{sin alpha} cos((n + 1)alpha)]Substituting this result back into the expressions for the cosine and sine squared sums, we get:[sum_{k=1}^{n} cos^{2}(kalpha) = frac{n}{2} + frac{1}{2} left( frac{sin(nalpha) cos((n + 1)alpha)}{sin alpha} right) = frac{n}{2} + frac{sin(nalpha) cos((n + 1)alpha)}{2 sin alpha}][sum_{k=1}^{n} sin^{2}(kalpha) = frac{n}{2} - frac{1}{2} left( frac{sin(nalpha) cos((n + 1)alpha)}{sin alpha} right) = frac{n}{2} - frac{sin(nalpha) cos((n + 1)alpha)}{2 sin alpha}]Thus, the simplified expressions are:[cos^{2} alpha + cos^{2} 2alpha + ldots + cos^{2} nalpha = boxed{frac{n}{2} + frac{sin(nalpha) cos((n + 1)alpha)}{2 sin alpha}}][sin^{2} alpha + sin^{2} 2alpha + ldots + sin^{2} nalpha = boxed{frac{n}{2} - frac{sin(nalpha) cos((n + 1)alpha)}{2 sin alpha}}]]

🔑:325. Let's use the fact that cos ^{2} x=frac{1+cos 2 x}{2}. From this, using the result of the previous problem, we get:begin{aligned}& cos ^{2} alpha+cos ^{2} 2 alpha+ldots+cos ^{2} n alpha= & quad=frac{1}{2}[cos 2 alpha+cos 4 alpha+ldots+cos 2 n alpha+n]= & =frac{1}{2}left[frac{sin (n+1) alpha cos n alpha}{sin alpha}-1right]+frac{n}{2}=frac{sin (n+1) alpha cos n alpha}{2 sin alpha}+frac{n-1}{2} & text { And since } sin ^{2} x=1-cos ^{2} x, text { then } & sin ^{2} alpha+sin ^{2} 2 alpha+ldots+sin ^{2} n alpha= & =n-frac{sin (n+1) alpha cos n alpha}{2 sin alpha}-frac{n-1}{2}=frac{n+1}{2}-frac{sin (n+1) alpha cos n alpha}{2 sin alpha}end{aligned}

🧠:[<think>Okay, let me try to solve this problem. So, we have ω is a complex cube root of unity. I remember that cube roots of unity satisfy the equation ω³ = 1. Also, there are three cube roots of unity: 1, ω, and ω². Since ω is not equal to 1, we know that 1 + ω + ω² = 0. That might come in handy.The expression we need to compute is (1 - ω)(1 - ω²)(1 - ω⁴)(1 - ω⁸). The options are A. ω, B. ω̄ (which is the complex conjugate of ω, so that's ω²), C. 9, D. 6. Hmm.First, maybe I can simplify the exponents of ω. Since ω³ = 1, any exponent can be reduced modulo 3. So ω⁴ = ω^(3 + 1) = ω^1 * ω^3 = ω * 1 = ω. Similarly, ω⁸ = ω^(6 + 2) = (ω³)^2 * ω² = 1² * ω² = ω². So ω⁴ is ω, and ω⁸ is ω². Therefore, the original expression simplifies to (1 - ω)(1 - ω²)(1 - ω)(1 - ω²). So that's [(1 - ω)(1 - ω²)]².Wait, so the product becomes [(1 - ω)(1 - ω²)] squared. Let me compute (1 - ω)(1 - ω²) first. Let's expand that.Multiply (1 - ω)(1 - ω²):1*1 = 11*(-ω²) = -ω²(-ω)*1 = -ω(-ω)*(-ω²) = ω³But ω³ = 1, so that term is 1.So adding up all the terms: 1 - ω² - ω + 1 = (1 + 1) + (-ω - ω²) = 2 - (ω + ω²)But earlier, I remember that 1 + ω + ω² = 0, so ω + ω² = -1. Therefore, 2 - (-1) = 2 + 1 = 3. So (1 - ω)(1 - ω²) = 3.Therefore, the original expression is [3]² = 9. So the answer is C. 9. Wait, but let me double-check. Wait, the original expression is (1 - ω)(1 - ω²)(1 - ω⁴)(1 - ω⁸). We substituted ω⁴ with ω and ω⁸ with ω², so it becomes (1 - ω)(1 - ω²)(1 - ω)(1 - ω²) which is [ (1 - ω)(1 - ω²) ]² = 3² = 9. So yeah, 9 is an option here, option C. But wait, the answer options don't have 3, so 9 is there. So C is the answer. Let me check again.Alternatively, maybe there's another way to compute it. Let me recall that for cube roots of unity, the minimal polynomial is x² + x + 1 = 0. So we have ω² = -ω -1. Maybe that can help in computation. Let's see.But we already did it by expanding. Alternatively, (1 - ω)(1 - ω²) can be computed using the formula for the product. Alternatively, since (x - ω)(x - ω²) = x² - (ω + ω²)x + ω³. Since ω³ = 1, this becomes x² - (-1)x + 1 = x² + x + 1. So if we substitute x = 1, then (1 - ω)(1 - ω²) = 1² + 1 + 1 = 3. That's another way to see it.So indeed, (1 - ω)(1 - ω²) = 3, so the square is 9. Therefore, the answer should be C. 9. Wait, but let me check if I made a mistake in the exponents.Original problem: (1 - ω)(1 - ω²)(1 - ω⁴)(1 - ω⁸). So ω⁴: since 4 divided by 3 is 1 remainder 1, so ω⁴ = ω^(3 + 1) = ω. Similarly, ω⁸: 8 divided by 3 is 2 remainder 2, so ω⁸ = ω^(6 + 2) = (ω³)^2 * ω² = 1² * ω² = ω². So yes, ω⁴ is ω, ω⁸ is ω². So indeed, the expression becomes (1 - ω)(1 - ω²)(1 - ω)(1 - ω²) = [ (1 - ω)(1 - ω²) ]² = 3² = 9. So answer is C.Alternatively, maybe there's a different approach. Let me think. The cube roots of unity satisfy ω² = conjugate of ω. So ω̄ = ω². So (1 - ω)(1 - ω²) is the same as (1 - ω)(1 - ω̄). The product of (1 - ω)(1 - ω̄) is |1 - ω|². Wait, modulus squared. But since ω is a complex number on the unit circle, |1 - ω| is the distance between 1 and ω on the complex plane. But perhaps this is overcomplicating. However, if we compute |1 - ω|², that is (1 - ω)(1 - ω̄) = 1 - ω - ω̄ + |ω|². But since |ω| = 1, because it's a root of unity, so |ω|² = 1. So this becomes 1 - ω - ω̄ + 1 = 2 - (ω + ω̄). But ω̄ = ω², so ω + ω̄ = ω + ω² = -1, as before. So again, 2 - (-1) = 3. Therefore, |1 - ω|² = 3, so modulus is sqrt(3). But maybe that's not needed here.But regardless, the product (1 - ω)(1 - ω²) = 3, so the original expression is 3*3 = 9. So answer is C. Therefore, the correct answer is C. 9.Final Answerboxed{C}</think>Given that (omega) is a complex cube root of unity, we need to find the value of ((1 - omega)(1 - omega^2)(1 - omega^4)(1 - omega^8)).First, we simplify the exponents of (omega) using the property (omega^3 = 1). This gives:[omega^4 = omega^{3 + 1} = omega quad text{and} quad omega^8 = omega^{6 + 2} = (omega^3)^2 cdot omega^2 = 1^2 cdot omega^2 = omega^2]Thus, the expression simplifies to:[(1 - omega)(1 - omega^2)(1 - omega)(1 - omega^2) = [(1 - omega)(1 - omega^2)]^2]Next, we compute the product ((1 - omega)(1 - omega^2)). Expanding this product:[(1 - omega)(1 - omega^2) = 1 - omega - omega^2 + omega^3]Since (omega^3 = 1) and (omega + omega^2 = -1) (from the equation (1 + omega + omega^2 = 0)), we substitute these values:[1 - omega - omega^2 + 1 = 2 - (omega + omega^2) = 2 - (-1) = 3]Therefore, the product ((1 - omega)(1 - omega^2) = 3). Squaring this result gives:[[(1 - omega)(1 - omega^2)]^2 = 3^2 = 9]Thus, the value of ((1 - omega)(1 - omega^2)(1 - omega^4)(1 - omega^8)) is (boxed{C}).,<think>Okay, so I have this problem here: Let ω be a complex cube root of unity. Then I need to find the value of the product (1 - ω)(1 - ω²)(1 - ω⁴)(1 - ω⁸), and the options are A. ω, B. ω̄ (which is the complex conjugate of ω), C. 9, D. 6. Hmm, let me think step by step.First, I remember that cube roots of unity satisfy the equation ω³ = 1. So ω³ = 1. Also, the cube roots of unity are 1, ω, and ω². So there are three distinct cube roots of unity. Since ω is a primitive root here (not equal to 1), we know that 