2016 AIME I Problems/Problem 1: Difference between revisions
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==Solution== | ==Solution== | ||
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>. | 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>. | ||
== Video Solution by OmegaLearn == | |||
https://youtu.be/3wNLfRyRrMo?t=153 | |||
~ pi_is_3.14 | |||
== 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 06:53, 4 November 2022
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. Error creating thumbnail: File missing
