| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the problem, we need to determine the value of \( a_{2023} \) in the unique sequence of integers \( a_1, a_2, \ldots, a_{2023} \) such that the equation \[ \tan 2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x} \] holds whenever \( \tan 2023x \) is defined. We start by using the properties of the tangent function and its multiple-angle formulas. Recall that the tangent function satisfies the following identity for any integer \( n \): \[ \tan (nx) = \frac{P_n(\tan x)}{Q_n(\tan x)} \] where \( P_n(y) \) and \( Q_n(y) \) are polynomials in \( y = \tan x \). Specifically, for \( n = 2023 \), we have: \[ \tan 2023x = \frac{P_{2023}(\tan x)}{Q_{2023}(\tan x)} \] Given the structure of the problem, we need to identify the coefficients of the highest degree terms in \( P_{2023}(y) \) and \( Q_{2023}(y) \). The polynomials \( P_n(y) \) and \( Q_n(y) \) are derived from the recursive relations: \[ \tan (n+1)x = \frac{\tan nx + \tan x}{1 - \tan nx \tan x} \] This implies: \[ P_{n+1}(y) = y P_n(y) + Q_n(y) \] \[ Q_{n+1}(y) = Q_n(y) - y P_n(y) \] Starting from \( P_1(y) = y \) and \( Q_1(y) = 1 \), we can see that the highest degree term in \( P_n(y) \) is \( y^n \) and in \( Q_n(y) \) is \( -y^{n-1} \). For \( n = 2023 \), the highest degree term in \( P_{2023}(y) \) is \( y^{2023} \), and the highest degree term in \( Q_{2023}(y) \) is \( -y^{2022} \). Thus, the coefficient of \( \tan^{2023} x \) in \( P_{2023}(\tan x) \) is 1, and the coefficient of \( \tan^{2022} x \) in \( Q_{2023}(\tan x) \) is -1. Therefore, the coefficient \( a_{2023} \) in the given equation must be: \[ a_{2023} = 1 \] Hence, the answer is: \[ \boxed{D} \] |
C |
| deepseek-r1 | To determine the value of \( a_{2023} \) in the given expansion of \( \tan(2023x) \), we need to consider the general form of the tangent multiple-angle formula. For an odd integer \( n \), the expansion of \( \tan(nx) \) can be written as a rational function where the numerator is a polynomial of degree \( n \) and the denominator is a polynomial of degree \( n-1 \). The key steps are as follows: 1. For odd \( n \), the leading term in the numerator of the expansion of \( \tan(nx) \) is \( (-1)^{(n-1)/2} \tan^n x \). This pattern is derived from known expansions for small \( n \) (e.g., \( \tan(3x) \), \( \tan(5x) \), \( \tan(7x) \)) where the leading coefficient alternates between \( 1 \) and \( -1 \). 2. For \( n = 2023 \), which is an odd number, we need to determine the sign of the leading term. Since \( 2023 \equiv 3 \mod 4 \), we use the formula \( (-1)^{(n-1)/2} \) to find the coefficient. 3. Calculating \( (-1)^{(2023-1)/2} = (-1)^{1011} = -1 \). Thus, the coefficient \( a_{2023} \) in the numerator of the expansion of \( \tan(2023x) \) is \( -1 \). The final answer is \(\boxed{CCCCC}\). |
C |
| gpto1 | CCCCC | C |
| gpt-4o-2024-08-06 | To solve this problem, we need to understand the identity for the tangent of a multiple angle. The identity for \(\tan(nx)\) in terms of \(\tan(x)\) is given by: \[ \tan(nx) = \frac{P_n(\tan x)}{Q_n(\tan x)} \] where \(P_n\) and \(Q_n\) are polynomials in \(\tan x\) of degree \(n\) and \(n-1\) respectively. Specifically, for \(\tan(2023x)\), we have: \[ \tan(2023x) = \frac{a_1 \tan x + a_3 \tan^3 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x + \cdots + a_{2022} \tan^{2022} x} \] The coefficients \(a_k\) are determined by the expansion of \(\tan(nx)\) using the tangent addition formula: \[ \tan((n+1)x) = \frac{\tan(nx) + \tan x}{1 - \tan(nx) \tan x} \] This recursive formula can be used to derive the coefficients of the polynomials \(P_n\) and \(Q_n\). The key observation is that the highest degree term in the numerator of \(\tan(nx)\) is \((-1)^{n-1} \tan^n x\). This is because the recursive formula introduces a negative sign in the denominator, which affects the highest degree term. For \(n = 2023\), the highest degree term in the numerator is \((-1)^{2022} \tan^{2023} x\). Since \(2022\) is even, \((-1)^{2022} = 1\). Therefore, the coefficient \(a_{2023}\) of \(\tan^{2023} x\) is: \[ a_{2023} = (-1)^{2022} = 1 \] Thus, the answer is \(\boxed{1}\). Therefore, the correct answer is \(\boxed{1}\), and the corresponding string is: \[ \text{DDDDD} \] |
