Difference between revisions of "2016 AIME I Problems/Problem 1"
(Added problem and solution) |
(→Solution) |
||
Line 10: | Line 10: | ||
<math>S(a)+S(-a)=12/(1-a)+12/(1+a)</math> | <math>S(a)+S(-a)=12/(1-a)+12/(1+a)</math> | ||
<math>=12(1+a)/(1-a^2)+12(1-a)/(1-a^2)=24/(1-a^2)=24*1/(1-a^2)=24*14=336</math> | <math>=12(1+a)/(1-a^2)+12(1-a)/(1-a^2)=24/(1-a^2)=24*1/(1-a^2)=24*14=336</math> | ||
+ | |||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2016|n=I|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Revision as of 17:51, 4 March 2016
Problem 1
For , let
denote the sum of the geometric series
Let
between
and
satisfy
. Find
.
Solution
Therefore,
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.