| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The given product is a well-known Wallis product which converges to $\frac{\pi}{2}$. The expression can be rewritten as: $$\frac{2}{\pi} \cdot \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{2}{\pi} \cdot \prod_{k=1}^{\infty} \frac{4k^2}{4k^2-1} = \frac{2}{\pi} \cdot \frac{\pi}{2} = 1.$$ Thus, the value of the given expression is $1$. |
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| deepseekr1 | The problem asks for the value of \( \frac{2}{\pi} \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} \). First, we recognize that the product part of the expression can be simplified. Each term in the product is: \[ \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{4k^2}{4k^2 - 1} \] This product is known as the Wallis product for \( \pi/2 \). The Wallis product formula states: \[ \frac{\pi}{2} = \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} \] Thus, the given product \( \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} \) equals \( \frac{\pi}{2} \). Multiplying this result by \( \frac{2}{\pi} \) gives: \[ \frac{2}{\pi} \times \frac{\pi}{2} = 1 \] Therefore, the value of the expression is: \[ \boxed{1} \] |
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| gpto1 | To find the value of the given infinite product: \[ P = \frac{2}{\pi} \times \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} \] We can recognize that this product is closely related to Wallis' formula for \(\pi\). Let's first simplify the terms of the product. Note that: \[ (2k)^2 = 4k^2 \] \[ (2k - 1)(2k + 1) = (2k)^2 - 1 = 4k^2 - 1 \] So, each term of the product simplifies to: \[ \frac{4k^2}{4k^2 - 1} \] Therefore, the product becomes: \[ P = \frac{2}{\pi} \times \prod_{k=1}^{\infty} \frac{4k^2}{4k^2 - 1} \] Now, recall Wallis' product, which states: \[ \frac{\pi}{2} = \prod_{k=1}^{\infty} \frac{4k^2}{4k^2 - 1} \] Comparing this with our product, we see that: \[ \prod_{k=1}^{\infty} \frac{4k^2}{4k^2 - 1} = \frac{\pi}{2} \] Substituting back into our expression for \(P\): \[ P = \frac{2}{\pi} \times \frac{\pi}{2} = 1 \] **Answer: 1** |
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| GPT-4o | To find the value of the given infinite product, we start with the expression: \[ P = \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} \] Let's break down the general term of the product: \[ \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{4k^2}{(2k-1)(2k+1)} \] Observe that: \[ (2k-1)(2k+1) = 4k^2 - 1 \] Thus, the term can be rewritten as: \[ \frac{4k^2}{4k^2 - 1} \] Next, we can factor and simplify the denominator: \[ 4k^2 - 1 = (2k-1)(2k+1) \] So the term becomes: \[ \frac{4k^2}{4k^2 - 1} = \frac{4k^2}{(2k-1)(2k+1)} \] Now, consider the infinite product: \[ P = \prod_{k=1}^{\infty} \frac{4k^2}{4k^2 - 1} \] This can be split into partial fractions to give: \[ \frac{4k^2}{4k^2 - 1} = 1 + \frac{1}{4k^2 - 1} \] However, a more insightful approach involves recognizing this as a telescoping product by rewriting each term: \[ \frac{4k^2}{4k^2 - 1} = \frac{4k^2}{(2k-1)(2k+1)} \] We can rewrite this as a ratio of factorials: \[ \frac{4k^2}{(2k-1)(2k+1)} = \frac{2k}{2k-1} \cdot \frac{2k}{2k+1} \] The product then becomes: \[ P = \prod_{k=1}^{\infty} \left( \frac{2k}{2k-1} \cdot \frac{2k}{2k+1} \right) \] Observe that the product has a telescoping nature. Let's write out the first few terms to see the cancellation: \[ P = \left( \frac{2 \cdot 1}{1 \cdot 3} \right) \cdot \left( \frac{2 \cdot 2}{3 \cdot 5} \right) \cdot \left( \frac{2 \cdot 3}{5 \cdot 7} \right) \cdot \ldots \] Most terms in the numerator and denominator will cancel out except for the initial few terms, leading to: \[ P = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \ldots \] Now, we simplify and identify the pattern. The product simplifies to: \[ P = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \ldots = \frac{2 \cdot 4 \cdot 6 \cdot \ldots}{1 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot \ldots} \] To evaluate this, we use known results from infinite product representations of \(\pi\). It is known that: \[ \prod_{k=1}^{\infty} \frac{4k^2}{4k^2 - 1} = \frac{\pi}{2} \] Therefore, substituting back into our original problem: \[ 2/\pi * \frac{\pi}{2} = 1 \] Hence, the value is: \[ \boxed{1} \] |
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| llama3.1-70B-Instruct | A nice infinite product! Let's break it down step by step: $$\prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} = \prod_{k=1}^{\infty} \frac{4k^2}{4k^2 - 1}$$ Now, we can use the fact that $\frac{4k^2}{4k^2 - 1} = \frac{1}{1 - \frac{1}{4k^2}}$. So, we have: $$\prod_{k=1}^{\infty} \frac{1}{1 - \frac{1}{4k^2}}$$ This is a telescoping product, meaning that each term cancels with the next one, leaving only the first and last terms. To see this, let's write out the first few terms: $$\frac{1}{1 - \frac{1}{4}} \cdot \frac{1}{1 - \frac{1}{16}} \cdot \frac{1}{1 - \frac{1}{36}} \cdot \frac{1}{1 - \frac{1}{64}} \cdot \ldots$$ Now, notice that the denominator of each term cancels with the numerator of the next term: $$\frac{1}{1 - \frac{1}{4}} \cdot \frac{1}{1 - \frac{1}{16}} = \frac{1}{1 - \frac{1}{4}} \cdot \frac{16}{15} = \frac{4}{3} \cdot \frac{16}{15} = \frac{64}{45}$$ $$\frac{64}{45} \cdot \frac{1}{1 - \frac{1}{36}} = \frac{64}{45} \cdot \frac{36}{35} = \frac{2304}{1575}$$ And so on. In the limit, all the terms cancel, leaving only the first term: $$\frac{1}{1 - \frac{1}{4}} = \frac{4}{3}$$ So, we have: $$\prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{4}{3}$$ Now, we can multiply by $2/\pi$ to get the final answer: $$\frac{2}{\pi} \cdot \frac{4}{3} = \frac{8}{3\pi}$$ Therefore, the value of the given expression is $\boxed{\frac{8}{3\pi}}$. |
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| llama3.1-405B-Instruct | The value of the given product can be evaluated as follows: $$\prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} = \prod_{k=1}^{\infty} \frac{4k^2}{(2k-1)(2k+1)}$$ We can rewrite the product as: $$\prod_{k=1}^{\infty} \frac{4k^2}{(2k-1)(2k+1)} = \prod_{k=1}^{\infty} \frac{4k^2}{(2k-1)(2k+1)} \cdot \frac{2k-1}{2k-1} \cdot \frac{2k+1}{2k+1}$$ This allows us to cancel out the terms: $$\prod_{k=1}^{\infty} \frac{4k^2}{(2k-1)(2k+1)} \cdot \frac{2k-1}{2k-1} \cdot \frac{2k+1}{2k+1} = \prod_{k=1}^{\infty} \frac{4k^2(2k-1)(2k+1)}{(2k-1)^2(2k+1)^2}$$ Simplifying further, we get: $$\prod_{k=1}^{\infty} \frac{4k^2(2k-1)(2k+1)}{(2k-1)^2(2k+1)^2} = \prod_{k=1}^{\infty} \frac{4k^2}{(2k-1)(2k+1)}$$ Now, we can use the fact that: $$\prod_{k=1}^{\infty} \frac{4k^2}{(2k-1)(2k+1)} = \frac{\pi}{2}$$ Therefore, the value of the original product is: $$\frac{2}{\pi} \cdot \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{2}{\pi} \cdot \frac{\pi}{2} = 1$$ So, the value of the given product is 1. |
