Difference between revisions of "2016 AIME I Problems/Problem 1"
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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== | ||
− | 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 | + | 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 == | == 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}} |
Latest revision as of 20:09, 7 February 2020
Problem 1
For , let denote the sum of the geometric series 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 |
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