| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the coefficient of \(x^2y^5\) in the expansion of \((x + 2y)^7\), we can use the binomial theorem. The binomial theorem states that for any real numbers \(a\) and \(b\) and non-negative integer \(n\), \[ (a + b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k \] In this case, \(a = x\), \(b = 2y\), and \(n = 7\). We want the term where \(a^{n-k}b^k = x^{7-k}(2y)^k = x^2y^5\). This means \(k=5\) and \(n-k=2\), so we substitute these values into the binomial theorem: \[ \binom{7}{5}x^{7-5}(2y)^5 = \binom{7}{5}x^2(32y^5) \] Now, we calculate \(\binom{7}{5}\): \[ \binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7 \cdot 6}{2 \cdot 1} = 21 \] So the coefficient of \(x^2y^5\) is: \[ 21 \cdot 32 = 672 \] Therefore, the coefficient of \(x^2y^5\) in the expansion of \((x + 2y)^7\) is \(\boxed{672}\). |
672 |
| deepseekr1 | The coefficient of \(x^2y^5\) in the expansion of \((x + 2y)^7\) can be found using the binomial theorem. The theorem states that the term corresponding to \(x^{n-k}(2y)^k\) in \((x + 2y)^n\) is given by \(\binom{n}{k} x^{n-k} (2y)^k\). For the term \(x^2y^5\): - \(n = 7\) - The exponent of \(x\) is 2, so \(n - k = 2 \Rightarrow k = 5\). Calculate the binomial coefficient and the constants: 1. **Binomial Coefficient**: \(\binom{7}{5} = \binom{7}{2} = \frac{7!}{2!5!} = 21\). 2. **Power of \(2y\)**: \((2)^5 = 32\). Multiply these results: \[ 21 \times 32 = 672 \] **Answer:** \(\boxed{672}\) |
672 |
| gpto1 | To find the coefficient of \( x^2 y^5 \) in the expansion of \( (x + 2y)^7 \), we'll use the **Binomial Theorem**, which states: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n - k} b^k \] In this case: - \( a = x \) - \( b = 2y \) - \( n = 7 \) The general term of the expansion is: \[ T(k) = \binom{7}{k} x^{7 - k} (2y)^k \] We need to find the term where the exponents of \( x \) and \( y \) match \( x^2 y^5 \). **Step 1: Find the value of \( k \) that gives \( x^2 y^5 \)** - The exponent of \( x \) in the term is \( 7 - k \). - The exponent of \( y \) in the term is \( k \) (from \( (2y)^k \)). Set up the equations based on the desired exponents: 1. For \( x \): \[ 7 - k = 2 \quad \Rightarrow \quad k = 5 \] 2. For \( y \): \[ k = 5 \] Both conditions agree when \( k = 5 \). **Step 2: Compute the coefficient using \( k = 5 \)** Compute the binomial coefficient: \[ \binom{7}{5} = \binom{7}{2} = \frac{7!}{5! \, 2!} = \frac{7 \times 6}{2 \times 1} = 21 \] Compute \( (2y)^5 \): \[ (2y)^5 = 2^5 y^5 = 32 y^5 \] Multiply the binomial coefficient, \( x^{7 - k} \), and \( (2y)^k \): \[ T(5) = \binom{7}{5} x^{7 - 5} (2y)^5 = 21 \times x^2 \times 32 y^5 = 21 \times 32 \times x^2 y^5 \] Compute the numerical coefficient: \[ 21 \times 32 = 672 \] **Conclusion:** The coefficient of \( x^2 y^5 \) in the expansion of \( (x + 2y)^7 \) is **672**. **Answer:** 672 |
672 |
| GPT-4o | To find the coefficient of \( x^2y^5 \) in the expansion of \((x + 2y)^7\), we can use the binomial theorem. The binomial theorem states that: \[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \] In this problem, \( x = x \), \( y = 2y \), and \( n = 7 \). We are specifically looking for the term that includes \( x^2 y^5 \). In the general term of the binomial expansion, \((x + 2y)^7\) is: \[ \binom{7}{k} x^{7-k} (2y)^k \] We need the exponents of \( x \) and \( y \) to match \( x^2 y^5 \). Therefore, we set up the following system of equations: \[ 7 - k = 2 \quad \text{and} \quad k = 5 \] Solving for \( k \): \[ 7 - k = 2 \implies k = 5 \] Now, substitute \( k = 5 \) into the general term: \[ \binom{7}{5} x^{7-5} (2y)^5 \] Simplify the exponents and the binomial coefficient: \[ \binom{7}{5} x^2 (2y)^5 = \binom{7}{5} x^2 (2^5 y^5) = \binom{7}{5} x^2 (32 y^5) \] Since \(\binom{7}{5} = \binom{7}{2} = \frac{7!}{2!5!} = \frac{7 \times 6}{2 \times 1} = 21\): \[ = 21 x^2 (32 y^5) = 21 \times 32 x^2 y^5 = 672 x^2 y^5 \] Therefore, the coefficient of \( x^2 y^5 \) in the expansion of \((x + 2y)^7\) is: \[ \boxed{672} \] |