C |
| DeepSeek-Coder-V2-Lite-Instruct | To determine \( a_{2023} \) in the given sequence, we need to use the tangent addition formula and properties of trigonometric functions. The given equation is: \[ \tan 2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x} \] We can use the multiple angle formula for tangent. Specifically, for \( n = 2023 \), the tangent of a multiple angle is given by: \[ \tan(2023x) = \frac{\binom{2023}{1} \tan x - \binom{2023}{3} \tan^3 x + \binom{2023}{5} \tan^5 x - \cdots + (-1)^{2023} \binom{2023}{2023} \tan^{2023} x}{1 - \binom{2023}{2} \tan^2 x + \binom{2023}{4} \tan^4 x - \cdots + (-1)^{2023} \binom{2023}{2022} \tan^{2022} x} \] Here, \( \binom{2023}{k} \) represents the binomial coefficient. The coefficients \( a_k \) in the given equation correspond to the binomial coefficients in the expansion of \( \tan(2023x) \). The general term in the numerator of the given equation is: \[ \sum_{k \text{ odd}} \binom{2023}{k} (-1)^{(k-1)/2} \tan^k x \] The term \( \tan^{2023} x \) corresponds to \( k = 2023 \), which is an odd number. The coefficient of \( \tan^{2023} x \) in the numerator is: \[ \binom{2023}{2023} (-1)^{(2023-1)/2} = \binom{2023}{2023} (-1)^{1011} = 1 \cdot (-1) = -1 \] Thus, \( a_{2023} = -1 \). Therefore, the answer is: \[ \boxed{C} \] |
C |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Recall the formula for the tangent of a multiple angle The formula for the tangent of a multiple angle $n\theta$ can be expressed in terms of the tangent of the angle $\theta$ using the following formula: $\tan n\theta = \frac{\binom{n}{1}\tan \theta - \binom{n}{3}\tan^3 \theta + \binom{n}{5}\tan^5 \theta - \cdots}{1 - \binom{n}{2}\tan^2 \theta + \binom{n}{4}\tan^4 \theta - \cdots}$. ## Step 2: Apply the formula to the given problem Using the formula for the tangent of a multiple angle, we can express $\tan 2023x$ in terms of $\tan x$ as follows: $\tan 2023x = \frac{\binom{2023}{1}\tan x - \binom{2023}{3}\tan^3 x + \binom{2023}{5}\tan^5 x - \cdots}{1 - \binom{2023}{2}\tan^2 x + \binom{2023}{4}\tan^4 x - \cdots}$. ## Step 3: Identify the coefficient of the highest power of $\tan x$ Comparing the given sequence with the formula, we can see that the coefficient of the highest power of $\tan x$ in the numerator is $a_{2023}$. In the formula, the coefficient of the highest power of $\tan x$ is $\binom{2023}{2023} = 1$ if $n$ is odd. ## Step 4: Determine the value of $a_{2023}$ Since the power of $\tan x$ in the numerator is 2023, which is an odd number, the value of $a_{2023}$ is $\binom{2023}{2023} = 1$ if the pattern in the formula holds. However, the problem statement does not explicitly state that the sequence follows the binomial expansion pattern, so we need to verify this. ## Step 5: Verify the pattern Looking at the formula, we can see that the sign of the terms alternates between positive and negative. Since 2023 is an odd number, the last term in the numerator should have a positive sign. Therefore, the value of $a_{2023}$ is indeed 1. The final answer is: $\boxed{1}$ |
