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

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== Problem ==
 
== Problem ==
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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 <math> \frac mn </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[integer]]s. Find <math> m+n. </math>
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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 <math> \frac mn </math> where <math> m </math> and <math> n </math> are relatively prime integers. Find <math> m+n. </math>
 
  
 
== Solution ==
 
== Solution ==
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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>. 
  
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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 integer | odd]], so <math>\displaystyle \gcd(403, 399) = 1</math>.)
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(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 integer | odd]], so <math>\displaystyle \gcd(403, 399) = 1</math>.)
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== See Also ==
 
  
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*[[2005 AIME II Problems/Problem 2| Previous problem]]
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== See also ==
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*[[2005 AIME II Problems/Problem 4| Next problem]]
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{{AIME box|year=2005|n=II|num-b=2|num-a=4}}
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*[[2005 AIME II Problems]]
 
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 18:43, 21 March 2007

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 $\frac mn$ where $m$ and $n$ are relatively prime integers. Find $m+n.$

Solution

Let's call the first term of the original geometric series $a$ and the common ratio $r$, so $2005 = a + ar + ar^2 + \ldots$. Using the sum formula for infinite geometric series, we have $(*)\;\;\frac a{1 -r} = 2005$. Then we form a new series, $a^2 + a^2 r^2 + a^2 r^4 + \ldots$. We know this series has sum $20050 = \frac{a^2}{1 - r^2}$. Dividing this equation by $\displaystyle (*)$, we get $\displaystyle 10 = \frac a{1 + r}$. Then $\displaystyle a = 2005 - 2005r$ and $\displaystyle a = 10 + 10r$ so $\displaystyle 2005 - 2005r = 10 + 10r$, $\displaystyle 1995 = 2015r$ and finally $\displaystyle r = \frac{1995}{2015} = \frac{399}{403}$, so the answer is $\displaystyle 399 + 403 = 802$.

(We know this last fraction is fully reduced by the Euclidean algorithm -- because $\displaystyle4 = 403 - 399$, $\displaystyle \gcd(403, 399) | 4$. But 403 is odd, so $\displaystyle \gcd(403, 399) = 1$.)

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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