Problem:
Define the sequence a1β,a2β,a3β,β¦ by anβ=βk=1nβsin(k), where k represents radian measure. Find the index of the 100th term for which anβ<0.
Solution:
First notice that
anββ=k=1βnβsin(21β)sin(21β)sin(k)β=k=1βnβ2sin(21β)cos(kβ21β)βcos(k+21β)β=2sin(21β)cos(21β)βcos(n+21β)ββ
Therefore anβ<0 if and only if cos(21β)<cos(n+21β). Because the cosine function has period 2Ο, and cosx=cos(2Οβx), this inequality holds if and only if n is between 2Οmβ1 and 2Οm for some positive integer m. In other words, the index of the m th negative term in the given sequence is the greatest integer less than 2Οm. Because 3.14<Ο<3.145, it follows that 628<200Ο<629. Thus the index of the 100th negative term is 628β.
The problems on this page are the property of the MAA's American Mathematics Competitions