C |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Recall the formula for the tangent of a sum of two angles The formula for the tangent of a sum of two angles $x$ and $y$ is given by $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$. ## Step 2: Apply the formula for the tangent of a sum of two angles to $\tan 2023x$ We can express $\tan 2023x$ as $\tan (2022x + x)$ and apply the formula for the tangent of a sum of two angles. ## Step 3: Derive the expression for $\tan 2023x$ using the formula Applying the formula, we get $\tan 2023x = \frac{\tan 2022x + \tan x}{1 - \tan 2022x \tan x}$. ## Step 4: Express $\tan 2022x$ in terms of $\tan x$ We can express $\tan 2022x$ as $\tan (2021x + x)$ and apply the formula for the tangent of a sum of two angles. ## Step 5: Derive the expression for $\tan 2022x$ using the formula Applying the formula, we get $\tan 2022x = \frac{\tan 2021x + \tan x}{1 - \tan 2021x \tan x}$. ## Step 6: Continue expressing $\tan kx$ in terms of $\tan x$ for $k = 2021, 2020, ..., 2$ We can continue applying the formula for the tangent of a sum of two angles to express $\tan kx$ in terms of $\tan x$ for $k = 2021, 2020, ..., 2$. ## Step 7: Derive the expression for $\tan 2023x$ in terms of $\tan x$ After expressing $\tan kx$ in terms of $\tan x$ for $k = 2021, 2020, ..., 2$, we can derive the expression for $\tan 2023x$ in terms of $\tan x$. ## Step 8: Identify the coefficient of $\tan^{2023} x$ in the expression for $\tan 2023x$ The coefficient of $\tan^{2023} x$ in the expression for $\tan 2023x$ is $a_{2023}$. ## Step 9: Determine the value of $a_{2023}$ After deriving the expression for $\tan 2023x$ in terms of $\tan x$, we can determine the value of $a_{2023}$. ## Step 10: Recall the pattern of the coefficients of the tangent function The coefficients of the tangent function follow a pattern, and we can use this pattern to determine the value of $a_{2023}$. ## Step 11: Determine the value of $a_{2023}$ using the pattern Using the pattern of the coefficients of the tangent function, we can determine that $a_{2023} = (-1)^{2023/2} \cdot 2023$. ## Step 12: Simplify the expression for $a_{2023}$ Since $2023$ is an odd number, $2023/2$ is not an integer, but we can simplify the expression for $a_{2023}$ as $a_{2023} = (-1)^{1011} \cdot 2023$. ## Step 13: Evaluate the expression for $a_{2023}$ Evaluating the expression for $a_{2023}$, we get $a_{2023} = -2023$. The final answer is: $\boxed{A}$ |
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| Qwen2-72B-Instruct | The given expression is related to the expansion of $\tan(n\theta)$ in terms of $\tan(\theta)$. The coefficients in this expansion can be found using the Chebyshev polynomials of the second kind. However, for large values like $n=2023$, it's not practical to compute these directly. For the highest power term, $a_{2023}\tan^{2023}x$, we can use the fact that the coefficient of the highest power term in the expansion of $\tan(n\theta)$ in terms of $\tan(\theta)$ is $n$ when $n$ is odd (which 2023 is). This is because the highest power term comes from the binomial expansion of $(\tan(\theta) + \tan(\theta))^n$ which simplifies to $2^n\tan^n(\theta)$ for odd $n$, and since $\tan(2\theta) = \frac{2\tan(\theta)}{1-\tan^2(\theta)}$, the coefficient of the highest power term in the expansion of $\tan(n\theta)$ is $n$ for odd $n$. Therefore, $a_{2023} = 2023$. The answer is $\boxed{\textbf{(E) } 2023}$, so the string is EEEEE. |
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| baichuan4 | To find $a_{2023}$, we can use the fact that the Taylor series expansion of $\tan(x)$ around $x=0$ is given by: $$\tan(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} 2^{2n} (2^{2n}-1) B_{2n}}{(2n)!} x^{2n-1}$$ where $B_{2n}$ are the Bernoulli numbers. This series can also be written as: $$\tan(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} 4^n (4^n-1) B_{2n}}{(2n)!} x^{2n-1}$$ We can compare this series with the given expression: $$\tan(2023x) = \frac{a_1 \tan(x) + a_3 \tan^3(x) + a_5 \tan^5(x) + \cdots + a_{2023} \tan^{2023}(x)}{1 + a_2 \tan^2(x) + a_4 \tan^4(x) + \cdots + a_{2022} \tan^{2022}(x)}$$ Since $\tan(2023x)$ is a polynomial in $\tan(x)$ of degree $2023$, we can match the coefficients of the highest power of $\tan(x)$ in both expressions. In the Taylor series, the coefficient of $x^{2023}$ is: $$\frac{(-1)^{1011} 4^{1012} (4^{1012}-1) B_{2024}}{(2024)!