672 |
| llama3.1-70B-Instruct | The formula $(x + 2y)^7$ can be expanded using the binomial theorem, which states that for any positive integer $n$, $$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$ where $\binom{n}{k}$ is the binomial coefficient. In this case, we have $a = x$, $b = 2y$, and $n = 7$. We want to find the coefficient of $x^2y^5$, which means we need to find the term in the expansion where $x$ has a power of 2 and $y$ has a power of 5. Using the binomial theorem, we can write the expansion of $(x + 2y)^7$ as: $$(x + 2y)^7 = \sum_{k=0}^{7} \binom{7}{k} x^{7-k} (2y)^k$$ We can simplify this expression by noting that $\binom{7}{k} = \frac{7!}{k!(7-k)!}$, and that $(2y)^k = 2^k y^k$. Now, we need to find the term in the expansion where $x$ has a power of 2 and $y$ has a power of 5. This means we need to find the term where $k = 5$, since $x^{7-k} = x^2$ and $(2y)^k = 2^5 y^5$. The term we are looking for is: $$\binom{7}{5} x^2 (2y)^5 = \frac{7!}{5!(7-5)!} x^2 2^5 y^5 = \frac{7!}{5!2!} x^2 2^5 y^5$$ Simplifying this expression, we get: $$\frac{7!}{5!2!} x^2 2^5 y^5 = \frac{7 \cdot 6 \cdot 5!}{5!2!} x^2 2^5 y^5 = \frac{7 \cdot 6}{2} x^2 2^5 y^5 = 21 x^2 2^5 y^5$$ Therefore, the coefficient of $x^2y^5$ in the formula $(x + 2y)^7$ is $21 \cdot 2^5 = 672$. |
672 |
| llama3.1-405B-Instruct | The coefficient of $x^2y^5$ in the expansion of $(x + 2y)^7$ can be found using the binomial theorem. The binomial theorem states that for any positive integer $n$, $$(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$$ where $\binom{n}{k}$ is the binomial coefficient, which is defined as: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ In this case, we have $a = x$, $b = 2y$, and $n = 7$. We want to find the coefficient of $x^2y^5$, which means we need to find the term in the expansion where $x$ has a power of 2 and $y$ has a power of 5. Using the binomial theorem, we can write the expansion of $(x + 2y)^7$ as: $$(x + 2y)^7 = \sum_{k=0}^7 \binom{7}{k} x^{7-k} (2y)^k$$ The term we are interested in is the one where $k = 5$, which gives us: $$\binom{7}{5} x^{7-5} (2y)^5 = \binom{7}{5} x^2 (2y)^5$$ Now, we can calculate the binomial coefficient $\binom{7}{5}$: $$\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = \frac{7 \cdot 6}{2} = 21$$ So, the coefficient of $x^2y^5$ is: $$21 \cdot 2^5 = 21 \cdot 32 = 672$$ Therefore, the coefficient of $x^2y^5$ in the expansion of $(x + 2y)^7$ is 672. |
672 |
| DeepSeek-Coder-V2 | To find the coefficient of \(x^2y^5\) in the expansion of \((x + 2y)^7\), we can use the Binomial Theorem. The Binomial Theorem states that: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \] In this case, \(a = x\), \(b = 2y\), and \(n = 7\). We are looking for the term that contains \(x^2y^5\). This corresponds to the term where the exponent of \(x\) is 2 and the exponent of \(2y\) is 5. According to the Binomial Theorem, the general term in the expansion is given by: \[ \binom{7}{k} x^{7-k} (2y)^k \] We need to find \(k\) such that the exponents match \(x^2y^5\). This means: \[ 7 - k = 2 \quad \text{and} \quad k = 5 \] Solving for \(k\): \[ 7 - k = 2 \implies k = 5 \] So, the term we are interested in is: \[ \binom{7}{5} x^{7-5} (2y)^5 \] Simplifying this: \[ \binom{7}{5} x^2 (2y)^5 = \binom{7}{5} x^2 2^5 y^5 \] Next, we calculate the binomial coefficient \(\binom{7}{5}\): \[ \binom{7}{5} = \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \] Now, we calculate \(2^5\): \[ 2^5 = 32 \] Putting it all together, the term is: \[ 21 x^2 32 y^5 = 21 \times 32 x^2 y^5 = 672 x^2 y^5 \] Thus, the coefficient of \(x^2y^5\) is: \[ \boxed{672} \] |
