Difference between revisions of "2020 AIME II Problems/Problem 6"
(→Solution 3) |
(→Solution 3) |
||
Line 16: | Line 16: | ||
Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given <math>t_1 = 20</math> and <math>t_2 = 21</math>, so now we are able to determine the numerical value of <math>t_3</math> using this information: | Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given <math>t_1 = 20</math> and <math>t_2 = 21</math>, so now we are able to determine the numerical value of <math>t_3</math> using this information: | ||
<cmath>t_3 = \frac{5t_{3-1}+1}{25t_{3-2}} = \frac{5t_{2}+1}{25t_{1}} = \frac{5(21) + 1}{25(20)} = \frac{105 + 1}{500}t_3 = \frac{106}{500} = \frac{53}{250}</cmath> | <cmath>t_3 = \frac{5t_{3-1}+1}{25t_{3-2}} = \frac{5t_{2}+1}{25t_{1}} = \frac{5(21) + 1}{25(20)} = \frac{105 + 1}{500}t_3 = \frac{106}{500} = \frac{53}{250}</cmath> | ||
− | |||
<cmath>t_4 = \frac{5t_{4-1}+1}{25t_{4-2}} = \frac{5t_{3}+1}{25t_{2}} = \frac{5(\frac{53}{250}) + 1}{25(21)} = \frac{\frac{53}{50} + 1}{525} = \frac{\frac{103}{50}}{525} = \frac{103}{26250}</cmath> | <cmath>t_4 = \frac{5t_{4-1}+1}{25t_{4-2}} = \frac{5t_{3}+1}{25t_{2}} = \frac{5(\frac{53}{250}) + 1}{25(21)} = \frac{\frac{53}{50} + 1}{525} = \frac{\frac{103}{50}}{525} = \frac{103}{26250}</cmath> | ||
− | |||
<cmath>t_5 = \frac{5t_{5-1}+1}{25t_{5-2}} = \frac{5t_{4}+1}{25t_{3}} = \frac{5(\frac{103}{26250}) + 1}{25(\frac{53}{250})} = \frac{\frac{103}{5250} + 1}{\frac{53}{10}} = \frac{\frac{5353}{5250}}{\frac{53}{10}} = \frac{101}{525}</cmath> | <cmath>t_5 = \frac{5t_{5-1}+1}{25t_{5-2}} = \frac{5t_{4}+1}{25t_{3}} = \frac{5(\frac{103}{26250}) + 1}{25(\frac{53}{250})} = \frac{\frac{103}{5250} + 1}{\frac{53}{10}} = \frac{\frac{5353}{5250}}{\frac{53}{10}} = \frac{101}{525}</cmath> | ||
− | |||
<cmath>t_6 = \frac{5t_{6-1}+1}{25t_{6-2}} = \frac{5t_{5}+1}{25t_{4}} = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})} = \frac{\frac{101}{105} + 1}{\frac{103}{1050}} = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 20</cmath> | <cmath>t_6 = \frac{5t_{6-1}+1}{25t_{6-2}} = \frac{5t_{5}+1}{25t_{4}} = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})} = \frac{\frac{101}{105} + 1}{\frac{103}{1050}} = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 20</cmath> | ||
Revision as of 11:48, 31 March 2022
Contents
[hide]Problem
Define a sequence recursively by , , andfor all . Then can be expressed as , where and are relatively prime positive integers. Find .
Solution
Let . Then, we have where and . By substitution, we find , , , , and . So has a period of . Thus . So, . ~mn28407
Solution 2 (Official MAA)
More generally, let the first two terms be and and replace and in the recursive formula by and , respectively. Then some algebraic calculation shows that so the sequence is periodic with period . Therefore The requested sum is .
Solution 3
Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given and , so now we are able to determine the numerical value of using this information:
Alas, we have figured this sequence is period 5! Thus, let us take , which is , and therefore . According to the original problem statement, our answer is essentially . ~ nikenissan
Video Solution
https://youtu.be/_JTWJxbDC1A ~ CNCM
Video Solution 2
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.