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 1== |
We know that <math>S(r)=\frac{12}{1-r}</math>, and <math>S(-r)=\frac{12}{1+r}</math>. Therefore, <math>S(a)S(-a)=\frac{144}{1-a^2}</math>, so <math>2016=\frac{144}{1-a^2}</math>. We can divide out <math>144</math> to get <math>\frac{1}{1-a^2}=14</math>. We see <math>S(a)+S(-a)=\frac{12}{1-a}+\frac{12}{1+a}=\frac{12(1+a)}{1-a^2}+\frac{12(1-a)}{1-a^2}=\frac{24}{1-a^2}=24*14=\fbox{336}</math> | We know that <math>S(r)=\frac{12}{1-r}</math>, and <math>S(-r)=\frac{12}{1+r}</math>. Therefore, <math>S(a)S(-a)=\frac{144}{1-a^2}</math>, so <math>2016=\frac{144}{1-a^2}</math>. We can divide out <math>144</math> to get <math>\frac{1}{1-a^2}=14</math>. We see <math>S(a)+S(-a)=\frac{12}{1-a}+\frac{12}{1+a}=\frac{12(1+a)}{1-a^2}+\frac{12(1-a)}{1-a^2}=\frac{24}{1-a^2}=24*14=\fbox{336}</math> | ||
Revision as of 21:17, 1 March 2018
Contents
Problem 1
For , let denote the sum of the geometric series Let between and satisfy . Find .
Solution 1
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 |
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