1998 AIME Problems/Problem 5

Problem

Given that $A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,$ find $|A_{19} + A_{20} + \cdots + A_{98}|.$

Solution

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:

$\left|\sum_{i=19}^{98} A_i\right| = \displaystyle - \frac{19(18)}{2} + \frac{20(19)}{2} + \frac{21(20)}{2} - \frac{22(21)}{2} - \frac{23(22)}{2} + \frac{24(23)}{2} \cdots - \frac{98(97)}{2}$

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 $\displaystyle |-2 \cdot 20| = 040$ .

1998 AIME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1998 AIME Problems/Problem 5

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