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| DeepSeek-Coder-V2 | To solve the given problem, we need to evaluate the expression: \[ \frac{2}{\pi} \cdot \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} \] First, let's analyze the infinite product part: \[ \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} \] We can rewrite each term in the product: \[ \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{4k^2}{(2k-1)(2k+1)} \] Now, let's consider the product of these terms from \( k = 1 \) to \( k = n \): \[ \prod_{k=1}^n \frac{4k^2}{(2k-1)(2k+1)} \] This can be written as: \[ \frac{4 \cdot 1^2}{1 \cdot 3} \cdot \frac{4 \cdot 2^2}{3 \cdot 5} \cdot \frac{4 \cdot 3^2}{5 \cdot 7} \cdots \frac{4 \cdot n^2}{(2n-1)(2n+1)} \] Notice that many terms cancel out. Specifically, the numerator of each term cancels with part of the denominator of the next term. This pattern continues, and we are left with: \[ \frac{2 \cdot 2}{1 \cdot 3} \cdot \frac{4 \cdot 4}{3 \cdot 5} \cdot \frac{6 \cdot 6}{5 \cdot 7} \cdots \frac{(2n) \cdot (2n)}{(2n-1)(2n+1)} \] As \( n \) approaches infinity, the product simplifies to: \[ \frac{2 \cdot 2}{1 \cdot 3} \cdot \frac{4 \cdot 4}{3 \cdot 5} \cdot \frac{6 \cdot 6}{5 \cdot 7} \cdots = \frac{2}{\pi} \] Thus, the infinite product converges to: \[ \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{\pi}{2} \] Now, we multiply this result by the factor \( \frac{2}{\pi} \): \[ \frac{2}{\pi} \cdot \frac{\pi}{2} = 1 \] Therefore, the value of the given expression is: \[ \boxed{1} \] |
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| Qwen2-72B-Instruct | The given product is a well-known Wallis product which converges to $\frac{\pi}{2}$. The expression can be rewritten as: $$\frac{2}{\pi} \cdot \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{2}{\pi} \cdot \prod_{k=1}^{\infty} \frac{4k^2}{4k^2-1} = \frac{2}{\pi} \cdot \frac{\pi}{2} = 1.$$ Thus, the value of the given expression is $1$. |
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| baichuan4 | To solve this problem, we can analyze the infinite product and make some observations. First, let's write the product more clearly: $$2/\pi \cdot \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)}$$ We can factor out a 4 from the numerator: $$2/\pi \cdot \prod_{k=1}^{\infty} \frac{4k^2}{(2k-1)(2k+1)}$$ Now, let's look at the fraction inside the product: $$\frac{4k^2}{(2k-1)(2k+1)} = \frac{4k^2}{4k^2 - 1}$$ This can be rewritten as: $$\frac{4k^2}{(2k)^2 - 1^2}$$ Using the formula $a^2 - b^2 = (a - b)(a + b)$, we get: $$\frac{4k^2}{(2k - 1)(2k + 1)}$$ Now, let's multiply the first few terms of the product to see if there's a pattern: $$\frac{4 \cdot 1^2}{(2 \cdot 1 - 1)(2 \cdot 1 + 1)} \cdot \frac{4 \cdot 2^2}{(2 \cdot 2 - 1)(2 \cdot 2 + 1)} \cdot \frac{4 \cdot 3^2}{(2 \cdot 3 - 1)(2 \cdot 3 + 1)} \cdots$$ Simplifying the first few terms: $$\frac{4}{1 \cdot 3} \cdot \frac{16}{3 \cdot 5} \cdot \frac{36}{5 \cdot 7} \cdots$$ We can see that the denominators are cancelling out in a telescoping fashion: $$\frac{4}{3} \cdot \frac{16}{5} \cdot \frac{36}{7} \cdots$$ The product will converge to some value. To find the exact value, we need to use a known result. This product is actually a variant of Wallis's product for $\pi$, which states: $$\frac{\pi}{2} = \frac{2 \cdot 2}{2 \cdot 1} \cdot \frac{2 \cdot 4}{4 \cdot 3} \cdot \frac{4 \cdot 6}{6 \cdot 5} \cdot \frac{6 \cdot 8}{8 \cdot 7} \cdots$$ If we divide our product by Wallis's product, we get: $$\frac{2/\pi \cdot 4/3 \cdot 16/5 \cdot 36/7 \cdots}{2 \cdot 2/2 \cdot 1 \cdot 2 \cdot 4/4 \cdot 3 \cdot 4 \cdot 6/6 \cdot 5 \cdot 6 \cdot 8/8 \cdot 7 \cdots}$$ This simplifies to: $$\frac{2/\pi}{2} = \frac{1}{\pi}$$ So, the value of the original product is $\frac{1}{\pi}$. |
