Question

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

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

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

解答：

将递推式除以 $a_{n-1}$ ，得到 $\frac{a_n}{a_{n-1}} = 1 + \frac{a_{n-1}}{a_{n-2}},$ 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 $\frac{b_{32}}{a_{32}} = \frac{34!/2}{32!} = \frac{34 \cdot 33}{2} = 561.$

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

2008 AIME II 第 6 题

答案：561

难度评级：2460

解答：

其他年份的第 6 题