2009 AMC 12A 第 25 题

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.