672 |
| Qwen2-72B-Instruct | To find the coefficient of $x^2y^5$ in the expansion of $(x + 2y)^7$, we can use the binomial theorem. The binomial theorem states that for any real numbers $a$ and $b$ and non-negative integer $n$, \[ (a + b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k \] In this case, $a = x$, $b = 2y$, and $n = 7$. We want the term where $a^{n-k}b^k = x^{7-k}(2y)^k = x^2y^5$. This means $k=5$ and $n-k=2$, so we substitute these values into the binomial theorem: \[ \binom{7}{5}x^{7-5}(2y)^5 = \binom{7}{5}x^2(32y^5) \] Now, we calculate $\binom{7}{5}$: \[ \binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7 \cdot 6}{2 \cdot 1} = 21 \] So the coefficient of $x^2y^5$ is: \[ 21 \cdot 32 = 672 \] Therefore, the coefficient of $x^2y^5$ in the expansion of $(x + 2y)^7$ is $\boxed{672}$. |
672 |
| baichuan4 | To find the coefficient of $x^2y^5$ in the expansion of $(x + 2y)^7$, we can use the binomial theorem. The binomial theorem states that for any positive integer $n$, $$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k,$$ where $\binom{n}{k}$ is the binomial coefficient, which represents the number of ways to choose $k$ elements from a set of $n$ elements. In our case, $a = x$, $b = 2y$, and $n = 7$. We want to find the term corresponding to $x^{7-k}(2y)^k = x^2y^5$, which means we need $k = 5$. Therefore, we have: $$\binom{7}{5} x^{7-5} (2y)^5 = \binom{7}{5} x^2 (2y)^5.$$ Now, we just need to calculate the binomial coefficient $\binom{7}{5}$ and multiply it by the appropriate factors: $$\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7 \cdot 6}{2 \cdot 1} = 21.$$ So, the coefficient of $x^2y^5$ is: $$21 \cdot 2^5 = 21 \cdot 32 = 672.$$ Therefore, the coefficient of $x^2y^5$ in the expansion of $(x + 2y)^7$ is $672$. |
672 |
| doubao | We can use the binomial theorem to expand $(x+2y)^7$. The general term in the expansion is $C_7^r x^{7-r}(2y)^r$. We want to find the term where $r=5$ (since that will give $y^5$), so the term is $C_7^5 x^{7-5}(2y)^5 = C_7^5 x^2 (2^5 y^5) = 21\times x^2\times 32y^5 = 672x^2y^5$. So the coefficient of $x^2y^5$ is 672. | 672 |
| minimax | To find the coefficient of $x^2y^5$ in the expansion of $(x + 2y)^7$, we can use the binomial theorem. The binomial theorem states that: $$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$ where $\binom{n}{k}$ is the binomial coefficient, which can be calculated as: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ In our case, $a = x$, $b = 2y$, and $n = 7$. We want to find the term where $x$ is squared and $y$ is raised to the fifth power, which means we need the term where $k = 5$ (since $b^k$ corresponds to $y^5$). Using the binomial coefficient formula, we calculate the coefficient for $k = 5$: $$\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = \frac{7 \times 6}{2 \times 1} = 21$$ Now, we multiply this binomial coefficient by the corresponding powers of $a$ and $b$: $$\binom{7}{5} x^{7-5} (2y)^5 = 21 x^2 (2^5 y^5) = 21 x^2 (32 y^5) = 672 x^2 y^5$$ Therefore, the coefficient of $x^2y^5$ in the expansion of $(x + 2y)^7$ is $672$. |
672 |
| kimi | To find the coefficient of \(x^2y^5\) in the expansion of \((x + 2y)^7\), we can use the Binomial Theorem. The Binomial Theorem states that: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \] where \(\binom{n}{k}\) is a binomial coefficient, calculated as: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] In our case, \(a = x\), \(b = 2y\), and \(n = 7\). We want to find the term where the power of \(x\) is 2 and the power of \(y\) is 5. This corresponds to \(k = 5\), since \(x\) will be raised to the power of \(7 - k\) and \(y\) to the power of \(k\). So, we have: \[ \binom{7}{5} x^{7-5} (2y)^5 \] Calculating the binomial coefficient: \[ \binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7 \times 6}{2 \times 1} = 21 \] Now, we calculate the term: \[ 21 x^{2} (2y)^5 = 21 x^2 \cdot 32y^5 = 672 x^2 y^5 \] Therefore, the coefficient of \(x^2y^5\) is 672. |