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| doubao | Let's consider the infinite product $\prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)}$. We can rewrite each term as: \[ \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{4k^2}{4k^2-1} = 1 + \frac{1}{4k^2-1} \] So the product becomes: \[ \prod_{k=1}^{\infty} \left(1 + \frac{1}{4k^2-1}\right) \] Now, consider the function $\sin x$. Its Taylor series expansion is $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$. We can rewrite $\sin x$ as: \[ \sin x = x \prod_{k=1}^{\infty} \left(1 - \frac{x^2}{(k\pi)^2}\right) \] Letting $x = \frac{\pi}{2}$, we have: \[ 1 = \frac{\pi}{2} \prod_{k=1}^{\infty} \left(1 - \frac{(\frac{\pi}{2})^2}{(k\pi)^2}\right) = \frac{\pi}{2} \prod_{k=1}^{\infty} \left(1 - \frac{1}{4k^2}\right) \] Multiplying both sides by $2$ gives: \[ 2 = \pi \prod_{k=1}^{\infty} \left(1 - \frac{1}{4k^2}\right) \] Now, notice that: \[ \prod_{k=1}^{\infty} \left(1 - \frac{1}{4k^2}\right) = \frac{1}{\prod_{k=1}^{\infty} \left(1 + \frac{1}{4k^2-1}\right)} \] So, $\prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{1}{2/\pi} = \frac{\pi}{2}$. Finally, $2/\pi * \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} = 2/\pi * \frac{\pi}{2} = 1$. Therefore, the answer is $1$. |
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| minimax | The given problem involves an infinite product that can be simplified using Wallis's product for π. Wallis's product is given by: $$ \frac{\pi}{2} = \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} $$ The problem asks for the value of $2/\pi$ times this product. Let's first write down the Wallis's product and then manipulate it to match the given expression: $$ \frac{\pi}{2} = \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} $$ Now, we want to find the value of: $$ \frac{2}{\pi} \cdot \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} $$ Notice that the product we want to evaluate is exactly the reciprocal of the Wallis's product. Therefore, we can write: $$ \frac{2}{\pi} \cdot \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{2}{\pi} \cdot \frac{1}{\frac{\pi}{2}} $$ Simplifying this expression, we get: $$ \frac{2}{\pi} \cdot \frac{2}{\pi} = \frac{4}{\pi^2} $$ So the value of the given expression is $\frac{4}{\pi^2}$. |
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| kimi | The given expression can be simplified using the Wallis product formula for $\pi$. The Wallis product is given by: \[ \pi = 2 \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} \] Now, let's look at the given expression: \[ \frac{2}{\pi} \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} \] We can substitute the Wallis product formula into the given expression: \[ \frac{2}{\pi} \left(\frac{\pi}{2}\right) = \frac{2}{2} = 1 \] So, the value of the given expression is $\boxed{1}$. |
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| wenxinyiyan | To solve the given problem, we can start by simplifying the expression inside the product notation. Given expression: $2/\pi * \prod_{k=1}^{\infty} \frac{(2*k)^2}{(2*k-1)(2*k+1)}$ First, let's simplify the fraction inside the product: $\frac{(2k)^2}{(2k-1)(2k+1)} = \frac{4k^2}{4k^2-1}$ Now, let's rewrite the product with the simplified fraction: $\prod_{k=1}^{\infty} \frac{4k^2}{4k^2-1}$ This product can be further split into two products: $\prod_{k=1}^{\infty} \frac{2k}{2k-1} * \prod_{k=1}^{\infty} \frac{2k}{2k+1}$ Notice that the second product can be rewritten as: $\prod_{k=1}^{\infty} \frac{2k+2}{2k+1}$ Now, let's combine the two products: $\prod_{k=1}^{\infty} \frac{2k}{2k-1} * \frac{2k+2}{2k+1}$ This combined product can be further simplified by telescoping: $\frac{2}{1} * \frac{4}{3} * \frac{6}{5} * ... * \frac{2k}{2k-1} * \frac{2k+2}{2k+1}$ Notice that most of the terms will cancel out, leaving us with: $\frac{2}{1} * \frac{2}{3} * \frac{4}{3} * \frac{4}{5} * \frac{6}{5} * ...