2015 AIME II Problems/Problem 13

Problem

Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$ .

Solution

If $n = 1$ , $a_n = sin(1) > 0$ . Then if $n$ satisfies $a_n < 0$ , $n \ge 2$ , and $a_n =\cfrac{1}{\sin{1}} \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].$ Since $\sin 1$ is positive, it does not affect the sign of $a_n$ . Let $b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)$ . Now since $\cos(0) + \cos(1) = 2\cos(\cfrac{1}{2})\cos(\cfrac{1}{2})$ and $\cos(n) + \cos(n + 1) = 2\cos(n + \cfrac{1}{2})\cos(\cfrac{1}{2})$ , $b_n$ is negative if and only if $\cos(\cfrac{1}{2}) < \cos(n + \cfrac{1}{2})$ , or when $n \in [2k\pi - 1, 2k\pi]$ . Since $\pi$ is irrational, there is always only one integer in the range, so there are values of $n$ such that $a_n < 0$ at $2\pi, 4\pi, \cdots$ . Then the hundredth such value will be when $k = 100$ and $n = \lfloor 200\pi \rfloor = \lfloor 6.28318 \rfloor = \boxed{628}$ .

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2015 AIME II Problems/Problem 13

Problem

Solution