AMC12 2019 A

AMC12 2019 A · Q9

AMC12 2019 A · Q9. It mainly tests Sequences & recursion (algebra), Manipulating equations.

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

一个数列由 $a_1 = 1$、$a_2 = 3/7$ 和对于所有 $n \geq 3$，$a_n = (a_{n-2} \cdot a_{n-1})/(2a_{n-2} - a_{n-1})$ 递归定义。那么 $a_{2019}$ 可以写成 $p/q$，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 $1,\ \frac{3}{7},\ \frac{3}{11},\ \frac{3}{15},\ \frac{3}{19},\ \cdots$. 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 $a_{2019}=\frac{3}{4\cdot 2019-1}=\frac{3}{8075}$, and the requested sum is $3+8075=8078$.

答案（E）：该数列从 $1,\ \frac{3}{7},\ \frac{3}{11},\ \frac{3}{15},\ \frac{3}{19},\ \cdots$ 开始。由此规律可猜想 $a_n=\frac{3}{4n-1}$。检验初始条件 $n=1$ 与 $n=2$，并注意当 $n\ge 3$ 时， $$ \frac{\frac{3}{4(n-2)-1}\cdot \frac{3}{4(n-1)-1}}{2\cdot \frac{3}{4(n-2)-1}-\frac{3}{4(n-1)-1}} = \frac{\frac{3}{4n-9}\cdot \frac{3}{4n-5}}{\frac{6}{4n-9}-\frac{3}{4n-5}} = \frac{9}{6(4n-5)-3(4n-9)} = \frac{9}{12n-3} = \frac{3}{4n-1}, $$ 从而验证了该猜想。因此 $a_{2019}=\frac{3}{4\cdot 2019-1}=\frac{3}{8075}$，所求和为 $3+8075=8078$。

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