2016
Problem - 2965
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 $S(a)+S(-a)$.
Because $S(r) = \frac{12}{1-r}$ and, thus, $S(-r)=\frac{12}{1+r}$. It follows that $$2016=S(a)S(-a)=\frac{144}{1-a^2}\implies \frac{1}{1-a^2}=14$$
Then $$S(a)+S(-a)=\frac{12}{1-a}+\frac{12}{1+a}=\frac{24}{1-a^2}=24\times 14=\boxed{336}$$