Problem:
Define the sequence a1,a2,a3,… by an=∑k=1nsin(k), where k represents radian measure. Find the index of the 100th term for which an<0.
Solution:
First notice that
an=k=1∑nsin(21)sin(21)sin(k)=k=1∑n2sin(21)cos(k−21)−cos(k+21)=2sin(21)cos(21)−cos(n+21)
Therefore an<0 if and only if cos(21)<cos(n+21). Because the cosine function has period 2π, and cosx=cos(2π−x), this inequality holds if and only if n is between 2πm−1 and 2πm 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 2πm. Because 3.14<π<3.145, it follows that 628<200π<629. Thus the index of the 100th negative term is 628.
The problems on this page are the property of the MAA's American Mathematics Competitions