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Difference between revisions of "2020 AIME II Problems/Problem 6"

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Let <math>t_n=\frac{s_n}{5}</math>. Then, we have <math>s_n=\frac{s_{n-1}+1}{s_{n-2}}</math> where <math>s_1 = 100</math> and <math>s_2 = 105</math>. By substitution, we find <math>s_3 = \frac{53}{50}</math>, <math>s_4=\frac{103}{105\cdot50}</math>, <math>s_5=\frac{101}{105}</math>, <math>s_6=100</math>, and <math>s_7=105</math>. So <math>s_n</math> has a period of <math>5</math>. Thus <math>s_{2020}=s_5=\frac{101}{105}</math>. So, <math>\frac{101}{105\cdot 5}\implies 101+525=\boxed{626}</math>.
 
Let <math>t_n=\frac{s_n}{5}</math>. Then, we have <math>s_n=\frac{s_{n-1}+1}{s_{n-2}}</math> where <math>s_1 = 100</math> and <math>s_2 = 105</math>. By substitution, we find <math>s_3 = \frac{53}{50}</math>, <math>s_4=\frac{103}{105\cdot50}</math>, <math>s_5=\frac{101}{105}</math>, <math>s_6=100</math>, and <math>s_7=105</math>. So <math>s_n</math> has a period of <math>5</math>. Thus <math>s_{2020}=s_5=\frac{101}{105}</math>. So, <math>\frac{101}{105\cdot 5}\implies 101+525=\boxed{626}</math>.
 
~mn28407
 
~mn28407
 +
 +
==Solution 2 (Official MAA)==
 +
More generally, let the first two terms be <math>a</math> and <math>b</math> and replace <math>5</math> and <math>25</math> in the recursive formula by <math>k</math> and <math>k^2</math>, respectively. Then some algebraic calculation shows that
 +
<cmath>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,</cmath>so the sequence is periodic with period <math>5</math>. Therefore
 +
<cmath>t_{2020} = t_{5} =  \frac{20\cdot 5 +1}{21\cdot  25} = \frac{101}{525}.</cmath>The requested sum is <math>101+525=626</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 12:24, 8 June 2020

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 $101+525=626$.

Video Solution

https://youtu.be/_JTWJxbDC1A ~ CNCM

Video Solution 2

https://youtu.be/__B3pJMpfSk

~IceMatrix

See Also

2020 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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