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

Revision as of 16:54, 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

$S(r)=12/(1-r)$ $S(-r)=12/(1+r)$ Therefore, $S(a)S(-a)=144/(1-a^2)$ $2016=144/(1-a^2)$ $1/(1-a^2)=14$ $S(a)+S(-a)=12/(1-a)+12/(1+a)$ $=12(1+a)/(1-a^2)+12(1-a)/(1-a^2)=24/(1-a^2)=24*1/(1-a^2)=24*14=336$

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 == See also ==
-{{AIME box|year=2016|n=I|num-b=10|num-a=12}}
+{{AIME box|year=2016|n=I|num-b=0|num-a=2}}
 {{MAA Notice}}

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

Revision as of 16:54, 4 March 2016

Problem 1

Solution

See also