Question

递归定义数列 $t_1 = 20$ 、 $t_2 = 21$ ，并且对所有 $n \ge 3$ ， $t_n = \frac{5t_{n-1} + 1}{25t_{n-2}}$ 那么 $t_{2020}$ 可写成 $\frac{p}{q}$ ，其中 $p$ 与 $q$ 是互质正整数。求 $p + q$ 。

Define a sequence recursively by $t_1 = 20,$ $t_2 = 21,$ and $t_n = \frac{5t_{n-1} + 1}{25t_{n-2}}$ for all $n \ge 3.$ Then $t_{2020}$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

答案

答案：626

知识点：递推找规律

难度评级：2340

解答：

精确计算各项： $\begin{aligned} t_3 &= \frac{5 \cdot 21 + 1}{25 \cdot 20} = \frac{53}{250}, \\ t_4 &= \frac{5 \cdot \frac{53}{250} + 1}{25 \cdot 21} = \frac{103}{26250}, \\ t_5 &= \frac{5 \cdot \frac{103}{26250} + 1}{25 \cdot \frac{53}{250}} \\ &= \frac{5353}{5250} \cdot \frac{10}{53} = \frac{101}{525}, \end{aligned}$ 其中用到 $5353 = 53 \cdot 101$ 。接着 $t_6 = \frac{5 \cdot \frac{101}{525} + 1}{25 \cdot \frac{103}{26250}}$ $= \frac{206}{105} \cdot \frac{1050}{103}$ $= 20 = t_1$ ，且 $t_7 = \frac{101}{101/21} = 21 = t_2$ 。

因为每一项只由前两项决定，数列以 $5$ 为周期。由于 $2020$ 是 $5$ 的倍数，得到 $t_{2020} = t_5 = \frac{101}{525}$ 。因为 $101$ 是质数且不整除 $525$ ，该分数已经约分，所以 $p + q = 101 + 525 = 626$ 。

Computing terms exactly: $\begin{aligned} t_3 &= \frac{5 \cdot 21 + 1}{25 \cdot 20} = \frac{53}{250}, \\ t_4 &= \frac{5 \cdot \frac{53}{250} + 1}{25 \cdot 21} = \frac{103}{26250}, \\ t_5 &= \frac{5 \cdot \frac{103}{26250} + 1}{25 \cdot \frac{53}{250}} \\ &= \frac{5353}{5250} \cdot \frac{10}{53} = \frac{101}{525}, \end{aligned}$ using $5353 = 53 \cdot 101.$ Then $t_6 = \frac{5 \cdot \frac{101}{525} + 1}{25 \cdot \frac{103}{26250}}$ $= \frac{206}{105} \cdot \frac{1050}{103}$ $= 20 = t_1$ and $t_7 = \frac{101}{101/21} = 21 = t_2.$

Since each term depends only on the two preceding terms, the sequence repeats with period $5.$ Because $2020$ is a multiple of $5,$ we get $t_{2020} = t_5 = \frac{101}{525}.$ As $101$ is prime and does not divide $525,$ the fraction is reduced, and $p + q = 101 + 525 = 626.$

2020 AIME II 第 6 题

答案：626

难度评级：2340

解答：

其他年份的第 6 题