Question

一个数列的前两项为 $a_1 = 1$ 和 $a_2 = \dfrac{1}{\sqrt{3}}$ 。对于 $n \ge 1$ ， $a_{n+2} = \dfrac{a_n + a_{n+1}}{1 - a_n a_{n+1}}\text{。}$ 求 $|a_{2009}|$ ？

The first two terms of a sequence are $a_1 = 1$ and $a_2 = \dfrac{1}{\sqrt{3}}.$ For $n \ge 1,$ $a_{n+2} = \dfrac{a_n + a_{n+1}}{1 - a_n a_{n+1}}.$ What is $|a_{2009}|?$

$0$

$2 - \sqrt{3}$

$\dfrac{1}{\sqrt{3}}$

$1$

$2 + \sqrt{3}$

答案：A

知识点：三角恒等式递推模运算

难度评级：2520

小提示：

递推式对应正切加法公式 $\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$ 。

The recursion matches the tangent addition formula $\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$

大提示：

因为 $a_1 = \tan\dfrac{\pi}{4}$ 且 $a_2 = \tan\dfrac{\pi}{6}$ ，写成 $a_n = \tan\dfrac{\pi c_n}{12}$ ，并追踪 $c_n \bmod 12$ 。

Since $a_1 = \tan\dfrac{\pi}{4}$ and $a_2 = \tan\dfrac{\pi}{6},$ write $a_n = \tan\dfrac{\pi c_n}{12}$ and track $c_n \bmod 12$

解答：

递推式正是正切加法公式，并且 $a_1 = \tan\dfrac{\pi}{4}$ ， $a_2 = \tan\dfrac{\pi}{6}$ 。

写成 $a_n = \tan\dfrac{\pi c_n}{12}$ ，其中 $c_1 = 3$ 、 $c_2 = 2$ ，且 $c_{n+2} \equiv c_n + c_{n+1} \pmod{12}$ 。序列 $c_n$ 为 $\begin{gathered} 3, 2, 5, 7, 0, 7, 7, 2, \\ 9, 11, 8, 7, 3, 10, 1, 11, \\ 0, 11, 11, 10, 9, 7, 4, 11, \ldots \end{gathered}$ 其周期为 $24$ 。

因为 $2009 = 24\cdot 83 + 17$ ，所以 $c_{2009} = c_{17} = 0$ ，从而 $a_{2009} = \tan 0 = 0$ ， $|a_{2009}| = 0$ 。

因此，正确答案是 A。

The recursion is exactly the tangent addition formula, and $a_1 = \tan\dfrac{\pi}{4},$ $a_2 = \tan\dfrac{\pi}{6}.$

Writing $a_n = \tan\dfrac{\pi c_n}{12}$ with $c_1 = 3,$ $c_2 = 2,$ and $c_{n+2} \equiv c_n + c_{n+1} \pmod{12},$ the sequence $c_n$ is $\begin{gathered} 3, 2, 5, 7, 0, 7, 7, 2, \\ 9, 11, 8, 7, 3, 10, 1, 11, \\ 0, 11, 11, 10, 9, 7, 4, 11, \ldots \end{gathered}$ which is periodic with period $24.$

Since $2009 = 24\cdot 83 + 17,$ $c_{2009} = c_{17} = 0,$ so $a_{2009} = \tan 0 = 0$ and $|a_{2009}| = 0.$

Thus, the correct answer is A.

2009 AMC 12A 第 25 题

答案：A

难度评级：2520

小提示：

大提示：

解答：

其他年份的第 25 题