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

Revision as of 18:07, 4 March 2016

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

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 == See also ==
-{{AIME box|year=2016|n=I|num-b=0|num-a=2}}
+{{AIME box|year=2016|n=I|before=First Problem|num-a=2}}
 {{MAA Notice}}

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

Revision as of 18:07, 4 March 2016

Problem 1

Solution

See also