Question

递归定义数列： $x_0=5$ ，且

$x_{n+1}=\dfrac{x_n^2+5x_n+4}{x_n+6}$

对所有非负整数 $n$ 成立。设 $m$ 为满足

$x_m\le4+\dfrac{1}{2^{20}}.$

的最小正整数。 $m$ 位于下列哪个区间？

Define a sequence recursively by $x_0=5$ and

$x_{n+1}=\dfrac{x_n^2+5x_n+4}{x_n+6}$

for all nonnegative integers $n.$ Let $m$ be the least positive integer such that

$x_m\le4+\dfrac{1}{2^{20}}.$

In which of the following intervals does $m$ lie?

$[9,26]$

$[27,80]$

$[81,242]$

$[243,728]$

$[729,\infty)$

答案：C

知识点：递推裂项相消极限情形界定

难度评级：2330

解答：

令 $a_n=x_n-4$ 。简单计算得到 $\begin{gathered} a_{n+1}=x_{n+1}-4 \\ =\dfrac{(x_n+5)(x_n-4)}{x_n+6} \\ =a_n\cdot\dfrac{x_n+5}{x_n+6}. \end{gathered}$

从 $a_0=1$ 开始，各项保持为正并递减。因为 $x_n$ 从 $5$ 递减趋向 $4$ ，每个比值 $\dfrac{x_n+5}{x_n+6}$ 都严格介于 $\dfrac{9}{10}$ 和 $\dfrac{10}{11}$ 之间。

因此 $a_m$ 被夹在 $\left(\dfrac{9}{10}\right)^m$ 与 $\left(\dfrac{10}{11}\right)^m$ 之间。求解 $a_m\le2^{-20}$ 可知 $m$ 大约在 $132$ 到 $146$ 之间，落在 $[81,242]$ 中。

所以正确答案是 C。

Let $a_n=x_n-4.$ A short computation gives $\begin{gathered} a_{n+1}=x_{n+1}-4 \\ =\dfrac{(x_n+5)(x_n-4)}{x_n+6} \\ =a_n\cdot\dfrac{x_n+5}{x_n+6}. \end{gathered}$

Starting from $a_0=1,$ the terms stay positive and decrease. Because $x_n$ decreases from $5$ toward $4,$ each ratio $\dfrac{x_n+5}{x_n+6}$ lies strictly between $\dfrac{9}{10}$ and $\dfrac{10}{11}.$

Hence $a_m$ is squeezed between $\left(\dfrac{9}{10}\right)^m$ and $\left(\dfrac{10}{11}\right)^m.$ Solving $a_m\le2^{-20}$ puts $m$ between about $132$ and $146,$ which lies in $[81,242].$

Thus, C is the correct answer.

2019 AMC 12B 第 22 题

答案：C

难度评级：2330

解答：

其他年份的第 22 题