}$$ This should match the coefficient of $\tan^{2023}(x)$ in the given expression, which is simply $a_{2023}$. Therefore, we have: $$a_{2023} = \frac{(-1)^{1011} 4^{1012} (4^{1012}-1) B_{2024}}{(2024)!}$$ The sign of this coefficient is negative because $(-1)^{1011}$ is negative. The magnitude of the coefficient is not important for determining the sign of $a_{2023}$. Hence, $a_{2023}$ is negative. Looking at the options, the only negative choices are $\textbf{(A) } -2023$ and $\textbf{(B) } -2022$. Without further information to determine the exact value of $a_{2023}$, we can only guess based on the available options. However, the magnitude of the coefficient in the Taylor series is likely to be very large due to the factorials in the denominator, so $\textbf{(A) } -2023$ seems more plausible than $\textbf{(B) } -2022$. Therefore, the best guess for $a_{2023}$ is $\textbf{(A) } -2023$. Duplicating the letter A five times gives: AAAAA. |
C |
| doubao | EEEEEEE | C |
| gpto1 | CCCCC | C |
| kimi | To determine the value of \(a_{2023}\) in the given sequence, we start by considering the tangent addition formula. The tangent of a sum of angles can be expressed as: \[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] We need to generalize this to the tangent of a multiple angle, specifically \(\tan(2023x)\). We can use the fact that \(\tan(2023x)\) can be expressed in terms of \(\tan x\) using the Chebyshev polynomials of the first kind, \(T_n(t)\), which satisfy the relation: \[ T_n(\tan x) = \tan(nx) \] The Chebyshev polynomial of the first kind for \(n = 2023\) is given by: \[ T_{2023}(t) = \cos(2023 \arccos(t)) \] Using the multiple angle formula for cosine, we have: \[ \cos(2023 \theta) = T_{2023}(\cos \theta) \] For \(\theta = \arccos(t)\), we get: \[ T_{2023}(t) = \cos(2023 \arccos(t)) \] The polynomial \(T_{2023}(t)\) can be expanded in terms of powers of \(t\): \[ T_{2023}(t) = a_0 + a_1 t + a_2 t^2 + \cdots + a_{2023} t^{2023} \] Given that \(\tan(2023x) = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x + \cdots + a_{2022} \tan^{2022} x}\), we compare this with the polynomial form of \(T_{2023}(t)\). The leading term in the numerator of the tangent formula corresponds to the highest power of \(\tan x\), which is \(\tan^{2023} x\). This term must match the leading term in the polynomial expansion of \(T_{2023}(t)\). The leading coefficient of \(t^{2023}\) in \(T_{2023}(t)\) is known to be \(2^{2022}\) because the Chebyshev polynomials of the first kind have the property that the leading coefficient of \(T_n(t)\) is \(2^{n-1}\). Therefore, for \(n = 2023\): \[ a_{2023} = 2^{2022} \] However, we need to consider the sign. The tangent function has a periodicity and symmetry that affects the sign of the coefficients. Specifically, for odd \(n\), the leading coefficient of \(\tan(nx)\) in terms of \(\tan x\) is negative. Since 2023 is odd, the leading coefficient \(a_{2023}\) must be negative. Thus, we have: \[ a_{2023} = -2^{2022} \] Given the options, we need to find the closest match. The correct answer is: \(\boxed{-1}\) |
C |
| minimax | To find the unique sequence of integers $a_1, a_2, \cdots a_{2023}$, we can use the tangent addition formula in a recursive manner. The tangent addition formula states that for any two angles $x$ and $y$: $$\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$$ We can use this formula to express $\tan 2023x$ in terms of $\tan x$ and $\tan 2022x$, then $\tan 2022x$ in terms of $\tan x$ and $\tan 2021x$, and so on, until we express $\tan 2x$ in terms of $\tan x$ and $\tan x$. Let's start with the first few steps: $$\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$$ $$\tan 3x = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x} = \frac{\frac{2\tan x}{1 - \tan^2 x} + |
C |
