Question

一个数列递归定义为 $a_1 = 1$ ， $a_2 = \dfrac{3}{7}$ ，且

$a_n = \dfrac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}$

对所有 $n \ge 3$ 成立。那么 $a_{2019}$ 可写成 $\dfrac{p}{q}$ ，其中 $p$ 和 $q$ 是互质的正整数。求 $p + q$ ？

A sequence of numbers is defined recursively by $a_1 = 1,$ $a_2 = \dfrac{3}{7},$ and

$a_n = \dfrac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}$

for all $n \ge 3.$ Then $a_{2019}$ can be written as $\dfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $p + q?$

$2020$

$4039$

$6057$

$6061$

$8078$

答案：E

知识点：递推换元法等差数列

难度评级：1500

解答：

取倒数， $\begin{aligned} \dfrac{1}{a_n} &= \dfrac{2a_{n-2} - a_{n-1}}{a_{n-2}a_{n-1}} \\ &= \dfrac{2}{a_{n-1}} - \dfrac{1}{a_{n-2}}. \end{aligned}$

令 $b_n = \dfrac{1}{a_n}$ 则 $b_n = 2b_{n-1} - b_{n-2}$ ，所以 $b_n$ 是等差数列，且 $b_1 = 1$ ， $b_2 = \dfrac{7}{3}$ ，公差为 $\dfrac{4}{3}$ 。

因此 $b_{2019} = 1 + 2018 \cdot \dfrac{4}{3} = \dfrac{8075}{3}$ ，所以 $a_{2019} = \dfrac{3}{8075}$ 。由于它们互质， $p + q = 8078$ 。

所以正确答案是 E。

Taking reciprocals, $\begin{aligned} \dfrac{1}{a_n} &= \dfrac{2a_{n-2} - a_{n-1}}{a_{n-2}a_{n-1}} \\ &= \dfrac{2}{a_{n-1}} - \dfrac{1}{a_{n-2}}. \end{aligned}$

Let $b_n = \dfrac{1}{a_n}.$ Then $b_n = 2b_{n-1} - b_{n-2},$ so $b_n$ is arithmetic with $b_1 = 1,$ $b_2 = \dfrac{7}{3},$ and common difference $\dfrac{4}{3}.$

Thus $b_{2019} = 1 + 2018 \cdot \dfrac{4}{3} = \dfrac{8075}{3},$ so $a_{2019} = \dfrac{3}{8075}.$ Since these are relatively prime, $p + q = 8078.$

Thus, the correct answer is E.

2019 AMC 12A 第 9 题

答案：E

难度评级：1500

解答：

其他年份的第 9 题