Question

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

Solution:

Computing terms exactly: $t_3 = \frac{5 \cdot 21 + 1}{25 \cdot 20} = \frac{53}{250}, \qquad t_4 = \frac{5 \cdot \frac{53}{250} + 1}{25 \cdot 21} = \frac{103}{26250}, \qquad t_5 = \frac{5 \cdot \frac{103}{26250} + 1}{25 \cdot \frac{53}{250}} = \frac{5353}{5250} \cdot \frac{10}{53} = \frac{101}{525},$ 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 Problem 6

Solution:

Problem 6 in Other Years