2005 AIME II Problems/Problem 3

Revision as of 20:47, 9 March 2009 by Meta Knight (talk | contribs) (Solution: Added a box around the final answer.)

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 $(*)$, we get $10 = \frac a{1 + r}$. Then $a = 2005 - 2005r$ and $a = 10 + 10r$ so $2005 - 2005r = 10 + 10r$, $1995 = 2015r$ and finally $r = \frac{1995}{2015} = \frac{399}{403}$, so the answer is $399 + 403 = \boxed{802}$.

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

See also

2005 AIME II (ProblemsAnswer KeyResources)
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