Difference between revisions of "2016 AIME I Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
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= \pm \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>. | 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= \pm \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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== 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 23:44, 11 April 2016
Contents
Problem 1
For , let denote the sum of the geometric series Let between and satisfy . Find .
Solution
We know that , and . Therefore, , so . We can divide out to get . We see
Solution 2
The sum of an infinite geometric series is . The product so dividing by gives . , 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.