Question

存在唯一的整数序列 $a_1,a_2,\cdots a_{2023}$ ，使得只要 $\tan 2023x$ 有定义，就有求 $a_{2023}$ ？ $\begin{gathered} \tan 2023x\\ {}=\tiny\dfrac{a_1\tan x+a_3\tan^3 x+a_5\tan^5 x+\cdots+a_{2023}\tan^{2023}x}{1+a_2\tan^2 x+a_4\tan^4 x\cdots+a_{2022}\tan^{2022}x} \end{gathered}$

There is a unique sequence of integers $a_1,a_2,\cdots a_{2023}$ such that $\begin{gathered} \tan 2023x\\ {}=\tiny\dfrac{a_1\tan x+a_3\tan^3 x+a_5\tan^5 x+\cdots+a_{2023}\tan^{2023}x}{1+a_2\tan^2 x+a_4\tan^4 x\cdots+a_{2022}\tan^{2022}x} \end{gathered}$ whenever $\tan 2023x$ is defined. What is $a_{2023}?$

$-2023$

$-2022$

$-1$

$1$

$2023$

答案：C

知识点：棣莫弗定理二项式定理

难度评级：2650

解答：

由 De Moivre 公式， $(\cos x+i\sin x)^{2023}$ $=\cos 2023x+i\sin 2023x$ 。展开左边并取虚部与实部之比，再把分子、分母同除以 $\cos^{2023}x$ ，即可把 $\tan 2023x$ 写成题中关于 $\tan x$ 的有理函数。

系数 $a_{2023}$ 是分子中 $\tan^{2023}x$ 的系数，来自 $k=2023$ 项： $\begin{gathered} a_{2023}=(-1)^{(2023-1)/2}\binom{2023}{2023}\\ {}=(-1)^{1011}\\ {}=-1. \end{gathered}$

所以正确答案是 C。

By De Moivre, $(\cos x+i\sin x)^{2023}$ $=\cos 2023x+i\sin 2023x.$ Expanding the left side and taking the ratio of imaginary to real parts gives $\tan 2023x$ as the stated rational function of $\tan x$ after dividing numerator and denominator by $\cos^{2023}x.$

The coefficient $a_{2023}$ is the coefficient of $\tan^{2023}x$ in the numerator, which comes from the $k=2023$ term: $\begin{gathered} a_{2023}=(-1)^{(2023-1)/2}\binom{2023}{2023}\\ {}=(-1)^{1011}\\ {}=-1. \end{gathered}$

Thus, the correct answer is C.

2023 AMC 12A 第 25 题

答案：C

难度评级：2650

解答：

其他年份的第 25 题