2001 AIME II Problems/Problem 3: Difference between revisions
No edit summary |
Dhillonr25 (talk | contribs) |
||
| (4 intermediate revisions by 3 users not shown) | |||
| Line 23: | Line 23: | ||
<cmath>x_{531} + x_{753} + x_{975} = (-1)^{106} x_1 + (-1)^{150} x_3 + (-1)^{194} x_5 = 211 + 420 + 267 = \boxed{898}.</cmath> | <cmath>x_{531} + x_{753} + x_{975} = (-1)^{106} x_1 + (-1)^{150} x_3 + (-1)^{194} x_5 = 211 + 420 + 267 = \boxed{898}.</cmath> | ||
== Solution Variant == | |||
The recursive formula suggests telescoping. Indeed, if we add <math>x_n</math> and <math>x_{n-1}</math>, we have <math>x_n + x_{n-1} = (x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4}) + (x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) = x_{n-1} - x_{n-5}</math>. | |||
Subtracting <math>x_{n-1}</math> yields <math>x_n = -x_{n-5} \implies x_n = -(-(x_{n-10})) = x_{n-10}</math>. | |||
Thus, | |||
<cmath>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}.</cmath> | |||
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 00:28, 12 November 2022
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
