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

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Revision as of 17:38, 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, $S(a)S(-a)=144/(1-a^2)$ $2016=144/(1-a^2)$ $1/(1-a^2)=14$ $S(a)+S(-a)=12/(1-a)+12/(1+a)$ $=12(1+a)/(1-a^2)+12(1-a)/(1-a^2)=24/(1-a^2)=24*1/(1-a^2)=24*14=336$

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