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 1

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 $S(a)+S(-a)=\frac{12}{1-a}+\frac{12}{1+a}=\frac{12(1+a)}{1-a^2}+\frac{12(1-a)}{1-a^2}=\frac{24}{1-a^2}=24 \cdot 14=\fbox{336}$

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

2016 AIME I (Problems • Answer Key • Resources)
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

Page

Toolbox

Search

2016 AIME I Problems/Problem 1

Contents

Problem 1

Solution 1

Solution 2

See also