AMC12 2019 B

AMC12 2019 B · Q22

AMC12 2019 B · Q22. It mainly tests Sequences & recursion (algebra), Inequalities (AM-GM etc. basic).

Define a sequence recursively by $x_0 = 5$ and $$ x_{n+1} = \frac{x_n^2 + 5x_n + 4}{x_n + 6} $$ for all nonnegative integers $n$. Let $m$ be the least positive integer such that $$ x_m \le 4 + \frac{1}{220}. $$ In which of the following intervals does $m$ lie?

递归定义数列 $x_0 = 5$，且 $$ x_{n+1} = \frac{x_n^2 + 5x_n + 4}{x_n + 6} $$ 对所有非负整数 $n$ 成立。设 $m$ 为最小的正整数使得 $$ x_m \le 4 + \frac{1}{220}. $$ $m$ 位于下列哪个区间？

(A) [9, 26] [9, 26]

(B) [27, 80] [27, 80]

(D) [243, 728] [243, 728]

(E) [729, $\infty$) [729, $\infty$)

Answer

Correct choice: (C)

正确答案：（C）

Solution

It suffices to study $y_n = x_n - 4$ and find the least positive integer $m$ such that $y_m \le 1/250$. We have $y_0 = 1$ and $$ y_{n+1} = \frac{y_n(y_n + 9)}{y_n + 10}. $$ The sequence $(y_n)$ is strictly decreasing and positive. Because $$ \frac{y_{n+1}}{y_n} = 1 - \frac{1}{y_n + 10}, $$ it follows that $$ \frac{9}{10} \le \frac{y_{n+1}}{y_n} \le \frac{10}{11}, $$ and because $y_0 = 1$, $$ \left(\frac{9}{10}\right)^k \le y_k \le \left(\frac{10}{11}\right)^k \quad \text{for all integers } k \ge 2. $$ Now note that $$ \left(\frac{1}{2}\right)^{1/4} < \frac{9}{10} $$ because this is equivalent to $0.5 < (0.9)^4 = (0.81)^2$. Therefore $$ \left(\frac{1}{2}\right)^{m/4} < y_m \le \frac{1}{220} \implies m > 80. $$ Also, $$ \left(\frac{11}{10}\right)^{10} > 2 \implies \frac{10}{11} < \left(\frac{1}{2}\right)^{1/10}. $$ Thus $$ \frac{1}{220} < y_{m-1} < \left(\frac{10}{11}\right)^{m-1} < \left(\frac{1}{2}\right)^{(m-1)/10} \implies m < 201. $$ Hence $80 < m < 201$, so the correct choice is **(C)**. (Numerical computation shows $m = 133$.

研究 $y_n = x_n - 4$ 即可，找到最小的正整数 $m$ 使得 $y_m \le 1/250$。有 $y_0 = 1$ 和 $$ y_{n+1} = \frac{y_n(y_n + 9)}{y_n + 10}. $$ 数列 $(y_n)$ 严格递减且正。因为 $$ \frac{y_{n+1}}{y_n} = 1 - \frac{1}{y_n + 10}, $$ 故 $$ \frac{9}{10} \le \frac{y_{n+1}}{y_n} \le \frac{10}{11}, $$ 且因 $y_0 = 1$， $$ \left(\frac{9}{10}\right)^k \le y_k \le \left(\frac{10}{11}\right)^k \quad \text{对所有整数 } k \ge 2.$$ 注意 $$ \left(\frac{1}{2}\right)^{1/4} < \frac{9}{10} $$ 因为等价于 $0.5 < (0.9)^4 = (0.81)^2$。因此 $$ \left(\frac{1}{2}\right)^{m/4} < y_m \le \frac{1}{220} \implies m > 80. $$ 又 $$ \left(\frac{11}{10}\right)^{10} > 2 \implies \frac{10}{11} < \left(\frac{1}{2}\right)^{1/10}. $$ 故 $$ \frac{1}{220} < y_{m-1} < \left(\frac{10}{11}\right)^{m-1} < \left(\frac{1}{2}\right)^{(m-1)/10} \implies m < 201. $$ 因此 $80 < m < 201$，故正确选项为 **(C)**。（数值计算显示 $m = 133$）。

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