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

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 $\frac{1}{1-a^2}=14\implies a=\sqrt{\frac{13}{14}}$ . $\frac{12}{1-a}+\frac{12}{1+a}=\frac{24}{1-a^2}$ , so the answer is $14\cdot 24=\boxed{336}$ .

-- $a=\sqrt{\frac{13}{14}}$ is not necessary in this case.

2016 AIME I (Problems • Answer Key • Resources)
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2016 AIME I Problems/Problem 1

Contents

Problem 1

Solution

Solution 2

See also