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> | ||
| + | |||
| + | ~Dhillonr25 | ||
| + | |||
== 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
Problem
Given that
find the value of 
.
Solution
We find that 
by the recursive formula. Summing the recursions
yields 
. Thus 
. Since 
, it follows that
![\[x_{531} + x_{753} + x_{975} = (-1)^{106} x_1 + (-1)^{150} x_3 + (-1)^{194} x_5 = 211 + 420 + 267 = \boxed{898}.\]](http://latex.artofproblemsolving.com/c/d/f/cdf59f4694003dd3213f49cc4c7086ac6a437ab7.png)
Solution Variant
The recursive formula suggests telescoping. Indeed, if we add 
and 
, we have 
.
Subtracting yields 
.
Thus,
![\[x_{531} + x_{753} + x_{975} = x_1 + x_3 + x_5 = x_1 + x_3 + (x_4 - x_3 + x_2 - x_1) = x_2 + x_4 = 375 + 523 = \boxed{898}.\]](http://latex.artofproblemsolving.com/5/d/5/5d591452aa2842e8fb0d6c669560a007e57e5c87.png)
Notice that we didn't need to use the values of 
or 
at all.
Non-Rigorous Solution
Calculate the first few terms:
![\[211,375,420,523,267,-211,-375,-420,-523,\dots\]](http://latex.artofproblemsolving.com/f/1/f/f1f8c92c567a04a4a9d057c707bb3854a3fdb8c6.png)
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. 
