Difference between revisions of "2016 AIME I Problems/Problem 1"
(Added problem and solution) |
(→Solution) |
||
| (17 intermediate revisions by 9 users not shown) | |||
| Line 2: | Line 2: | ||
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> | + | 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>. |
| − | + | ||
| − | + | == See also == | |
| − | <math>S(a)S(-a)=144 | + | {{AIME box|year=2016|n=I|before=First Problem|num-a=2}} |
| − | <math> | + | {{MAA Notice}} |
| − | |||
| − | <math> | ||
| − | |||
Revision as of 20:09, 7 February 2020
Problem 1
For 
, let 
denote the sum of the geometric series ![\[12+12r+12r^2+12r^3+\cdots .\]](http://latex.artofproblemsolving.com/3/2/f/32f5063cd768644e208da71eb8e0f6ead9a57248.png)
Let 
between 
and 
satisfy 
. Find 
.
Solution
The sum of an infinite geometric series is 
. The product 
. 
, so the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
