2016 AIME I Problems/Problem 1

Revision as of 15:17, 17 February 2019 by Firebolt360 (talk | contribs) (Solution 2)

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

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 $14\cdot 24=\boxed{336}$.

See also

2016 AIME I (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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