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

## Problem 1

For $-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

We know that $S(r)=\frac{12}{1-r}$, and $S(-r)=\frac{12}{1+r}$. Therefore, $S(a)S(-a)=\frac{144}{1-a^2}$, so $2016=\frac{144}{1-a^2}$. We can divide out $144$ to get $\frac{1}{1-a^2}=14$. We see $S(a)+S(-a)=\frac{12}{1-a}+\frac{12}{1+a}=\frac{12(1+a)}{1-a^2}+\frac{12(1-a)}{1-a^2}=\frac{24}{1-a^2}=24*14=\fbox{336}$

## Solution 2

The sum of an infinite geometric series is $\frac{a}{1-r}\rightarrow \frac{12}{1\mp a}$. The product $S(a)S(-a)=\frac{144}{1-a^2}=2016$ so dividing by $144$ gives $\frac{1}{1-a^2}=14\implies a= \pm \sqrt{\frac{13}{14}}$. $\frac{12}{1-a}+\frac{12}{1+a}=\frac{24}{1-a^2}$, so the answer is $14\cdot 24=\boxed{336}$.

Critique:

This solution is the same as the one above. Also, $a= \pm \sqrt{\frac{13}{14}}$, not $a=\sqrt{\frac{13}{14}}$

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