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Difference between revisions of "2016 AIME I Problems/Problem 1"

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<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>
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== See also ==
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{{AIME box|year=2016|n=I|num-b=10|num-a=12}}
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{{MAA Notice}}

Revision as of 17:51, 4 March 2016

Problem 1

For $-1<r<1$, let $S(r)$ denote the sum of the geometric series \[12+12r+12r^2+12r^3+\cdots .\] Let $a$ between $-1$ and $1$ satisfy $S(a)S(-a)=2016$. Find $S(a)+S(-a)$.

Solution

$S(r)=12/(1-r)$ $S(-r)=12/(1+r)$ Therefore, $S(a)S(-a)=144/(1-a^2)$ $2016=144/(1-a^2)$ $1/(1-a^2)=14$ $S(a)+S(-a)=12/(1-a)+12/(1+a)$ $=12(1+a)/(1-a^2)+12(1-a)/(1-a^2)=24/(1-a^2)=24*1/(1-a^2)=24*14=336$


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

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