AMC10 2019 A

AMC10 2019 A · Q15

AMC10 2019 A · Q15. It mainly tests Sequences & recursion (algebra), Manipulating equations.

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

一个数列由 $a_1 = 1$，$a_2 = \frac{3}{7}$，且对于所有 $n \ge 3$，$a_n = \frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}$ 递推定义。那么 $a_{2019}$ 可以写成 $\frac{p}{q}$，其中 $p$ 和 $q$ 是互质的正整数。$p + q$ 是多少？

(A) 2020 2020

(B) 4039 4039

(D) 6061 6061

(E) 8078 8078

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): The sequence begins 1, $\frac{3}{7}$, $\frac{3}{11}$, $\frac{3}{15}$, $\frac{3}{19}$, $\ldots$ This pattern leads to the conjecture that $a_n=\frac{3}{4n-1}$. Checking the initial conditions $n=1$ and $n=2$, and observing that for $n\ge 3$, \[ \frac{\frac{3}{4(n-2)-1}\cdot\frac{3}{4(n-1)-1}}{2\cdot\frac{3}{4(n-2)-1}-\frac{3}{4(n-1)-1}} = \frac{\frac{3}{4n-9}\cdot\frac{3}{4n-5}}{\frac{6}{4n-9}-\frac{3}{4n-5}} = \frac{9}{6(4n-5)-3(4n-9)} = \frac{9}{12n-3} = \frac{3}{4n-1}, \] confirms the conjecture. Therefore $a_{2019}=\frac{3}{4\cdot2019-1}=\frac{3}{8075}$, and the requested sum is $3+8075=8078$.

答案（E）：该数列开始为 1、$\frac{3}{7}$、$\frac{3}{11}$、$\frac{3}{15}$、$\frac{3}{19}$、$\ldots$ 这一规律引导我们猜想 $a_n=\frac{3}{4n-1}$。检验初始条件 $n=1$ 与 $n=2$，并注意到当 $n\ge 3$ 时， \[ \frac{\frac{3}{4(n-2)-1}\cdot\frac{3}{4(n-1)-1}}{2\cdot\frac{3}{4(n-2)-1}-\frac{3}{4(n-1)-1}} = \frac{\frac{3}{4n-9}\cdot\frac{3}{4n-5}}{\frac{6}{4n-9}-\frac{3}{4n-5}} = \frac{9}{6(4n-5)-3(4n-9)} = \frac{9}{12n-3} = \frac{3}{4n-1}, \] 从而证实该猜想。因此 $a_{2019}=\frac{3}{4\cdot2019-1}=\frac{3}{8075}$，所求的和为 $3+8075=8078$。

Topics