Problem:
Given that Ak=2k(k−1)cos2k(k−1)π, find ∣A19+A20+⋯+A98∣.
Solution:
Notice that 2k(k−1) is even when k=4m or k=4m+1, and odd otherwise. It follows that
A4m−1+A4m=−2(4m−1)(4m−2)+24m(4m−1)=4m−1
and
A4m+1+A4m+2=2(4m+1)4m−2(4m+2)(4m+1)=−4m−1
hence A4m−1+A4m+A4m+1+A4m+2=−2. Thus A19+A20+A21+⋯+A98, which is a sum of eighty terms, equals 20(−2)=−40.
The problems on this page are the property of the MAA's American Mathematics Competitions