2015 AIME II Problem 13

Question

Define the sequence

a_1,

a_2,

a_3,

\ldots

by

a_n = \sum_{k=1}^{n} \sin(k),

where

k

represents radian measure. Find the index of the

100

th term for which

a_n \lt 0.

Answer

Multiply the sum by

2\sin\frac{1}{2}

and use a product-to-sum identity to make it telescope

a_n \lt 0

exactly when

2\pi m - 1 \lt n \lt 2\pi m

for some positive integer

m,

and each such interval contains exactly one integer

Multiplying each term by

2\sin\frac{1}{2}

and using

2\sin k \sin\frac{1}{2}

= \cos\left(k - \frac{1}{2}\right)

- \cos\left(k + \frac{1}{2}\right),

the sum telescopes:

a_n = \frac{\cos\frac{1}{2} - \cos\left(n + \frac{1}{2}\right)}{2\sin\frac{1}{2}}.

So

a_n \lt 0

exactly when

\cos\left(n + \frac{1}{2}\right) \gt \cos\frac{1}{2},

which happens exactly when

n + \frac{1}{2}

is within

\frac{1}{2}

of a multiple of

2\pi:

\begin{aligned} &2\pi m - \tfrac{1}{2} \lt n + \tfrac{1}{2} \\ &\lt 2\pi m + \tfrac{1}{2}, \end{aligned}

i.e.

2\pi m - 1 \lt n \lt 2\pi m.

Each interval

(2\pi m - 1,\, 2\pi m)

has length

1

and contains exactly one integer, namely

\lfloor 2\pi m \rfloor.

Hence the

100

th negative term has index

\lfloor 200\pi \rfloor.

Since

3.14 \lt \pi \lt 3.145,

we have

628 \lt 200\pi \lt 629,

so the index is

628.

Problem 13 in other years