2020 AIME II Problems/Problem 6

Problem

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 $p+q$ .

Solution

Let $t_n=\frac{s_n}{5}$ . Then, we have $s_n=\frac{s_{n-1}+1}{s_{n-2}}$ where $s_1 = 100$ and $s_2 = 105$ . By substitution, we find $s_3 = \frac{53}{50}$ , $s_4=\frac{103}{105\cdot50}$ , $s_5=\frac{101}{105}$ , $s_6=100$ , and $s_7=105$ . So $s_n$ has a period of $5$ . Thus $s_{2020}=s_5=\frac{101}{105}$ . So, $\frac{101}{105\cdot 5}\implies 101+525=\boxed{626}$ . ~mn28407

Video Solution

https://youtu.be/_JTWJxbDC1A ~ CNCM

Video Solution 2

https://youtu.be/__B3pJMpfSk

~IceMatrix

2020 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2020 AIME II Problems/Problem 6

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Problem

Solution

Video Solution

Video Solution 2

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