2016 AIME I Problems/Problem 1

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

2016 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 0	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2016 AIME I Problems/Problem 1

Problem 1

Solution

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