# Difference between revisions of "2005 AIME II Problems/Problem 3"

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== Solution == | == Solution == | ||

− | Let's call the first term of the original [[geometric series]] <math>a</math> and the common ratio <math>r</math>, so <math>2005 = a + ar + ar^2 + \ldots</math>. Using the sum formula for [[infinite]] geometric series, we have <math>(*)\;\;\frac a{1 -r} = 2005</math>. Then we form a new series, <math>a^2 + a^2 r^2 + a^2 r^4 + \ldots</math>. We know this series has sum <math>20050 = \frac{a^2}{1 - r^2}</math>. Dividing this equation by <math>\displaystyle (*)</math>, we get <math>\displaystyle 10 = \frac a{1 + r}</math>. Then <math>a = 2005 - 2005r</math> and <math>a = 10 + 10r</math> so <math>2005 - 2005r = 10 + 10r</math>, <math>1995 = 2015r</math> and finally <math>r = \frac{1995}{2015} = \frac{399}{403}</math>, so the answer is <math>399 + 403 = 802</math>. (We know this last fraction is fully reduced by the [[Euclidean algorithm]] -- because <math> | + | Let's call the first term of the original [[geometric series]] <math>a</math> and the common ratio <math>r</math>, so <math>2005 = a + ar + ar^2 + \ldots</math>. Using the sum formula for [[infinite]] geometric series, we have <math>(*)\;\;\frac a{1 -r} = 2005</math>. Then we form a new series, <math>a^2 + a^2 r^2 + a^2 r^4 + \ldots</math>. We know this series has sum <math>20050 = \frac{a^2}{1 - r^2}</math>. Dividing this equation by <math>\displaystyle (*)</math>, we get <math>\displaystyle 10 = \frac a{1 + r}</math>. Then <math>\displaystyle a = 2005 - 2005r</math> and <math>\displaystyle a = 10 + 10r</math> so <math>\displaystyle 2005 - 2005r = 10 + 10r</math>, <math>\displaystyle 1995 = 2015r</math> and finally <math>\displaystyle r = \frac{1995}{2015} = \frac{399}{403}</math>, so the answer is <math>\displaystyle 399 + 403 = 802</math>. (We know this last fraction is fully reduced by the [[Euclidean algorithm]] -- because <math>\displaystyle4 = 403 - 399</math>, <math>\displaystyle \gcd(403, 399) | 4</math>. But 403 is odd, so <math>\displaystyle \gcd(403, 399) = 1</math>.) |

== See Also == | == See Also == |

## Revision as of 22:09, 9 September 2006

## Problem

An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is where and are relatively prime integers. Find

## Solution

Let's call the first term of the original geometric series and the common ratio , so . Using the sum formula for infinite geometric series, we have . Then we form a new series, . We know this series has sum . Dividing this equation by , we get . Then and so , and finally , so the answer is . (We know this last fraction is fully reduced by the Euclidean algorithm -- because , . But 403 is odd, so .)