2002 AIME I Problems/Problem 4
Problem
Consider the sequence defined by for . Given that , for positive integers and with , find .
Solution 1
. Thus,
Which means that
Since we need a factor of 29 in the denominator, we let .* Substituting, we get
so
Since is an integer, , so . It quickly follows that and , so .
*If , a similar argument to the one above implies and , which implies . This is impossible since .
Solution 2
Note that . This can be proven by induction. Thus, . Cross-multiplying yields , and adding to both sides gives . Clearly, and . Hence, , , and .
~ keeper1098
Video Solution by OmegaLearn
https://youtu.be/lH-0ul1hwKw?t=134
~ pi_is_3.14
See also
2002 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.