Problem:
For −1<r<1, let S(r) denote the sum of the geometric series
12+12r+12r2+12r3+⋯
Let a between −1 and 1 satisfy S(a)S(−a)=2016. Find S(a)+S(−a).
Solution:
S(a)+S(−a)=1−a12+1+a12=122⋅1−a2122=61S(a)S(−a)=61⋅2016=336
The problems on this page are the property of the MAA's American Mathematics Competitions