1 + ω + ω² = 0. That's a key identity often used in these problems.Now, the given expression is (1 - ω)(1 - ω²)(1 - ω⁴)(1 - ω⁸). Let me see if I can simplify the exponents first. Since ω³ = 1, any exponent of ω can be reduced modulo 3. So, let's compute ω⁴ and ω⁸.Starting with ω⁴: 4 divided by 3 is 1 with a remainder of 1, so ω⁴ = ω^(3+1) = ω³ * ω = 1 * ω = ω. Similarly, ω⁸: 8 divided by 3 is 2 with a remainder of 2, so ω⁸ = ω^(3*2 + 2) = (ω³)² * ω² = 1² * ω² = ω². So actually, ω⁴ is ω and ω⁸ is ω². Therefore, the original expression simplifies to (1 - ω)(1 - ω²)(1 - ω)(1 - ω²). Wait, that's (1 - ω) squared times (1 - ω²) squared. So the product becomes [(1 - ω)(1 - ω²)]².Let me compute (1 - ω)(1 - ω²) first. Let's multiply those two factors:(1 - ω)(1 - ω²) = 1*(1 - ω²) - ω*(1 - ω²) = 1 - ω² - ω + ω³. But ω³ = 1, so ω³ becomes 1. Therefore, substituting that in: 1 - ω² - ω + 1 = (1 + 1) - ω - ω² = 2 - (ω + ω²). But from the identity 1 + ω + ω² = 0, we can solve for ω + ω² = -1. Therefore, substituting that in: 2 - (-1) = 2 + 1 = 3. So (1 - ω)(1 - ω²) = 3. Therefore, the entire expression is [3]^2 = 9. Wait, so is the answer 9? That's option C. Hmm, but let me check again, because maybe I made a mistake in simplifying the exponents.Wait, ω⁴: 4 divided by 3 is 1 with remainder 1, so ω⁴ = ω^(3+1) = ω^3 * ω = 1 * ω = ω. Similarly, ω⁸: 8 divided by 3 is 2 with remainder 2, so ω⁸ = ω^(6+2) = ω^6 * ω^2. But ω^6 is (ω³)^2 = 1² = 1, so ω⁸ = 1 * ω² = ω². So yes, ω⁴ is ω and ω⁸ is ω², so (1 - ω⁴) is (1 - ω) and (1 - ω⁸) is (1 - ω²). Therefore, the product is indeed (1 - ω)^2(1 - ω²)^2, which is [(1 - ω)(1 - ω²)]². As we found before, that's 3² = 9. So the answer should be C. 9. But wait, let me check again.Alternatively, maybe there's another approach. Let me recall that for cube roots of unity, the minimal polynomial is x² + x + 1 = 0. Therefore, 1 - ω and 1 - ω² can be related. Let me note that |1 - ω| and |1 - ω²| can be calculated as well, but since we are dealing with complex numbers, their product might be a real number.Alternatively, consider that (x - 1)(x - ω)(x - ω²) = x³ - 1. So if we set x = 1, then (1 - 1)(1 - ω)(1 - ω²) = 1 - 1 = 0, which checks out. But maybe we can use the factorization of x³ - 1 = (x - 1)(x - ω)(x - ω²). So, (x - ω)(x - ω²) = (x² - (ω + ω²)x + ω³). But ω + ω² = -1 and ω³ = 1, so (x - ω)(x - ω²) = x² + x + 1. Therefore, if we set x = 1, then (1 - ω)(1 - ω²) = 1² + 1 + 1 = 3, which matches the previous result. So indeed, (1 - ω)(1 - ω²) = 3. Hence, squaring that gives 9. Therefore, the answer is C. 9.Wait, but the original problem has four terms: (1 - ω)(1 - ω²)(1 - ω⁴)(1 - ω⁸). But since ω⁴ = ω and ω⁸ = ω², as established earlier, then it's (1 - ω)(1 - ω²)(1 - ω)(1 - ω²) = [(1 - ω)(1 - ω²)]² = 3² = 9. Therefore, the answer is C.But let me check once more, just to be sure. Maybe using another method. Suppose we compute each term:First term: (1 - ω)Second term: (1 - ω²)Third term: (1 - ω⁴) = (1 - ω), as ω⁴ = ω^(3 +1) = ω.Fourth term: (1 - ω⁸) = (1 - ω²), since ω⁸ = ω^(6 + 2) = (ω³)^2 * ω² = 1^2 * ω² = ω².Therefore, the product is (1 - ω)^2 (1 - ω²)^2. As before. Then, (1 - ω)(1 - ω²) = 3, so squared is 9. Alternatively, maybe calculate (1 - ω)^2 (1 - ω²)^2.Alternatively, expand (1 - ω)(1 - ω²) first:(1 - ω)(1 - ω²) = 1 - ω - ω² + ω³. Since ω³ = 1, this becomes 1 - ω - ω² + 1 = 2 - (ω + ω²). And since ω + ω² = -1, this becomes 2 - (-1) = 3. So squared is 9.Alternatively, compute (1 - ω)^2 (1 - ω²)^2 directly. Let's compute (1 - ω)^2 = 1 - 2ω + ω², and (1 - ω²)^2 = 1 - 2ω² + ω⁴. But ω⁴ = ω, so (1 - ω²)^2 = 1 - 2ω² + ω. Then, multiplying (1 - 2ω + ω²)(1 - 2ω² + ω). Let's multiply term by term:First term: 1*(1) = 11*(-2ω²) = -2ω²1*(ω) = ωNext, -2ω*(1) = -2ω-2ω*(-2ω²) = 4ω³ = 4*1 = 4-2ω*(ω) = -2ω²Then, ω²*(1) = ω²ω²*(-2ω²) = -2ω⁴ = -2ωω²*(ω) = ω³ = 1Now, let's collect all terms:1 - 2ω² + ω - 2ω + 4 - 2ω² + ω² - 2ω + 1Combine constants: 1 + 4 + 1 = 6Combine ω terms: ω - 2ω - 2ω = (1 - 2 - 2)ω = (-3ω)Combine ω² terms: -2ω² - 2ω² + ω² = (-2 -2 +1)ω² = (-3ω²)So the total expression is 6 - 3ω - 3ω². Factor out 3: 3(2 - ω - ω²). Again, since ω + ω² = -1, this becomes 3(2 - (-1)) = 3(3) = 9. So yes, same result. Therefore, the product is 9, so the answer is C.Alternatively, maybe using modulus and arguments. Since ω is a complex cube root of unity, it can be represented as e^(2πi/3) = cos(2π/3) + i sin(2π/3) = -1/2 + i√3/2. Similarly, ω² = e^(4πi/3) = cos(4π/3) + i sin(4π/3) = -1/2 - i√3/2.So 1 - ω is 1 - (-1/2 + i√3/2) = 1 + 1/2 - i√3/2 = 3/2 - i√3/2.Similarly, 1 - ω² is 1 - (-1/2 - i√3/2) = 1 + 1/2 + i√3/2 = 3/2 + i√3/2.Then, (1 - ω)(1 - ω²) is (3/2 - i√3/2)(3/2 + i√3/2). This is of the form (a - b)(a + b) = a² - b². So here, a = 3/2, b = i√3/2. Therefore, (3/2)² - (i√3/2)² = 9/4 - (i² * 3/4) = 9/4 - (-1 * 3/4) = 9/4 + 3/4 = 12/4 = 3. So again, this gives 3. Then squaring it gives 9. So the modulus of (1 - ω) is |1 - ω|. Let's compute that.The modulus of 1 - ω is sqrt((3/2)^2 + (√3/2)^2) = sqrt(9/4 + 3/4) = sqrt(12/4) = sqrt(3). Similarly, modulus of 1 - ω² is also sqrt(3). So the modulus of the product (1 - ω)(1 - ω²) is sqrt(3)*sqrt(3) = 3, which matches. Then squaring that modulus gives 9, which is a real number. Therefore, the answer is 9, which is option C.Alternatively, maybe the problem is designed to trick us into thinking it's ω or ω̄, but since the product is real, it can't be A or B. Then between C and D, since we computed 9, the answer is C.So, after multiple methods, I believe the answer is C. 9. Therefore, I think the correct option is C.Final Answerboxed{C}</think>Given that (omega) is a complex cube root of unity, we need to find the value of ((1 - omega)(1 - omega^2)(1 - omega^4)(1 - omega^8)).First, we note that (omega^3 = 1), so the exponents of (omega) can be reduced modulo 3. Therefore:[omega^4 = omega^{3+1} = omega quad text{and} quad omega^8 = omega^{6+2} = omega^2]This simplifies the given expression to:[(1 - omega)(1 - omega^2)(1 - omega)(1 - omega^2) = [(1 - omega)(1 - omega^2)]^2]Next, we compute the product ((1 - omega)(1 - omega^2)):[(1 - omega)(1 - omega^2) = 1 - omega - omega^2 + omega^3]Since (omega^3 = 1), this simplifies to:[1 - omega - omega^2 + 1 = 2 - (omega + omega^2)]Using the identity (1 + omega + omega^2 = 0), we know (omega + omega^2 = -1). Substituting this in, we get:[2 - (-1) = 3]Thus, the product ((1 - omega)(1 - omega^2) = 3). Squaring this result gives:[[(1 - omega)(1 - omega^2)]^2 = 3^2 = 9]Therefore, the value of ((1 - omega)(1 - omega^2)(1 - omega^4)(1 - omega^8)) is (boxed{C}).]

