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2016 AIME I Problems/Problem 1: Difference between revisions

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==Problem 1==
==Problem 1==
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==
<math>S(r)=12/(1-r)</math>
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>.
<math>S(-r)=12/(1+r)</math>
 
Therefore,
== Video Solution by OmegaLearn ==
<math>S(a)S(-a)=144/(1-a^2)</math>
https://youtu.be/3wNLfRyRrMo?t=153
<math>2016=144/(1-a^2)</math>
<math>1/(1-a^2)=14</math>
<math>S(a)+S(-a)=12/(1-a)+12/(1+a)</math>
<math>=12(1+a)/(1-a^2)+12(1-a)/(1-a^2)=24/(1-a^2)=24*1/(1-a^2)=24*14=336</math>


~ pi_is_3.14


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

Latest revision as of 15:45, 28 September 2025

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 $\frac{2016}{6}=\boxed{336}$.

Video Solution by OmegaLearn

https://youtu.be/3wNLfRyRrMo?t=153

~ pi_is_3.14

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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