AMC10 2019 B

AMC10 2019 B · Q24

AMC10 2019 B · Q24. 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?

递归定义序列 $x_0 = 5$ 且 $$x_{n+1} = \frac{x_n^2 + 5x_n + 4}{x_n + 6}$$ 对所有非负整数 $n$。设 $m$ 为最小的正整数使得 $$x_m \le 4 + \frac{1}{220}。$$ $m$ 位于以下哪个区间？

(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

24. Answer (C): First note that it suffices to study $y_n=x_n-4$ and find the least positive integer $m$ such that $y_m\le \frac{1}{2^{20}}$. Now $y_0=1$ and \[ y_{n+1}=\frac{y_n(y_n+9)}{y_n+10}. \] Observe that $(y_n)$ is a strictly decreasing sequence of positive numbers. 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 $y_0=1$, \[ \left(\frac{9}{10}\right)^k \le y_k \le \left(\frac{10}{11}\right)^k \] for all integers $k\ge 2$. Now note that \[ \left(\frac12\right)^{\frac14}<\frac{9}{10} \] because this is equivalent to $0.5<(0.9)^4=(0.81)^2$. Therefore \[ \left(\frac12\right)^{\frac{m}{4}}<y_m\le \frac{1}{2^{20}}, \] so $m>80$. Now note that \[ \left(\frac{11}{10}\right)^{10}=\left(1+\frac{1}{10}\right)^{10}>1+10\cdot\frac{1}{10}=2, \] so \[ \frac{10}{11}<\left(\frac12\right)^{\frac{1}{10}}. \] Thus \[ \frac{1}{2^{20}}<y_{m-1}<\left(\frac{10}{11}\right)^{m-1}<\left(\frac12\right)^{\frac{m-1}{10}}, \] so $m<201$. Thus $m$ lies in the range (C). (Numerical calculations will show that $m=133$.

24. 答案（C）：首先注意到，只需研究 $y_n=x_n-4$，并找出使得 $y_m\le \frac{1}{2^{20}}$ 的最小正整数 $m$。现有 $y_0=1$，且 \[ y_{n+1}=\frac{y_n(y_n+9)}{y_n+10}. \] 可见 $(y_n)$ 是一个严格递减的正数序列。因为 \[ \frac{y_{n+1}}{y_n}=1-\frac{1}{y_n+10}, \] 于是得到 \[ \frac{9}{10}\le \frac{y_{n+1}}{y_n}\le \frac{10}{11}, \] 又由于 $y_0=1$， \[ \left(\frac{9}{10}\right)^k \le y_k \le \left(\frac{10}{11}\right)^k \] 对所有整数 $k\ge 2$ 成立。再注意到 \[ \left(\frac12\right)^{\frac14}<\frac{9}{10}, \] 因为这等价于 $0.5<(0.9)^4=(0.81)^2$。因此 \[ \left(\frac12\right)^{\frac{m}{4}}<y_m\le \frac{1}{2^{20}}, \] 所以 $m>80$。另外注意 \[ \left(\frac{11}{10}\right)^{10}=\left(1+\frac{1}{10}\right)^{10}>1+10\cdot\frac{1}{10}=2, \] 故 \[ \frac{10}{11}<\left(\frac12\right)^{\frac{1}{10}}. \] 因此 \[ \frac{1}{2^{20}}<y_{m-1}<\left(\frac{10}{11}\right)^{m-1}<\left(\frac12\right)^{\frac{m-1}{10}}, \] 从而 $m<201$。于是 $m$ 落在范围（C）内。（数值计算会显示 $m=133$。

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