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>4 = 403 - 399</math>, <math>\gcd(403, 399) | 4</math>. But 403 is odd, so <math>\gcd(403, 399) = 1</math>.) | ||
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== See Also == | == See Also == | ||
*[[2005 AIME II Problems]] | *[[2005 AIME II Problems]] | ||
Revision as of 08:53, 21 July 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 
.)
