Difference between revisions of "2016 AIME I Problems/Problem 1"
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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== | ||
Latest revision as of 16:45, 28 September 2025
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 
.
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. 