🔑:6. mathrm{C} Original expression =(1-omega)^{2}left(1-omega^{2}right)^{2}=left(1-omega-omega^{2}+omega^{3}right)^{2}=(3)^{2}=9

🧠:[<think>Okay, so I have to find the number of 7-combinations of the multiset S, which has 4 a's, 4 b's, 3 c's, and 3 d's. Hmm, combinations in multisets... I remember this is about selecting elements where the order doesn't matter, but since it's a multiset, we can have multiple instances of each element, up to their given maximums. Let me recall the formula. For multisets, the number of k-combinations is similar to solving the equation x1 + x2 + ... + xn = k, where each xi is between 0 and the maximum number of that element. In this case, n is 4 (a, b, c, d) and the maxima are 4, 4, 3, 3. So we need non-negative integers x1, x2, x3, x4 such that x1 + x2 + x3 + x4 = 7, with x1 ≤4, x2 ≤4, x3 ≤3, x4 ≤3.Right, so the problem reduces to finding the number of non-negative integer solutions to that equation with those constraints. The standard way to do this without constraints is using stars and bars, which is C(n + k -1, k). But here we have upper limits, so we need to use inclusion-exclusion.Let me remember inclusion-exclusion for this case. The formula is:Number of solutions = C(7 + 4 -1, 7) - [C(7 - (4+1) + 4 -1, 7 - (4+1)) + ... for each variable exceeding their limit] + [terms for two variables exceeding] - ... and so on.Wait, maybe I should write it more systematically. The general formula when we have upper bounds is:Total = C(n + k -1, k) minus the sum over each variable exceeding its maximum, plus the sum for overlaps of two variables exceeding, etc.In this problem, n is 4 variables (a, b, c, d), k is 7. Each variable has a maximum: a and b up to 4, c and d up to 3.So first, calculate the total number of non-negative solutions without restrictions: C(7 + 4 -1, 7) = C(10,7) = 120.But then we need to subtract the solutions where at least one variable exceeds its maximum.Let me handle each variable:First, variables a and b can go up to 4, so if we subtract the cases where a ≥5 or b ≥5. Similarly, c and d can go up to 3, so subtract cases where c ≥4 or d ≥4.So, let's compute the number of solutions where a ≥5. Let’s set a' = a -5, then the equation becomes a' + b + c + d = 7 -5 = 2. So the number of solutions here is C(2 +4 -1, 2) = C(5,2)=10. Similarly for b ≥5, it's the same, 10.For c ≥4: set c' = c -4, then equation is a + b + c' + d =7 -4=3. Solutions: C(3 +4 -1,3)=C(6,3)=20. Same for d ≥4, so 20 each.So subtract these: 10 (a) +10 (b) +20 (c) +20 (d) = 60. So total so far: 120 -60=60.But now we need to add back the cases where two variables exceed their limits, since we subtracted them twice. So for overlaps:Cases where a ≥5 and b ≥5: a' + b' + c + d =7 -5 -5= -3. Which is impossible, so 0.Similarly, a ≥5 and c ≥4: a'=a-5, c'=c-4, then a' + b + c' + d =7 -5 -4= -2. Also impossible.Same for a ≥5 and d ≥4: same result. Similarly, b ≥5 and c ≥4, b ≥5 and d ≥4: impossible.Now, cases where c ≥4 and d ≥4: c' + d' + a + b =7 -4 -4= -1. Also impossible.So all two-variable overlaps result in negative remaining sums, which are 0. So no need to add anything here.What about overlaps of three variables? Similarly, subtracting even more would lead to more negative sums, so all zero.And the four-variable overlap is also impossible. So according to inclusion-exclusion, since all two-variable overlaps are zero, the total is 120 -60 +0 -0 +0=60.But wait, hold on. Wait, maybe I made a mistake here. Let me check.Wait, so for each of the variables exceeding, we subtracted 10,10,20,20. But when two variables exceed, even if their individual subtractions would lead to negative, but if two variables are subtracted, maybe in some cases the sum is not negative?Wait, for example, if we consider both a ≥5 and c ≥4: 7 -5 -4= -2, which is negative, so no solutions. Similarly, two variables from the same group: like c and d both exceeding. 7 -4 -4= -1, which is still negative. So indeed, all two-variable overlaps have negative sums, so they contribute zero. Therefore, inclusion-exclusion stops here, and the total is 120 -60=60.But wait, let me check with another approach to confirm. Maybe generating functions?The generating function for each element is:For a and b: 1 + x + x^2 + x^3 + x^4 (since up to 4)For c and d: 1 + x + x^2 + x^3 (up to 3)So the generating function is (1 + x + x^2 + x^3 + x^4)^2 * (1 + x + x^2 + x^3)^2.We need the coefficient of x^7 in this product.Let me compute that.First, note that (1 + x + x^2 + x^3 + x^4) is (1 - x^5)/(1 - x), and (1 + x + x^2 + x^3) is (1 - x^4)/(1 - x). So the generating function becomes [(1 - x^5)^2 * (1 - x^4)^2]/(1 - x)^4.Therefore, the coefficient of x^7 in this is the same as the coefficient of x^7 in (1 - 2x^5 -2x^4 + x^8 + 2x^9 + x^10) divided by (1 - x)^4. Wait, actually, expanding (1 - x^5)^2*(1 - x^4)^2:First, expand (1 - x^5)^2 = 1 - 2x^5 + x^{10}Then, (1 - x^4)^2 =1 -2x^4 + x^8Multiply them together: (1)(1 -2x^4 + x^8) -2x^5(1 -2x^4 +x^8) +x^{10}(1 -2x^4 +x^8)=1 -2x^4 +x^8 -2x^5 +4x^9 -2x^{13} +x^{10} -2x^{14} +x^{18}But since we are looking for the coefficient up to x^7, the terms with x^8 and higher can be ignored. So up to x^7, the expansion is 1 -2x^4 -2x^5.Therefore, the generating function up to x^7 is (1 -2x^4 -2x^5) / (1 - x)^4.So the coefficient of x^7 is the coefficient of x^7 in 1/(1 - x)^4 minus 2 times the coefficient of x^{7 -4}=x^3 in 1/(1 -x)^4 minus 2 times the coefficient of x^{7 -5}=x^2 in 1/(1 -x)^4.The coefficients for 1/(1 -x)^4 are C(n +3, 3). So:Coefficient of x^7: C(7 +3,3)=C(10,3)=120Minus 2*C(3 +3,3)=2*20=40Minus 2*C(2 +3,3)=2*10=20Total: 120 -40 -20=60.So same result as before. Therefore, the number is 60. But wait, let me check once more.Alternatively, maybe using complementary counting. But I think both inclusion-exclusion and generating functions lead to 60.Wait, but let me confirm with another approach.Alternatively, we can model this as integer solutions with x1 <=4, x2 <=4, x3 <=3, x4 <=3, sum to7.We can use stars and bars with inclusion-exclusion.So first, the total solutions without restrictions: C(7 +4 -1,4 -1)=C(10,3)=120.Then subtract the ones where x1 >=5, x2 >=5, x3 >=4, x4 >=4.For x1 >=5: set y1 =x1 -5, so y1 >=0. Then the equation is y1 +x2 +x3 +x4 =7 -5=2. Solutions: C(2 +4 -1,4 -1)=C(5,3)=10. Similarly for x2 >=5:10.For x3 >=4: set y3 =x3 -4, equation becomes x1 +x2 +y3 +x4=7 -4=3. Solutions: C(3 +4 -1,4 -1)=C(6,3)=20. Similarly for x4 >=4:20.Total subtract:10+10+20+20=60. So 120 -60=60.Now, add back the cases where two constraints are violated. But as before, when two variables exceed, the required sum becomes negative, so no solutions. Therefore, no addition. Hence, total is 60. So that's the same answer.Alternatively, maybe doing casework. Let's see.Case 1: Number of a's is 0 to4, same for b's, c's, d's. Let's partition 7 into these four variables with the constraints.But casework might be tedious. Let's see if we can find a pattern.Alternatively, let's consider the variables a and b (which can go up to4) and c and d (up to3). Let's fix the number of a's and b's, then see how many c's and d's are possible.Let me let a + b = t, so t can range from0 to8, but since total sum is7, t ranges from0 to7. Then c + d =7 -t. But c and d can each be at most3, so 7 -t <=6, which implies t >=1. Wait, no. If c and d can each be up to3, then c + d can be up to6. So 7 -t <=6, so t >=1. Wait, but t can be0. If t=0, then c + d=7, but since c and d can only sum to at most6, this is impossible. So t must be from1 to7. But since a and b can each be up to4, t can go from0 to8, but constrained by c + d=7 -t <=6, so t >=1.So t ranges from1 to7. For each t, compute the number of (a,b) pairs where a +b =t, a<=4, b<=4, and number of (c,d) pairs where c +d=7 -t, c<=3, d<=3.So for each t from1 to7:Compute N(t) = [number of a,b with a +b =t, a<=4, b<=4] * [number of c,d with c +d=7 -t, c<=3, d<=3]Then sum over t=1 to7.Let me compute each t:t=1:a +b=1. Possible pairs: (0,1),(1,0). But a and b can be 0. But the constraints are a<=4, b<=4. So yes, both are allowed. So 2 pairs.c +d=6. Since c and d <=3, maximum c +d=6 (3+3). So possible pairs: (3,3). So 1 pair.Thus N(1)=2*1=2.t=2:a +b=2. Possible pairs: (0,2),(1,1),(2,0). So 3 pairs.c +d=5. Since each <=3: possible pairs: (2,3),(3,2). So 2 pairs.N(2)=3*2=6.t=3:a +b=3. Possible pairs: (0,3),(1,2),(2,1),(3,0). 