Problem:
An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has sum 10 times the sum of the original series. The common ratio of the original series is m/n, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let a be the first term and r the ratio of the original series, and let S=2005. Then 1−ra​=S and 1−r2a2​=10S. Factor to obtain 10S=(1−ra​)(1+ra​)= S⋅1+ra​. Then 10=1+ra​ and S=1−ra​ imply that S(1−r)=10(1+r), so r=S+10S−10​=1995/2015=399/403, and m+n=802​.
The problems on this page are the property of the MAA's American Mathematics Competitions