2005 AIME II Problems/Problem 3

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 $\displaystyle \gcd(403, 399) = 1$ .)

2005 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2005 AIME II Problems/Problem 3

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