4 pairs.c +d=4. Possible pairs: (1,3),(2,2),(3,1). 3 pairs.N(3)=4*3=12.t=4:a +b=4. Possible pairs: (0,4),(1,3),(2,2),(3,1),(4,0). 5 pairs.c +d=3. Possible pairs: (0,3),(1,2),(2,1),(3,0). 4 pairs.N(4)=5*4=20.t=5:a +b=5. Now, a and b <=4. So possible pairs: (1,4),(2,3),(3,2),(4,1). 4 pairs.c +d=2. Possible pairs: (0,2),(1,1),(2,0). 3 pairs.N(5)=4*3=12.t=6:a +b=6. a and b <=4. So possible pairs: (2,4),(3,3),(4,2). 3 pairs.c +d=1. Possible pairs: (0,1),(1,0). 2 pairs.N(6)=3*2=6.t=7:a +b=7. a and b <=4. The maximum a +b=8 (4+4). But 7 is possible. Let's see: (3,4),(4,3). So 2 pairs.c +d=0. Only (0,0). So 1 pair.N(7)=2*1=2.Now, sum all N(t):t=1:2t=2:6 (total:8)t=3:12 (total:20)t=4:20 (total:40)t=5:12 (total:52)t=6:6 (total:58)t=7:2 (total:60)Total combinations:60. Same result. So this case-by-case approach also gives 60. Therefore, seems correct.Therefore, the number of 7-combinations is 60. So the answer is boxed{60}.Final Answerboxed{60}</think>To find the number of 7-combinations of the multiset ( S = {4 cdot a, 4 cdot b, 3 cdot c, 3 cdot d} ), we need to determine the number of non-negative integer solutions to the equation ( x_1 + x_2 + x_3 + x_4 = 7 ) where ( x_1 leq 4 ), ( x_2 leq 4 ), ( x_3 leq 3 ), and ( x_4 leq 3 ).Using the principle of inclusion-exclusion, we start with the total number of unrestricted solutions, which is given by the stars and bars method:[binom{7 + 4 - 1}{7} = binom{10}{7} = 120]Next, we subtract the cases where one or more variables exceed their maximum limits:- For ( x_1 geq 5 ), we set ( x_1' = x_1 - 5 ), leading to the equation ( x_1' + x_2 + x_3 + x_4 = 2 ). The number of solutions is ( binom{2 + 4 - 1}{2} = binom{5}{2} = 10 ). Similarly for ( x_2 geq 5 ), we get 10 solutions.- For ( x_3 geq 4 ), we set ( x_3' = x_3 - 4 ), leading to the equation ( x_1 + x_2 + x_3' + x_4 = 3 ). The number of solutions is ( binom{3 + 4 - 1}{3} = binom{6}{3} = 20 ). Similarly for ( x_4 geq 4 ), we get 20 solutions.Subtracting these cases:[120 - (10 + 10 + 20 + 20) = 120 - 60 = 60]We then check for overlaps where two variables exceed their limits, but all such cases result in negative sums, which contribute 0. Therefore, the total number of valid solutions remains:[60]This result is confirmed by generating functions and case-by-case analysis, both leading to the same conclusion.Thus, the number of 7-combinations of the multiset ( S ) is (boxed{60}).,<think>Okay, so I need to find the number of 7-combinations of this multiset S, which has 4 a's, 4 b's, 3 c's, and 3 d's. Hmm, let me recall what a combination in a multiset means. I think it's the number of ways to select 7 elements from S, considering the repetition limits of each element. Right, in combinations with repetition, if there were no limits on the number of each element, the formula would be C(n + k - 1, k), where n is the number of types of elements. But here, each element has a maximum number allowed. So this is a problem of combinations with repetition and restrictions. I remember that when there are upper bounds on the number of each element, we can use the inclusion-exclusion principle. The standard formula for combinations with repetition is C(n + k - 1, k), but we have to subtract the cases where we exceed the maximum allowed for each element. Let me structure this. The multiset has four types: a, b, c, d. Each with maximum counts 4, 4, 3, 3 respectively. We need to select 7 elements where the number of a's ≤ 4, b's ≤ 4, c's ≤ 3, and d's ≤ 3. So first, the total number of combinations without any restrictions would be C(7 + 4 - 1, 7) = C(10,7) = 120. Wait, no. Wait, the formula is C(n + k - 1, k), where n is the number of types. Here, n = 4 (a, b, c, d). So it should be C(4 + 7 - 1, 7) = C(10,7) = 120. But that's if there are unlimited quantities. But here, each element is restricted. So we need to adjust for the restrictions.Alternatively, maybe it's better to model it as integer solutions of the equation x1 + x2 + x3 + x4 = 7, where 0 ≤ x1 ≤4, 0 ≤x2 ≤4, 0 ≤x3 ≤3, 0 ≤x4 ≤3. Then, the number of solutions is what we need.Yes, that's another way to put it. So using the inclusion-exclusion principle. Let's recall how that works. The formula for the number of non-negative integer solutions to x1 + x2 + ... + xn = k, with xi ≤ mi for each i is equal to the inclusion-exclusion sum over all subsets of violating the constraints. But maybe step by step. First, compute the total number without restrictions: C(7 + 4 -1, 7) = C(10,7) = 120. Wait, no. Wait, the variables here are four variables (x1, x2, x3, x4), each representing the number of a, b, c, d. So n=4, k=7. So the total solutions without restrictions would be C(7 + 4 -1, 7) = C(10,7) = 120. Correct. Then subtract the solutions where x1 ≥5, x2 ≥5, x3 ≥4, x4 ≥4. So, inclusion-exclusion: subtract the cases where one variable exceeds its maximum, add back the cases where two variables exceed, subtract those where three variables exceed, and add back those where all four exceed. Let me define the variables:Let S be the total number without restrictions: 120.Let A1 be the set where x1 ≥5 (too many a's)A2: x2 ≥5 (too many b's)A3: x3 ≥4 (too many c's)A4: x4 ≥4 (too many d's)Then, the desired number is S - (|A1| + |A2| + |A3| + |A4|) + (|A1∩A2| + |A1∩A3| + |A1∩A4| + |A2∩A3| + |A2∩A4| + |A3∩A4|) - (|A1∩A2∩A3| + |A1∩A2∩A4| + |A1∩A3∩A4| + |A2∩A3∩A4|) + |A1∩A2∩A3∩A4|So we need to compute all these terms.First, compute |A1|: number of solutions where x1 ≥5. Let y1 = x1 -5, then y1 ≥0. Then the equation becomes y1 + x2 + x3 + x4 = 7 -5 = 2. So the number is C(2 +4 -1, 2) = C(5,2)=10. Similarly, |A2|=10, since x2 ≥5 is similar to x1 ≥5.