Difference between revisions of "2001 AIME II Problems/Problem 3"
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At this point it is pretty clear that the sequence is periodic with period 10 (one may prove it quite easily like in solution 1) so our answer is obviously <math>211+420+267=\boxed{898}</math> | At this point it is pretty clear that the sequence is periodic with period 10 (one may prove it quite easily like in solution 1) so our answer is obviously <math>211+420+267=\boxed{898}</math> | ||
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+ | ~Dhillonr25 | ||
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== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == | ||
https://youtu.be/lH-0ul1hwKw?t=870 | https://youtu.be/lH-0ul1hwKw?t=870 |
Latest revision as of 00:28, 12 November 2022
Contents
[hide]Problem
Given that
find the value of .
Solution
We find that by the recursive formula. Summing the recursions
yields . Thus . Since , it follows that
Solution Variant
The recursive formula suggests telescoping. Indeed, if we add and , we have .
Subtracting yields .
Thus,
Notice that we didn't need to use the values of or at all.
Non-Rigorous Solution
Calculate the first few terms:
At this point it is pretty clear that the sequence is periodic with period 10 (one may prove it quite easily like in solution 1) so our answer is obviously
~Dhillonr25
Video Solution by OmegaLearn
https://youtu.be/lH-0ul1hwKw?t=870
~ pi_is_3.14
See also
2001 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.