2008 AIME II 第 6 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

6.

数列

\{a_n\}

定义为

\begin{aligned} a_0 &= 1, \\ a_1 &= 1, \\ a_n &= a_{n-1} + \frac{a_{n-1}^2}{a_{n-2}} \quad (n \ge 2) \end{aligned}\text{。}

数列

\{b_n\}

定义为

\begin{aligned} b_0 &= 1, \\ b_1 &= 3, \\ b_n &= b_{n-1} + \frac{b_{n-1}^2}{b_{n-2}} \quad (n \ge 2) \end{aligned}\text{。}

求

\frac{b_{32}}{a_{32}}

。

The sequence

\{a_n\}

is defined by

\begin{aligned} a_0 &= 1, \\ a_1 &= 1, \\ a_n &= a_{n-1} + \frac{a_{n-1}^2}{a_{n-2}} \quad (n \ge 2). \end{aligned}

The sequence

\{b_n\}

is defined by

\begin{aligned} b_0 &= 1, \\ b_1 &= 3, \\ b_n &= b_{n-1} + \frac{b_{n-1}^2}{b_{n-2}} \quad (n \ge 2). \end{aligned}

Find

\frac{b_{32}}{a_{32}}.

答案

答案：561

知识点：递推阶乘裂项相消

难度评级：2460

小提示：

将递推式除以

a_{n-1}

可看出比值

\frac{a_n}{a_{n-1}}

每一步恰好增加

1

Divide the recurrence by

a_{n-1}

to see that the ratio

\frac{a_n}{a_{n-1}}

increases by exactly

1

at each step

大提示：

这些比值给出

a_n = n!

和

b_n = \frac{(n+2)!}{2}

；再求商

The ratios give

a_n = n!

and

b_n = \frac{(n+2)!}{2};

form the quotient

解答：

将递推式除以

a_{n-1}

，得到

\frac{a_n}{a_{n-1}} = 1 + \frac{a_{n-1}}{a_{n-2}}\text{，}

所以相邻项比值每一步恰好增加

1

。对

\{a_n\}

，第一个比值为

\frac{a_1}{a_0} = 1

，所以

\frac{a_n}{a_{n-1}} = n

，从而

a_n = n!

。同样的计算适用于

\{b_n\}

，它的第一个比值为

\frac{b_1}{b_0} = 3

，所以

\frac{b_n}{b_{n-1}} = n + 2

，且

b_n = \frac{(n+2)!}{2}

。

因此

\frac{b_{32}}{a_{32}} = \frac{\frac{34!}{2}}{32!} = \frac{34 \cdot 33}{2} = 561\text{。}

Dividing the recurrence by

a_{n-1}

gives

\frac{a_n}{a_{n-1}} = 1 + \frac{a_{n-1}}{a_{n-2}},

so the consecutive-term ratio increases by exactly

1

each step. For

\{a_n\}

the first ratio is

\frac{a_1}{a_0} = 1,

so

\frac{a_n}{a_{n-1}} = n

and

a_n = n!.

The same computation applies to

\{b_n\},

whose first ratio is

\frac{b_1}{b_0} = 3,

so

\frac{b_n}{b_{n-1}} = n + 2

and

b_n = \frac{(n+2)!}{2}.

Therefore

\frac{b_{32}}{a_{32}} = \frac{\frac{34!}{2}}{32!} = \frac{34 \cdot 33}{2} = 561.