Difference between revisions of "2001 AIME II Problems/Problem 3"
| Line 32: | Line 32: | ||
<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> | <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. | ||
== 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}} | ||
Revision as of 21:44, 30 December 2020
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.
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. 
