Question

数列递归定义为 $a_1 = 1$ ， $a_2 = \frac{3}{7}$ ，且对所有 $n \geq 3$ ，有 $a_{2019}$ 可写成 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 为互质正整数。求 $p+q$ 。 $a_n=\dfrac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}$

A sequence of numbers is defined recursively by $a_1 = 1,$ $a_2 = \frac{3}{7},$ and $a_n=\dfrac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}$ for all $n \geq 3.$ Then $a_{2019}$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $p+q ?$

$2020$

$4039$

$6057$

$6061$

$8078$

答案：E

知识点：递推等差数列换元法

难度评级：1540

解答：

$\begin{aligned} \frac1{a_n} &=\frac{2a_{n-2}-a_{n-1}}{a_{n-2}a_{n-1}}\\ &=\frac2{a_{n-1}}-\frac1{a_{n-2}}. \end{aligned}$

对递推式取倒数，得到 $\left\{\dfrac{1}{a_n}\right\}$ $\dfrac{1}{a_n} - \dfrac{1}{a_{n - 1}} = \dfrac{1}{a_{n - 1}} - \dfrac{1}{a_{n - 2}},$

这说明因而 $a_1$ 是等差数列。 $a_2$ ， $\dfrac{1}{\frac{3}{7}} - \dfrac{1}{1} = \dfrac{7}{3} - 1 = \dfrac{4}{3}.$

由 $a_{2019}=\frac3{8075}$ 和，可得公差为 $\frac1{a_{2019}}=1+2018\cdot\frac43=\frac{8075}{3},$

因此 $p + q$ 为 $8075 + 3 = 8078.$

所以正确答案是 E。

Taking reciprocals in the recursive formula gives $\begin{aligned} \frac1{a_n} &=\frac{2a_{n-2}-a_{n-1}}{a_{n-2}a_{n-1}}\\ &=\frac2{a_{n-1}}-\frac1{a_{n-2}}. \end{aligned}$

This means that $\dfrac{1}{a_n} - \dfrac{1}{a_{n - 1}} = \dfrac{1}{a_{n - 1}} - \dfrac{1}{a_{n - 2}},$ which tells us that $\left\{\dfrac{1}{a_n}\right\}$ is an arithmetic sequence.

Using $a_1$ and $a_2,$ we get that the common difference is $\dfrac{1}{\frac{3}{7}} - \dfrac{1}{1} = \dfrac{7}{3} - 1 = \dfrac{4}{3}.$

Therefore $\frac1{a_{2019}}=1+2018\cdot\frac43=\frac{8075}{3},$ so $a_{2019}=\frac3{8075}.$

$p + q$ is therefore $8075 + 3 = 8078.$

Thus, E is the correct answer.

2019 AMC 10A 第 15 题

答案：E

难度评级：1540

解答：

其他年份的第 15 题