672 |
| wenxinyiyan | 为了找到$(x + 2y)^7$展开后$x^2y^5$的系数，我们可以使用二项式定理。 二项式定理的通项公式是 $$T_{r+1} = C_n^r \cdot a^{n-r} \cdot b^r$$其中，$C_n^r$ 是组合数，表示从$n$个不同元素中取出$r$个元素的组合数，$a$和$b$是二项式中的两项，$n$是二项式的次数。 对于$(x + 2y)^7$，我们有$a = x$，$b = 2y$，$n = 7$。我们想要找到$x^2y^5$的系数，所以我们需要找到一个$r$值，使得$x$的指数为$2$，$y$的指数为$5$。 将$a$，$b$，和$n$代入通项公式，我们得到 $$T_{r+1} = C_7^r \cdot x^{7-r} \cdot (2y)^r = C_7^r \cdot 2^r \cdot x^{7-r} \cdot y^r$$为了找到$x^2y^5$的系数，我们需要$7-r=2$和$r=5$。解这个方程组，我们发现$r=5$是解。 现在我们可以找到$x^2y^5$的系数： $$C_7^5 \cdot 2^5 = \binom{7}{5} \cdot 32 = \frac{7!}{5!2!} \cdot 32 = 21 \cdot 32 = 672$$所以，$(x + 2y)^7$展开后$x^2y^5$的系数是$672$。 |
672 |
| zhipu | To find the coefficient of $x^2y^5$ in the expansion of $(x + 2y)^7$, we can use the Binomial Theorem. The Binomial Theorem states that: $$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$ where $\binom{n}{k}$ is the binomial coefficient, which is the number of ways to choose $k$ elements from a set of $n$ elements and is given by: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ In our problem, $a = x$, $b = 2y$, and $n = 7$. We want to find the term where the powers of $x$ and $y$ are $2$ and $5$, respectively. This means we are looking for the term where $k = 5$ because $b^k = (2y)^5$ will give us the $y^5$ term, and $a^{n-k} = x^{7-k}$ will give us the $x^2$ term when $k = 5$. Using the Binomial Theorem, the term we are interested in is: $$\binom{7}{5} x^{7-5} (2y)^5 = \binom{7}{5} x^2 (2y)^5$$ Now let's calculate the binomial coefficient $\binom{7}{5}$: $$\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = \frac{7 \times 6}{2 \times 1} = 21$$ Next, we calculate $(2y)^5$: $$(2y)^5 = 2^5 y^5 = 32y^5$$ Now we can put it all together: $$\binom{7}{5} x^2 (2y)^5 = 21 x^2 \cdot 32y^5 = 21 \cdot 32 x^2 y^5 = 672 x^2 y^5$$ So the coefficient of $x^2y^5$ in the expansion of $(x + 2y)^7$ is $672$. |
672 |
| gpto1 | To find the coefficient of \( x^2 y^5 \) in the expansion of \( (x + 2y)^7 \), we'll use the **Binomial Theorem**, which states: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n - k} b^k \] In this case: - \( a = x \) - \( b = 2y \) - \( n = 7 \) The general term of the expansion is: \[ T(k) = \binom{7}{k} x^{7 - k} (2y)^k \] We need to find the term where the exponents of \( x \) and \( y \) match \( x^2 y^5 \). **Step 1: Find the value of \( k \) that gives \( x^2 y^5 \)** - The exponent of \( x \) in the term is \( 7 - k \). - The exponent of \( y \) in the term is \( k \) (from \( (2y)^k \)). Set up the equations based on the desired exponents: 1. For \( x \): \[ 7 - k = 2 \quad \Rightarrow \quad k = 5 \] 2. For \( y \): \[ k = 5 \] Both conditions agree when \( k = 5 \). **Step 2: Compute the coefficient using \( k = 5 \)** Compute the binomial coefficient: \[ \binom{7}{5} = \binom{7}{2} = \frac{7!}{5! \, 2!} = \frac{7 \times 6}{2 \times 1} = 21 \] Compute \( (2y)^5 \): \[ (2y)^5 = 2^5 y^5 = 32 y^5 \] Multiply the binomial coefficient, \( x^{7 - k} \), and \( (2y)^k \): \[ T(5) = \binom{7}{5} x^{7 - 5} (2y)^5 = 21 \times x^2 \times 32 y^5 = 21 \times 32 \times x^2 y^5 \] Compute the numerical coefficient: \[ 21 \times 32 = 672 \] **Conclusion:** The coefficient of \( x^2 y^5 \) in the expansion of \( (x + 2y)^7 \) is **672**. **Answer:** 672 |
672 |
| deepseekv3 | server error: Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-cc1080c8-8483-9362-9c20-00d02763b52f', 'request_id': 'cc1080c8-8483-9362-9c20-00d02763b52f'} | 672 |