| wenxinyiyan | 首先，我们观察题目中给出的等式，它是一个关于$\tan x$的有理函数，分子是$\tan x$的奇数次幂的多项式，分母是$1$和$\tan x$的偶数次幂的多项式。这个等式在$\tan 2023x$有定义的时候都成立。 我们可以尝试将$x$替换为$-x$，由于$\tan(-x) = -\tan x$，我们可以得到： $$\tan(-2023x) = \frac{-a_1 \tan x - a_3 \tan^3 x - a_5 \tan^5 x - \cdots - a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}$$由于$\tan(-2023x) = -\tan 2023x$，我们可以将上面的等式和原来的等式相加，得到： $$0 = \frac{(a_1 - a_1) \tan x + (a_3 - a_3) \tan^3 x + \cdots + (a_{2023} - a_{2023}) \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}$$由于这个等式对所有$\tan x$都成立（除了使分母为0的$\tan x$值），我们可以得出$a_{2023} - a_{2023} = 0$，即$a_{2023} = 0$。 但是，这个结论显然与题目的选项不符。我们意识到，在上面的推导中，我们实际上没有使用到$\tan 2023x$的任何性质，因此这个推导是不正确的。 实际上，我们应该使用三角函数的加法定理。由于$\tan(a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$，我们可以将$\tan 2023x$表示为$\tan(2022x + x)$，然后反复应用加法定理，将$\tan 2023x$展开为$\tan x$的多项式。 但是，这种方法需要计算大量的项，并且很难直接得出$a_{2023}$的值。因此，我们需要寻找一种更简洁的方法。 观察选项，我们发现它们都是整数，并且只有选项C是$-1$。我们可以尝试将$a_{2023}$设为$-1$，然后验证等式是否成立。如果等式成立，那么$a_{2023}$就是$-1$。 为了验证等式，我们可以选择几个特定的$x$值，然后计算等式两边的值是否相等。如果对于所有选择的$x$值，等式两边的值都相等，那么我们可以合理地推测等式对于所有$x$值都成立。 但是，这种方法并不严谨，因为我们不能选择所有的$x$值进行验证。因此，我们需要寻找一种更严谨的方法。 最终，我们注意到题目中的等式是一个恒等式，即对于所有使等式有意义的$x$值，等式都成立。因此，我们可以尝试使用数学归纳法来证明等式。 假设当$n=k$时，等式成立，即 $$\tan kx = \frac{a_1 \tan x + a_3 \tan^3 x + \cdots + a_k \tan^k x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{k-1} \tan^{k-1} x}$$然后，我们考虑$n=k+1$的情况。使用三角函数的加法定理，我们可以将$\tan(k+1)x$表示为$\tan kx$和$\tan x$的函数。然后，我们将假设中的等式代入这个表达式，并进行化简。 经过化简，我们可以得到$n=k+1$时的等式，并且可以发现$a_{k+1}$的值与$k$有关。通过数学归纳法，我们可以得出$a_{2023}$的值。 但是，这种方法需要大量的计算和化简，并且很难直接得出$a_{2023}$的值。因此，在这里，我们并没有给出完整的解答。 然而，根据题目的来源和难度，我们可以推测题目中的等式可能与切比雪夫多项式有关。切比雪夫多项式是一类特殊的正交多项式，它们在许多领域都有广泛的应用。 如果我们将题目中的等式与切比雪夫多项式的性质 |
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| zhipu | To solve for $a_{2023}$ in the given expression for $\tan 2023x$, we can use the properties of Chebyshev polynomials of the first kind. These polynomials, $T_n(x)$, are defined such that $T_n(\cos \theta) = \cos(n\theta)$. For the tangent function, we have a related identity involving these polynomials. The identity for $\tan(nx)$ in terms of $\tan(x)$ can be expressed using Chebyshev polynomials of the first kind, but in a modified form to account for the tangent function's properties. Specifically, $\tan(nx)$ can be expressed as a rational function of $\tan(x)$, where the numerator and denominator are polynomials in $\tan(x)$. Given: \[ \tan 2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x + \cdots + a_{2022} \tan^{2022} x} \] We need to find the coefficient $a_{2023}$, which is the leading coefficient in the numerator's polynomial in $\tan x$. ### Step-by-step Analysis: 1. **Understanding the Polynomial Structure:** - The polynomial in the numerator is of degree 2023. - The polynomial in the denominator is of degree 2022. 2. **Chebyshev Polynomial Relation:** - The Chebyshev polynomial $T_n(x)$ for $\tan(nx)$ in terms of $\tan(x)$ involves specific coefficients that can be derived from the recurrence relations or direct computation. 3. **Leading Coefficient Insight:** - For $\tan(nx)$, the leading coefficient of the polynomial in $\tan(x)$ (in the numerator) is related to the sign and magnitude of $n$. - Specifically, for odd $n$, the leading coefficient of the polynomial representing $\tan(nx)$ is $n$. 4. **Application to the Problem:** - Here, $n = 2023$, which is odd. - The leading coefficient $a_{2023}$ in the numerator for $\tan(2023x)$ should thus be $2023$. However, the problem asks for the specific value of $a_{2023}$, and based on the typical structure of these polynomials and the given options, the sign is crucial. The correct sign for the leading coefficient in this context, considering the standard form of the polynomial expansion for tangent functions, is negative. Thus, $a_{2023} = -2023$. ### Conclusion: The correct answer is $\textbf{(A) } -2023$. Therefore, the string of duplicated letters is: \[ AAAAA \] |
C |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-e3e4fbe9-45b3-9fd1-a2d9-ed15486f3b7c', 'request_id': 'e3e4fbe9-45b3-9fd1-a2d9-ed15486f3b7c'} | C |