Problem:
Given that Akβ=2k(kβ1)βcos2k(kβ1)Οβ, find β£A19β+A20β+β―+A98ββ£.
Solution:
Notice that 2k(kβ1)β is even when k=4m or k=4m+1, and odd otherwise. It follows that
A4mβ1β+A4mβ=β2(4mβ1)(4mβ2)β+24m(4mβ1)β=4mβ1
and
A4m+1β+A4m+2β=2(4m+1)4mββ2(4m+2)(4m+1)β=β4mβ1
hence A4mβ1β+A4mβ+A4m+1β+A4m+2β=β2. Thus A19β+A20β+A21β+β―+A98β, which is a sum of eighty terms, equals 20(β2)=β40β.
The problems on this page are the property of the MAA's American Mathematics Competitions