1998 AIME Problems/Problem 5

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Problem

Though the problem may appear to be quite daunting, it is actually not that difficult. $\displaystyle \frac {k(k-1)}2$ always evaluates to an integer (triangular number), and the cosine of $\displaystyle n\pi$ where $n \in \mathbb{Z}$ is 1 if $\displaystyle n$ is even and -1 if $\displaystyle n$ is odd. $\displaystyle \frac {k(k-1)}2$ will be even if $\displaystyle 4|k$ or $\displaystyle 4|k-1$ , and odd otherwise.

So our sum looks something like:

If we group the terms in pairs, we see that we need a formula for $\frac{n(n-1)}{2} + \frac{n+1(n)}{2} = \left(\frac n2\right)(n+1 - (n-1)) = n$ . So the first two fractions add up to $\displaystyle 19$ , the next two to $\displaystyle -21$ , and so forth.

If we pair the terms again now, each pair adds up to $-2$ . There are $\displaystyle \frac{98-19+1}{2 \cdot 2} = 20$ such pairs, so our answer is $\displaystyle |-2 \cdot 20| = 040$ .