Difference between revisions of "2001 AIME II Problems/Problem 3"
Dhillonr25 (talk | contribs) (→Non-Rigorous Solution) |
|||
(2 intermediate revisions by 2 users not shown) | |||
Line 34: | Line 34: | ||
Notice that we didn't need to use the values of <math>x_1</math> or <math>x_3</math> at all. | Notice that we didn't need to use the values of <math>x_1</math> or <math>x_3</math> at all. | ||
+ | |||
+ | |||
+ | ==Non-Rigorous Solution== | ||
+ | |||
+ | Calculate the first few terms: | ||
+ | |||
+ | <cmath>211,375,420,523,267,-211,-375,-420,-523,\dots</cmath> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ~Dhillonr25 | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/lH-0ul1hwKw?t=870 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
== See also == | == See also == | ||
{{AIME box|year=2001|n=II|num-b=2|num-a=4}} | {{AIME box|year=2001|n=II|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:28, 12 November 2022
Contents
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.