|A3|: x3 ≥4. Let y3 = x3 -4, then the equation becomes x1 + x2 + y3 + x4 =7 -4=3. So number of solutions is C(3 +4 -1,3)=C(6,3)=20. Similarly, |A4|=20.Therefore, |A1| + |A2| + |A3| + |A4| =10+10+20+20=60.Now, compute the intersections.First, |A1∩A2|: x1 ≥5 and x2 ≥5. Let y1=x1-5, y2=x2-5. Then equation: y1 + y2 +x3 +x4=7 -5 -5= -3. Wait, negative. So no solutions. So |A1∩A2|=0.Similarly, |A1∩A3|: x1 ≥5, x3 ≥4. y1=x1-5, y3=x3-4. Then equation: y1 +x2 + y3 +x4=7 -5 -4= -2. Again, negative. So no solutions. Similarly, |A1∩A3|=0. Same for |A1∩A4|=0.Similarly, |A2∩A3|: x2 ≥5, x3 ≥4. y2=x2-5, y3=x3-4. Equation: x1 + y2 + y3 +x4=7 -5 -4= -2. Negative, so 0.Same for |A2∩A4|=0, |A3∩A4|: x3 ≥4 and x4 ≥4. Let y3=x3-4, y4=x4-4. Then equation: x1 +x2 + y3 + y4=7 -4 -4= -1. Negative. So 0.Therefore, all intersections of two sets are 0. So the second term in inclusion-exclusion is 0.Next, intersections of three sets. For example, |A1∩A2∩A3|: x1 ≥5, x2 ≥5, x3 ≥4. Then y1= x1-5, y2=x2-5, y3=x3-4. Equation: y1 + y2 + y3 +x4=7 -5 -5 -4= -7. Negative. So 0. Similarly, all intersections of three sets are 0. Similarly, intersection of four sets is also 0.Therefore, all higher intersections are 0. Hence, the inclusion-exclusion formula simplifies to S - (sum of |Ai|) + 0 -0 +0 = 120 -60=60.Wait, but that seems too straightforward. Let me check again.Wait, the calculation for |A1|, |A2| was 10 each, |A3|, |A4| was 20 each. Sum is 60. Then subtract that from 120 gives 60. But is that correct?Wait, but let me check an example. Suppose we have a smaller case where the answer is known. Let's see.Alternatively, maybe the formula was different. Wait, when you have upper bounds, the inclusion-exclusion is required, but in our case, the intersections of two or more sets are all zero, so the inclusion-exclusion stops after subtracting the single overcounts. Therefore, the answer is 120 -60=60.But let's test this with a different approach. Let's consider generating functions. The generating function for each element is:For a: 1 + x + x² + x³ + x⁴ (since up to 4 a's)Similarly for b: same as a.For c and d: 1 + x + x² + x³ (up to 3 each)Therefore, the generating function is (1 + x + x² + x³ + x⁴)² * (1 + x + x² + x³)²We need the coefficient of x⁷ in this product.Let me compute that. First, let's note that (1 + x + x² + x³ + x⁴) = (1 - x⁵)/(1 - x), and (1 + x + x² + x³) = (1 - x⁴)/(1 - x). Therefore, the generating function is [(1 - x⁵)² (1 - x⁴)²]/(1 - x)^4.So we can write this as (1 - 2x⁵ + x¹⁰)(1 - 2x⁴ + x⁸)/(1 - x)^4.Therefore, expanding the numerator:(1)(1 - 2x⁴ + x⁸) - 2x⁵(1 - 2x⁴ + x⁸) + x¹⁰(1 - 2x⁴ + x⁸)= 1 - 2x⁴ + x⁸ - 2x⁵ + 4x⁹ - 2x¹³ + x¹⁰ - 2x¹⁴ + x¹⁸So, the generating function becomes:[1 - 2x⁴ - 2x⁵ + x⁸ + 4x⁹ + x¹⁰ - 2x¹³ - 2x¹⁴ + x¹⁸] / (1 - x)^4Therefore, the coefficient of x⁷ in this generating function is equal to the coefficient of x⁷ in 1/(1 - x)^4 minus 2 times coefficient of x^(7 -4)=x³ in 1/(1 -x)^4 minus 2 times coefficient of x^(7 -5)=x² in 1/(1 -x)^4 plus coefficient of x^(7 -8)=x^{-1} which is 0, plus 4 times coefficient of x^(7 -9)=x^{-2} which is 0, etc. Similarly, the higher terms (x^8, x^10, etc.) would result in negative exponents for x^(7 - ...), so they contribute 0.Therefore, the coefficient of x⁷ is C(7 +4 -1,7) - 2C(3 +4 -1,3) -2C(2 +4 -1,2)Compute each term:C(10,7) - 2C(6,3) - 2C(5,2)C(10,7)=120, C(6,3)=20, C(5,2)=10.Therefore, 120 - 2*20 -2*10 =120 -40 -20=60.So this matches the inclusion-exclusion result. Therefore, the answer is 60.Wait, so that's confirming it. Therefore, the number of 7-combinations is 60. Hmm, so both methods give 60, so that must be the correct answer.But just to be thorough, let me think if there is another way. Maybe enumerating possible distributions.But with 7 elements and four types, enumerating all possible combinations would be tedious, but perhaps doable for smaller numbers. However, since both generating functions and inclusion-exclusion give 60, and cross-checked, it's likely correct.Alternatively, perhaps considering variables x1, x2 from 0 to4, and x3, x4 from 0 to3, such that x1 +x2 +x3 +x4=7.But even if I try to count, it's going to take time. Let me see.Let me fix x1 and x2, and see how much x3 +x4 can be.So x1 can range from 0 to4, x2 same. For each x1 and x2, x3 +x4 =7 -x1 -x2.But x3 and x4 can be at most 3 each, so x3 +x4 <=6. Therefore, 7 -x1 -x2 <=6 => x1 +x2 >=1.So x1 +x2 can be from 1 to7, but since x1 and x2 are each up to4, the maximum x1 +x2 is 8, but since x3 +x4 >=0, 7 -x1 -x2 >=0 => x1 +x2 <=7. So x1 +x2 ranges from 1 to7.But since x3 and x4 are each up to3, 7 -x1 -x2 must be <=6, so x1 +x2 >=1. But even if x1 +x2=0, x3 +x4=7, which is impossible, so x1 +x2 >=1.But maybe a better approach is to iterate over possible values of x1 +x2.Let me denote t =x1 +x2. Then t can be from max(0,7 -6)=1 to min(7,8)=7. Wait, but x3 +x4=7 -t, which must be between 0 and6. So t can be from 1 to7.But x1 and x2 each can be up to4, so t=x1 +x2 can be from 0 to8, but here x3 +x4=7 -t >=0 => t <=7. So t ranges from 0 to7. But x3 +x4=7 -t must be <=6, so t >=1. Therefore, t ranges from1 to7.For each t from1 to7, compute the number of (x1, x2) pairs where x1 <=4, x2 <=4, x1 +x2=t, and the number of (x3, x4) pairs where x3 <=3, x4 <=3, x3 +x4=7 -t.Multiply these two numbers for each t and sum over t.So let's compute this.First, for each t from1 to7:1. t=1:Number of (x1, x2): x1 +x2=1, x1 <=4, x2 <=4. Possible pairs: (0,1), (1,0). So 2.Number of (x3, x4): x3 +x4=6. But x3 <=3, x4 <=3. The maximum x3 +x4 is6 (3+3). So only one solution: x3=3, x4=3. So 1.So total for t=1: 2*1=2.2. t=2:(x1, x2): (0,2),(1,1),(2,0). So 3.x3 +x4=5. With x3, x4 <=3. Possible solutions:x3=2, x4=3; x3=3, x4=2. So 2.Total: 3*2=6.3. t=3:(x1, x2): (0,3),(1,2),(2,1),(3,0). So 4.x3 +x4=4. x3, x4 <=3. Solutions:(0,4) invalid, (1,3), (2,2), (3,1). But x4=4 is invalid, so valid ones: (1,3), (2,2), (3,1). So 3.Total: 4*3=12.4. t=4:(x1, x2): x1 +x2=4. x1, x2 <=4.Possible pairs: (0,4),(1,3),(2,2),(3,1),(4,0). So 5.x3 +x4=3. x3, x4 <=3. Solutions: (0,3),(1,2),(2,1),(3,0). So 4.Total:5*4=20.5. t=5:(x1, x2)=5, with x1, x2 <=4.Possible pairs: (1,4),(2,3),(3,2),(4,1). So 4.x3 +x4=2. Solutions: (0,2),(1,1),(2,0). So 3.Total:4*3=12.6. t=6:(x1, x2)=6, with x1, x2 <=4.Possible pairs: (2,4),(3,3),(4,2). So 3.x3 +x4=1. Solutions: (0,1),(1,0). So 2.Total:3*2=6.7. t=7:(x1, x2)=7, with x1, x2 <=4. Possible pairs: (3,4),(4,3). So 2.x3 +x4=0. Only one solution: (0,0). So 1.Total:2*1=2.Now sum all these up:t=1:2t=2:6 → total 8t=3:12 → total 20t=4:20 → total 40t=5:12 → total 52t=6:6 → total 58t=7:2 → total 60.So, total 60. Which matches the previous methods. Therefore, this also gives 60. So that's another confirmation. Therefore, I can be confident the answer is 60.Final AnswerThe number of 7-combinations is boxed{60}.</think>To find the number of 7-combinations of the multiset ( S = {4 cdot a, 4 cdot b, 3 cdot c, 3 cdot d} ), we need to consider the restrictions on the number of each element. This problem can be approached using the inclusion-exclusion principle and generating functions.First, we use the inclusion-exclusion principle. The total number of unrestricted 7-combinations of 4 elements is given by ( binom{7 + 4 - 1}{7} = binom{10}{7} = 120 ).Next, we subtract the cases where we exceed the maximum allowed for each element:- For ( x_1 geq 5 ) (too many ( a )'s), the number of solutions is ( binom{7 - 5 + 4 - 1}{7 - 5} = binom{5}{2} = 10 ). Similarly for ( x_2 geq 5 ), we get 10.- For ( x_3 geq 4 ) (too many ( c )'s), the number of solutions is ( binom{7 - 4 + 4 - 1}{7 - 4} = binom{6}{3} = 20 ). Similarly for ( x_4 geq 4 ), we get 20.Summing these, we subtract ( 10 + 10 + 20 + 20 = 60 ).We then check for intersections of these cases, but all intersections of two or more sets result in negative equations, leading to zero solutions. Thus, the inclusion-exclusion principle simplifies to ( 120 - 60 = 60 ).To confirm, we use generating functions. The generating function for each element is:- For ( a ) and ( b ): ( 1 + x + x^2 + x^3 + x^4 )- For ( c ) and ( d ): ( 1 + x + x^2 + x^3 )The generating function is ( (1 + x + x^2 + x^3 + x^4)^2 (1 + x + x^2 + x^3)^2 ). Expanding and finding the coefficient of ( x^7 ), we get the same result using inclusion-exclusion.Finally, enumerating possible distributions of elements confirms the result. Summing all valid combinations for each possible value of ( x_1 + x_2 ) and corresponding ( x_3 + x_4 ) gives a total of 60.Thus, the number of 7-combinations of the multiset ( S ) is (boxed{60}).]

