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

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For <math>-1<r<1</math>, let <math>S(r)</math> denote the sum of the geometric series <cmath>12+12r+12r^2+12r^3+\cdots .</cmath>  Let <math>a</math> between <math>-1</math> and <math>1</math> satisfy <math>S(a)S(-a)=2016</math>. Find <math>S(a)+S(-a)</math>.  
 
For <math>-1<r<1</math>, let <math>S(r)</math> denote the sum of the geometric series <cmath>12+12r+12r^2+12r^3+\cdots .</cmath>  Let <math>a</math> between <math>-1</math> and <math>1</math> satisfy <math>S(a)S(-a)=2016</math>. Find <math>S(a)+S(-a)</math>.  
 
==Solution==
 
==Solution==
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We know that <math>S(r)=\frac{12}{1-r}</math>, and <math>S(-r)=\frac{12}{1+r}</math>. Therefore, <math>S(a)S(-a)=\frac{144}{1-a^2}</math>, so <math>2016=\frac{144}{1-a^2}</math>. We can divide out <math>144</math> to get <math>\frac{1}{1-a^2}=14</math>. We see <math>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}</math>
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The sum of an infinite geometric series is <math>\frac{a}{1-r}\rightarrow \frac{12}{1\mp a}</math>. The product <math>S(a)S(-a)=\frac{144}{1-a^2}=2016</math>. <math>\frac{12}{1-a}+\frac{12}{1+a}=\frac{24}{1-a^2}</math>, so the answer is <math>\frac{2016}{6}=\boxed{336}</math>.
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==Solution 2==
 
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The sum of an infinite geometric series is <math>\frac{a}{1-r}\rightarrow \frac{12}{1\mp a}</math>. The product <math>S(a)S(-a)=\frac{144}{1-a^2}=2016</math> so dividing by <math>144</math> gives <math>\frac{1}{1-a^2}=14\implies a=\sqrt{\frac{13}{14}}</math>. <math>\frac{12}{1-a}+\frac{12}{1+a}=\frac{24}{1-a^2}</math>, so the answer is <math>14\cdot 24=\boxed{336}</math>.
 
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--  <math>a=\sqrt{\frac{13}{14}}</math> is not necessary in this case.
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2016|n=I|before=First Problem|num-a=2}}
 
{{AIME box|year=2016|n=I|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:09, 7 February 2020

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 $S(a)S(-a)=\frac{144}{1-a^2}=2016$. $\frac{12}{1-a}+\frac{12}{1+a}=\frac{24}{1-a^2}$, so the answer is $\frac{2016}{6}=\boxed{336}$.

See also

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

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