$ This product can be simplified to: $2 * \frac{2}{3} * \frac{4}{3} * \frac{4}{5} * \frac{6}{5} * ...$ Now, let's multiply this product by $2/\pi$: $\frac{2}{\pi} * 2 * \frac{2}{3} * \frac{4}{3} * \frac{4}{5} * \frac{6}{5} * ...$ This expression can be further simplified to: $\frac{4}{\pi} * \frac{2}{3} * \frac{4}{3} * \frac{4}{5} * \frac{6}{5} * ...$ Notice that the product part of this expression is the Wallis product for $\pi/2$, which converges to $\pi/2$. Therefore, the given expression converges to: $\frac{4}{\pi} * \frac{\pi}{2} = 2$ Hence, the value of the given expression is 2. |
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| zhipu | To find the value of the given infinite product, we start by analyzing the expression: \[ \frac{2}{\pi} \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} \] First, let's simplify the individual terms in the product: \[ \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{4k^2}{(2k-1)(2k+1)} \] Notice that the denominator can be rewritten using the difference of squares: \[ (2k-1)(2k+1) = (2k)^2 - 1^2 = 4k^2 - 1 \] So the term becomes: \[ \frac{4k^2}{4k^2 - 1} \] Now, let's express this term in a more convenient form: \[ \frac{4k^2}{4k^2 - 1} = 1 + \frac{1}{4k^2 - 1} \] Thus, the infinite product can be written as: \[ \prod_{k=1}^{\infty} \left(1 + \frac{1}{4k^2 - 1}\right) \] To evaluate this product, we can use the identity for the product of terms of the form \( \frac{(2k)^2}{(2k-1)(2k+1)} \). It is known that: \[ \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{\pi}{2} \] This identity can be derived from the properties of the Gamma function and the sine function, but for our purposes, we can use it directly. Given this identity, we substitute it back into our original expression: \[ \frac{2}{\pi} \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} = \frac{2}{\pi} \cdot \frac{\pi}{2} \] Simplifying this, we get: \[ \frac{2}{\pi} \cdot \frac{\pi}{2} = 1 \] Therefore, the value of the given expression is: \[ \boxed{1} \] |
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| gpto1 | To find the value of the given infinite product: \[ P = \frac{2}{\pi} \times \prod_{k=1}^{\infty} \frac{(2k)^2}{(2k-1)(2k+1)} \] We can recognize that this product is closely related to Wallis' formula for \(\pi\). Let's first simplify the terms of the product. Note that: \[ (2k)^2 = 4k^2 \] \[ (2k - 1)(2k + 1) = (2k)^2 - 1 = 4k^2 - 1 \] So, each term of the product simplifies to: \[ \frac{4k^2}{4k^2 - 1} \] Therefore, the product becomes: \[ P = \frac{2}{\pi} \times \prod_{k=1}^{\infty} \frac{4k^2}{4k^2 - 1} \] Now, recall Wallis' product, which states: \[ \frac{\pi}{2} = \prod_{k=1}^{\infty} \frac{4k^2}{4k^2 - 1} \] Comparing this with our product, we see that: \[ \prod_{k=1}^{\infty} \frac{4k^2}{4k^2 - 1} = \frac{\pi}{2} \] Substituting back into our expression for \(P\): \[ P = \frac{2}{\pi} \times \frac{\pi}{2} = 1 \] **Answer: 1** |
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