🔑:Solution: Let the required number be N, then N is the coefficient of t^{7} in the expansion ofA(t)=left(1+t+t^{2}+t^{3}+t^{4}right)^{2}left(1+t+t^{2}+t^{3}right)^{2}andbegin{aligned}A(t) & =left(frac{1-t^{5}}{1-t}right)^{2}left(frac{1-t^{4}}{1-t}right)^{2} & =left(1-t^{5}right)^{2}left(1-t^{4}right)^{2}(1-t)^{-4} & =left(1-2 t^{5}+t^{10}right)left(1-2 t^{4}+t^{8}right)(1-t)^{-4} & =left(1-2 t^{4}-2 t^{5}+t^{8}+cdotsright) sum_{k=0}^{infty}binom{k+3}{3} t^{k},end{aligned}Therefore,begin{aligned}N & =binom{7+3}{3}-2binom{3+3}{3}-2binom{2+3}{3} & =binom{10}{3}-2 cdotbinom{6}{3}-2 cdotbinom{5}{3} & =120-40-20=60 .end{aligned}

🧠:[<think>Okay, let's tackle this problem step by step. First, I need to understand what's being asked here. We have a sequence defined by a first term a₁ = 2/3, and each subsequent term is given by the recursive formula aₙ₊₁ = 2aₙ/(aₙ + 1). Part (1) asks us to prove that the sequence {1/aₙ - 1} is a geometric sequence. Hmm, geometric sequences have a common ratio, right? So if we can show that each term is a constant multiple of the previous term, then it's geometric. Let me write down what we need to do here.Let me define bₙ = 1/aₙ - 1 as given in part (2). Maybe part (1) is setting us up for part (2). Anyway, starting with part (1). Let's compute bₙ₊₁ in terms of bₙ. If we can express bₙ₊₁ = r * bₙ for some constant r, then we have a geometric sequence.Given aₙ₊₁ = 2aₙ/(aₙ + 1). Let's express bₙ₊₁ in terms of aₙ₊₁:bₙ₊₁ = 1/aₙ₊₁ - 1 = (aₙ + 1)/(2aₙ) - 1. Wait, let's do that step by step.First, substitute aₙ₊₁ into the expression for bₙ₊₁:bₙ₊₁ = 1/aₙ₊₁ - 1 = 1/(2aₙ/(aₙ + 1)) - 1.Dividing by a fraction is the same as multiplying by its reciprocal, so:1/(2aₙ/(aₙ + 1)) = (aₙ + 1)/(2aₙ)Therefore, bₙ₊₁ = (aₙ + 1)/(2aₙ) - 1Simplify this expression:= (aₙ + 1 - 2aₙ)/(2aₙ) = (1 - aₙ)/(2aₙ)But 1 - aₙ can be written as (1 - aₙ), but we have bₙ = 1/aₙ - 1. Let's see if we can express this in terms of bₙ.Given that bₙ = 1/aₙ - 1, then 1/aₙ = bₙ + 1. Therefore, substituting into the expression for bₙ₊₁:(1 - aₙ)/(2aₙ) = [1 - aₙ]/[2aₙ] = [ (1 - aₙ)/aₙ ] / 2 = [1/aₙ - 1]/2 = (bₙ)/2So, bₙ₊₁ = (1 - aₙ)/(2aₙ) = (bₙ)/2Therefore, we have bₙ₊₁ = (1/2) * bₙ. So this shows that the sequence {bₙ} is a geometric sequence with common ratio 1/2.Wait, but let me check if that's correct. Let me verify with the first term.Given a₁ = 2/3, so b₁ = 1/a₁ - 1 = 1/(2/3) - 1 = 3/2 - 1 = 1/2.Then, according to the recursion, b₂ should be (1/2)*b₁ = (1/2)*(1/2) = 1/4.Let's compute a₂ using the recursive formula:a₂ = 2a₁/(a₁ + 1) = 2*(2/3)/(2/3 + 1) = (4/3)/(5/3) = 4/5.Then, b₂ = 1/a₂ - 1 = 1/(4/5) - 1 = 5/4 - 1 = 1/4. Perfect, matches. So the common ratio is indeed 1/2, and the first term is 1/2. Therefore, {bₙ} is a geometric sequence with first term 1/2 and ratio 1/2. Hence, part (1) is proven.Moving on to part (2), which asks to find the sum Sₙ = sum_{k=1}^n [k * b_k].First, since we know that {bₙ} is a geometric sequence with b₁ = 1/2 and common ratio r = 1/2, the general term bₙ is b₁ * r^{n-1} = (1/2)*(1/2)^{n-1} = (1/2)^n. Wait, let's confirm that.Wait, geometric sequence term is bₙ = b₁ * r^{n-1} = (1/2) * (1/2)^{n-1} = (1/2)^n. So yes, bₙ = (1/2)^n. Let me check with n=1: b₁=1/2, which is (1/2)^1=1/2. n=2: (1/2)^2=1/4, which matches our earlier calculation. So that's correct.Therefore, the sequence {bₙ} is given by bₙ = (1/2)^n.Therefore, the term we need to sum is n * bₙ = n * (1/2)^n. So Sₙ = sum_{k=1}^n k*(1/2)^k.Hmm, so we need a formula for the sum S_n = sum_{k=1}^n k x^k, where x=1/2. There is a standard formula for this kind of sum.Recall that for the sum S = sum_{k=0}^infty k x^k, which converges for |x| < 1, the sum is x/(1 - x)^2. However, since we are dealing with a finite sum up to n terms, the formula is a bit different. Let me recall.The formula for the finite sum sum_{k=1}^n k x^{k} can be derived using generating functions or by manipulating the infinite sum formula. Let me recall the method.Let S = sum_{k=1}^n k x^k.We can write S = x + 2x^2 + 3x^3 + ... + n x^n.Multiply both sides by x:xS = x^2 + 2x^3 + 3x^4 + ... + n x^{n+1}Subtract the second equation from the first:S - xS = (x + x^2 + x^3 + ... + x^n) - n x^{n+1}The left side is S(1 - x). The right side is the sum of a geometric series from x to x^n minus n x^{n+1}.The sum x + x^2 + ... + x^n is x(1 - x^n)/(1 - x). Therefore,S(1 - x) = x(1 - x^n)/(1 - x) - n x^{n+1}Therefore,S = [x(1 - x^n)/(1 - x)^2 - n x^{n+1}/(1 - x)]So, simplifying,S = [x - (n + 1) x^{n + 1} + n x^{n + 2}]/(1 - x)^2Alternatively, written as:S = x(1 - (n + 1)x^n + n x^{n + 1}) / (1 - x)^2But let me check this with a small n. Let's take n=1:S = 1 * x = x. According to the formula:[x(1 - 2x^1 + 1 x^2)] / (1 - x)^2 = [x(1 - 2x + x^2)]/(1 - x)^2 = [x(1 - x)^2]/(1 - x)^2 = x. Correct.For n=2:S = x + 2x^2. According to the formula:[x(1 - 3x^2 + 2x^3)] / (1 - x)^2. Let's compute numerator:x - 3x^3 + 2x^4. Hmm, but denominator is (1 - x)^2.Wait, maybe I made a miscalculation. Let me recast the formula:Wait, from above:S(1 - x) = x(1 - x^n)/(1 - x) - n x^{n+1}Therefore,S = [x(1 - x^n)] / (1 - x)^2 - [n x^{n+1}] / (1 - x)So, writing that out:S = [x(1 - x^n) / (1 - x)^2] - [n x^{n + 1} / (1 - x)]To combine these terms, factor out x/(1 - x)^2:S = x/(1 - x)^2 [1 - x^n - n x^n (1 - x)]Simplify the bracket:1 - x^n - n x^n + n x^{n + 1}= 1 - x^n(1 + n) + n x^{n + 1}So, S = x/(1 - x)^2 [1 - (n + 1)x^n + n x^{n + 1}]Yes, that's correct. So, substituting x = 1/2.Therefore, S_n = (1/2)/(1 - 1/2)^2 [1 - (n + 1)(1/2)^n + n(1/2)^{n + 1}] - [n (1/2)^{n + 1}/(1 - 1/2)]Wait, actually, the formula is:S_n = [x(1 - x^n)/(1 - x)^2 - n x^{n + 1}/(1 - x)]Substituting x = 1/2:First term: (1/2)(1 - (1/2)^n)/(1 - 1/2)^2 = (1/2)(1 - 1/2^n)/(1/2)^2 = (1/2)(1 - 1/2^n)/(1/4) = (1/2)(4)(1 - 1/2^n) = 2(1 - 1/2^n)Second term: n*(1/2)^{n + 1}/(1 - 1/2) = n*(1/2)^{n + 1}/(1/2) = n*(1/2)^{n + 1}*2 = n*(1/2)^nTherefore, S_n = 2(1 - 1/2^n) - n*(1/2)^nSimplify:= 2 - 2/2^n - n/2^n= 2 - (2 + n)/2^nSo, S_n = 2 - (n + 2)/2^nLet me verify with n=1: S₁ = 1*(1/2) = 1/2. According to the formula: 2 - (1 + 2)/2^1 = 2 - 3/2 = 1/2. Correct.n=2: 1*(1/2) + 2*(1/4) = 1/2 + 1/2 = 1. Formula: 2 - (2 + 2)/4 = 2 - 4/4 = 2 - 1 = 1. Correct.n=3: 1*(1/2) + 2*(1/4) + 3*(1/8) = 1/2 + 1/2 + 3/8 = 1 + 3/8 = 11/8. Formula: 2 - (3 + 2)/8 = 2 - 5/8 = 16/8 - 5/8 = 11/8. Correct.Good, so the formula seems to hold. Therefore, the sum S_n is 2 - (n + 2)/2^n.Therefore, the answer to part (2) is S_n = 2 - (n + 2)/2^n.But let me write it in LaTeX notation.So, summarizing:Part (1): We defined bₙ = 1/aₙ - 1 and showed that bₙ₊₁ = (1/2) bₙ, hence it's a geometric sequence with ratio 1/2 and first term b₁ = 1/2.Part (2): The sum S_n = sum_{k=1}^n k b_k = sum_{k=1}^n k (1/2)^k = 2 - (n + 2)/2^n.Therefore, the answers are:(1) Proven as above.(2) S_n = 2 - (n + 2)/2^n.Final Answer(1) boxed{left{ frac{1}{a_n} - 1 right} text{ is a geometric sequence with common ratio } frac{1}{2}}.(2) The sum of the first ( n ) terms of the sequence ( {nb_n} ) is boxed{2 - frac{n + 2}{2^n}}.</think>Given the sequence {a_n} with the first term a_1= frac{2}{3}, and a_{n+1}= frac{2a_n}{a_n+1} for n=1,2,cdots:(1) To prove that the sequence left{ frac{1}{a_n}-1right} is a geometric sequence, we define b_n = frac{1}{a_n} - 1. We need to show that b_{n+1} = r cdot b_n for some constant r.Starting with the given recursive formula:[a_{n+1} = frac{2a_n}{a_n + 1}]we express b_{n+1} in terms of a_{n+1}:[b_{n+1} = frac{1}{a_{n+1}} - 1 = frac{a_n + 1}{2a_n} - 1 = frac{a_n + 1 - 2a_n}{2a_n} = frac{1 - a_n}{2a_n}]Since b_n = frac{1}{a_n} - 1, we have frac{1}{a_n} = b_n + 1. Substituting this into the expression for b_{n+1}:[b_{n+1} = frac{1 - a_n}{2a_n} = frac{frac{1}{a_n} - 1}{2} = frac{b_n}{2}]Thus, b_{n+1} = frac{1}{2} b_n, showing that {b_n} is a geometric sequence with common ratio frac{1}{2} and the first term b_1 = frac{1}{2}.(2) To find the sum of the first n terms of the sequence {nb_n}, denoted as S_n, we note that b_n = left(frac{1}{2}right)^n. Therefore, we need to find:[S_n = sum_{k=1}^n k left(frac{1}{2}right)^k]Using the formula for the sum of the series sum_{k=1}^n kx^k with x = frac{1}{2}, we have:[S_n = frac{x(1 - (n+1)x^n + nx^{n+1})}{(1 - x)^2}]Substituting x = frac{1}{2}:[S_n = frac{frac{1}{2}left(1 - (n+1)left(frac{1}{2}right)^n + nleft(frac{1}{2}right)^{n+1}right)}{left(1 - frac{1}{2}right)^2}]Simplifying this expression:[S_n = frac{frac{1}{2}left(1 - (n+1)left(frac{1}{2}right^nright) + nleft(frac{1}{2}right)^{n+1}}{left(frac{1}{2}right)^2} = 2 - frac{n + 2}{2^n}] Final Answer(1) boxed{left{ frac{1}{a_n} - 1 right} text{ is a geometric sequence with common ratio } frac{1}{2}}.(2) The sum of the first ( n ) terms of the sequence ( {nb_n} ) is boxed{2 - frac{n + 2}{2^n}}.,<think>Okay, so I need to solve this problem about the sequence {a_n}. Let me start by understanding what's given and what needs to be proven. First, the sequence is defined with a₁ = 2/3, and the recursive formula a_{n+1} = (2a_n)/(a_n + 1) for n ≥ 1. The first part asks me to prove that the sequence {1/a_n - 1} is a geometric sequence. Hmm, geometric sequences have a common ratio, right? So I need to show that each term of {1/a_n - 1} is a constant multiple of the previous term.Let me write down what's given and see if I can manipulate the recursive formula to get something useful. Let's denote b_n = 1/a_n - 1. Then, part (1) wants me to show that {b_n} is a geometric sequence. If I can express b_{n+1} in terms of b_n and show that it's a constant ratio, that would do it.So, starting with the recursive formula for a_n: a_{n+1} = (2a_n)/(a_n + 1). Let's try to express b_{n+1} in terms of a_{n+1} and then relate that to a_n.Given that b_n = 1/a_n - 1, then 1/a_n = b_n + 1. Therefore, a_n = 1/(b_n + 1). Maybe substituting this into the recursive formula for a_{n+1} would help.Let's compute a_{n+1}:a_{n+1} = (2a_n)/(a_n + 1) = [2*(1/(b_n + 1))]/[1/(b_n + 1) + 1]Let me simplify the denominator first:1/(b_n + 1) + 1 = [1 + (b_n + 1)]/(b_n + 1) = (b_n + 2)/(b_n + 1)So the entire expression for a_{n+1} becomes:[2/(b_n + 1)] / [(b_n + 2)/(b_n + 1)] = [2/(b_n + 1)] * [(b_n + 1)/(b_n + 2)] = 2/(b_n + 2)Therefore, a_{n+1} = 2/(b_n + 2)Now, let's compute b_{n+1} = 1/a_{n+1} - 1 = (b_n + 2)/2 - 1 = (b_n + 2 - 2)/2 = b_n/2So b_{n+1} = (1/2) * b_n. This shows that each term is half of the previous one, so {b_n} is a geometric sequence with common ratio 1/2. That proves part (1). Wait, let me double-check my steps to make sure I didn't make a mistake.Starting from a_{n+1} = 2a_n/(a_n + 1). Then substituting a_n = 1/(b_n + 1):a_{n+1} = [2*(1/(b_n + 1))]/[1/(b_n + 1) + 1]Calculating the denominator:1/(b_n + 1) + 1 = (1 + b_n + 1)/(b_n + 1) = (b_n + 2)/(b_n + 1). That seems right.Then the numerator is 2/(b_n + 1). So dividing numerator by denominator:[2/(b_n + 1)] / [(b_n + 2)/(b_n + 1)] = 2/(b_n + 2). Correct.Therefore, a_{n+1} = 2/(b_n + 2). Then 1/a_{n+1} = (b_n + 2)/2. Then subtract 1 to get b_{n+1}:(b_n + 2)/2 - 1 = (b_n + 2 - 2)/2 = b_n/2. Perfect, so b_{n+1} = (1/2) * b_n. So indeed, the sequence {b_n} is geometric with ratio 1/2. Great, part (1) is done.Now, part (2) says let b_n = 1/a_n -1, find the sum S_n of the first n terms of the sequence {n b_n}. So S_n = 1*b_1 + 2*b_2 + 3*b_3 + ... + n*b_n. Need to find a formula for this sum.First, since we know {b_n} is a geometric sequence with ratio 1/2, let's find its first term. Given a₁ = 2/3, so b₁ = 1/a₁ -1 = 1/(2/3) -1 = 3/2 -1 = 1/2. So b₁ = 1/2, and since the ratio is 1/2, the sequence {b_n} is (1/2), (1/2)^2, (1/2)^3, ..., so in general b_n = (1/2)^n.Therefore, nb_n = n*(1/2)^n. So S_n is the sum from k=1 to n of k*(1/2)^k.This is a standard sum. I remember there's a formula for the sum of