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

Solution 2 (Official MAA)

More generally, let the first two terms be $a$ and $b$ and replace $5$ and $25$ in the recursive formula by $k$ and $k^2$ , respectively. Then some algebraic calculation shows that $t_3 = \frac{b\,k+1}{a\, k^2},~~t_4 = \frac{a\, k + b\,k+1}{a\,b\, k^3},~~ t_5 = \frac{a\,k+1}{b\, k^2},~~ t_6 = a, \text{~ and ~}t_7 =b,$ so the sequence is periodic with period $5$ . Therefore $t_{2020} = t_{5} = \frac{20\cdot 5 +1}{21\cdot 25} = \frac{101}{525}.$ The requested sum is $101+525=626$ .

Solution 3

Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given $t_1 = 20$ and $t_2 = 21$ , so now we are able to determine the numerical value of $t_3$ using this information: $t_3 = \frac{5t_{3-1}+1}{25t_{3-2}}$ $\implies t_3 = \frac{5t_{2}+1}{25t_{1}}$ $\implies t_3 = \frac{5(21) + 1}{25(20)}$ $\implies t_3 = \frac{105 + 1}{500}$ $\implies t_3 = \frac{106}{500} \implies t_3 = \frac{53}{250}$ Now using this information, as well as the previous information, we are able to determine the value of $t_4$ : $t_4 = \frac{5t_{4-1}+1}{25t_{4-2}}$ $\implies t_4 = \frac{5t_{3}+1}{25t_{2}}$ $\implies t_4 = \frac{5(\frac{53}{250}) + 1}{25(21)}$ $\implies t_4 = \frac{\frac{53}{50} + 1}{525}$ $\implies t_4 = \frac{\frac{103}{50}}{525} \implies t_4 = \frac{103}{26250}$ Now using this information, as well as the previous information, we are able to determine the value of $t_5$ : $t_5 = \frac{5t_{5-1}+1}{25t_{5-2}}$ $\implies t_5 = \frac{5t_{4}+1}{25t_{3}}$ $\implies t_5 = \frac{5(\frac{103}{26250}) + 1}{25(\frac{53}{250})}$ $\implies t_5 = \frac{\frac{53}{5250} + 1}{\frac{53}{10}}$ $\implies t_5 = \frac{\frac{5353}{5250}}{\frac{53}{10}} \implies t_5 = \frac{101}{525}$ Now using this information, as well as the previous information, we are able to determine the value of $t_6$ : $t_6 = \frac{5t_{6-1}+1}{25t_{6-2}}$ $\implies t_6 = \frac{5t_{5}+1}{25t_{4}}$ $\implies t_5 = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})}$ $\implies t_5 = \frac{\frac{101}{105} + 1}{\frac{103}{1050}}$ $\implies t_5 = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 21$

Alas, we have figured this sequence is period 5! Thus, let us take $2020 \pmod 5$ , which is $5$ , and therefore $t_{2020} = t_5 = \frac{101}{525}$ . According to the original problem statement, our answer is essentially $\boxed{626}$ . ~ nikenissan

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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