k*r^k from k=1 to n. Let me recall how to derive it.Let S = sum_{k=1}^n k r^kThen, multiply both sides by r:rS = sum_{k=1}^n k r^{k+1} = sum_{k=2}^{n+1} (k-1) r^kSubtracting the two equations:S - rS = sum_{k=1}^n k r^k - sum_{k=2}^{n+1} (k-1) r^kLeft side: S(1 - r)Right side: sum_{k=1}^n k r^k - sum_{k=2}^{n+1} (k-1) r^kLet's adjust the indices. The first sum is from k=1 to n, and the second sum is from k=2 to n+1. Let's write out the terms:First sum: 1*r^1 + 2*r^2 + 3*r^3 + ... + n*r^nSecond sum: (2-1)r^2 + (3-1)r^3 + ... + (n -1) r^n + (n+1 -1) r^{n+1} = 1*r^2 + 2*r^3 + ... + (n-1)r^n + n r^{n+1}Subtracting the second sum from the first sum:[1*r^1 + 2*r^2 + 3*r^3 + ... + n*r^n] - [1*r^2 + 2*r^3 + ... + (n-1)r^n + n r^{n+1}]= 1*r^1 + (2*r^2 -1*r^2) + (3*r^3 -2*r^3) + ... + (n*r^n - (n-1)r^n) - n r^{n+1}= r + r^2 + r^3 + ... + r^n - n r^{n+1}That's a geometric series from k=1 to n of r^k, minus n r^{n+1}So S(1 - r) = [r(1 - r^n)/(1 - r)] - n r^{n+1}Therefore, solving for S:S = [r(1 - r^n)/(1 - r)^2] - [n r^{n+1}/(1 - r)]Alternatively, S = r/(1 - r)^2 - [r^{n+1}/(1 - r)^2 + n r^{n+1}/(1 - r)]But perhaps it's better to write it as:S = [r - (n + 1) r^{n + 1} + n r^{n + 2}] / (1 - r)^2Wait, maybe better to keep the two terms separate.But in any case, for our specific case, r = 1/2.So plugging r = 1/2 into the formula:S = [ (1/2)(1 - (1/2)^n ) / (1 - 1/2)^2 ] - [n (1/2)^{n + 1} / (1 - 1/2) ]First compute denominators:(1 - 1/2)^2 = (1/2)^2 = 1/41 - 1/2 = 1/2Therefore, first term:(1/2)(1 - (1/2)^n ) / (1/4) = (1/2)(1 - (1/2)^n ) * 4 = 2(1 - (1/2)^n )Second term:n (1/2)^{n + 1} / (1/2) = n (1/2)^{n + 1} * 2 = n (1/2)^nSo putting together:S = 2(1 - (1/2)^n ) - n (1/2)^n = 2 - 2*(1/2)^n - n*(1/2)^n = 2 - (2 + n)*(1/2)^nTherefore, S_n = 2 - (n + 2)(1/2)^nLet me check with small n to see if this works.For n =1: S_1 = 1*b_1 =1*(1/2) =1/2Plugging into formula: 2 - (1 + 2)(1/2)^1 = 2 - 3*(1/2) = 2 - 1.5 = 0.5. Correct.For n=2: S_2 =1*(1/2) +2*(1/4)=1/2 + 1/2=1Formula: 2 - (2 +2)(1/2)^2=2 -4*(1/4)=2 -1=1. Correct.n=3: S_3=1*(1/2)+2*(1/4)+3*(1/8)=1/2 +1/2 +3/8=1 + 3/8=11/8=1.375Formula: 2 - (3+2)(1/2)^3=2 -5*(1/8)=2 -5/8=11/8=1.375. Correct.n=4: S_4=1*(1/2)+2*(1/4)+3*(1/8)+4*(1/16)=1/2+1/2+3/8+1/4=1 + 3/8 + 1/4=1 + 3/8 + 2/8=1 +5/8=13/8=1.625Formula:2 - (4 +2)(1/2)^4=2 -6*(1/16)=2 -6/16=2 -3/8=13/8=1.625. Correct.Looks like the formula works. Therefore, the sum S_n=2 - (n+2)(1/2)^n.Alternatively, since (1/2)^n is the same as 1/2^n, so S_n=2 - (n + 2)/2^n.Thus, the answer to part (2) is 2 - (n + 2)/2^n.I need to present this in boxed notation. Let me recap:(1) We showed that b_{n+1} = (1/2) b_n, so {b_n} is geometric with ratio 1/2.(2) The sum S_n = sum_{k=1}^n k b_k = 2 - (n + 2)/2^n.Final Answer(1) boxed{left{ frac{1}{a_n} - 1 right} text{ is a geometric sequence with common ratio } frac{1}{2}}(2) boxed{2 - frac{n + 2}{2^n}}</think>Given the sequence {a_n} with the first term a_1 = frac{2}{3} and a_{n+1} = frac{2a_n}{a_n + 1} for n = 1, 2, cdots:(1) To prove that the sequence left{ frac{1}{a_n} - 1 right} is a geometric sequence, we define b_n = frac{1}{a_n} - 1. We need to show that b_{n+1} = frac{1}{2} b_n. Starting with the recursive formula for a_n:[a_{n+1} = frac{2a_n}{a_n + 1}]Substitute a_n = frac{1}{b_n + 1} into the formula:[a_{n+1} = frac{2 cdot frac{1}{b_n + 1}}{frac{1}{b_n + 1} + 1} = frac{frac{2}{b_n + 1}}{frac{1 + b_n + 1}{b_n + 1}} = frac{2}{b_n + 2}]Thus,[frac{1}{a_{n+1}} = frac{b_n + 2}{2}]Therefore,[b_{n+1} = frac{1}{a_{n+1}} - 1 = frac{b_n + 2}{2} - 1 = frac{b_n}{2}]This shows that b_{n+1} = frac{1}{2} b_n, proving that left{ frac{1}{a_n} - 1 right} is a geometric sequence with common ratio frac{1}{2}.(2) Let b_n = frac{1}{a_n} - 1. Since b_n is a geometric sequence with the first term b_1 = frac{1}{2} and common ratio frac{1}{2}, we have b_n = left( frac{1}{2} right)^n. We need to find the sum S_n of the first n terms of the sequence {n b_n}.The sum S_n is given by:[S_n = sum_{k=1}^n k b_k = sum_{k=1}^n k left( frac{1}{2} right)^k]Using the formula for the sum of k r^k from k=1 to n with r = frac{1}{2}:[S = frac{r(1 - r^n)}{(1 - r)^2} - frac{n r^{n+1}}{1 - r}]Substituting r = frac{1}{2}:[S_n = frac{frac{1}{2} left(1 - left( frac{1}{2} right)^n right)}{left(1 - frac{1}{2}right)^2} - frac{n left( frac{1}{2} right)^{n+1}}{1 - frac{1}{2}}]Simplifying the terms:[frac{frac{1}{2} left(1 - left( frac{1}{2} right)^n right)}{left( frac{1}{2} right)^2} = 2 left(1 - left( frac{1}{2} right)^n right)][frac{n left( frac{1}{2} right)^{n+1}}{frac{1}{2}} = n left( frac{1}{2} right)^n]Combining these results:[S_n = 2 left(1 - left( frac{1}{2} right)^n right) - n left( frac{1}{2} right)^n = 2 - left( n + 2 right) left( frac{1}{2} right)^n]Thus, the sum of the first n terms of the sequence {n b_n} is:[boxed{2 - frac{n + 2}{2^n}}] Final Answer(1) boxed{left{ frac{1}{a_n} - 1 right} text{ is a geometric sequence with common ratio } frac{1}{2}}(2) boxed{2 - frac{n + 2}{2^n}}]

🔑:(1) Proof: Since a_{n+1}= frac{2a_n}{a_n+1},therefore 1+ frac {1}{a_n}= frac {2}{a_{n+1}},therefore frac {1}{a_{n+1}}-1= frac {1}{2}( frac {1}{a_n}-1),Given a_1= frac{2}{3},thereforefrac{1}{a_1}-1= frac{1}{2}.therefore The sequence left{frac{1}{a_n} -1right} is a geometric sequence;(2) Solution: From (1), we have: frac{1}{a_n}-1= frac{1}{2}timesleft( frac{1}{2}right)^{n-1}=left( frac{1}{2}right)^n,therefore So nb_n= frac{n}{2^n}.Let S_n= frac {1}{2}+ frac {2}{2^2}+ frac {3}{2^3}+cdots+ frac {n}{2^n},frac{1}{2}S_n = frac {1}{2^2}+ frac {2}{2^3}+ frac {3}{2^4}+cdots+ frac {n-1}{2^n}+ frac {n}{2^{n+1}},thereforefrac{1}{2}S_n= frac{1}{2}+ frac{1}{2^2}+ frac{1}{2^3} +cdots+frac{1}{2^n}- frac{n}{2^{n+1}} =frac{ frac{1}{2}left(1- frac{1}{2^n}right)}{1- frac{1}{2}}- frac{n}{2^{n+1}}=1- frac{2+n}{2^{n+1}},therefore S_n=2- frac{2+n}{2^n}. Thus, the sum of the first n terms of the sequence {nb_n} is boxed{2- frac{2+n}{2^n}}.