Trigonometry

\(\frac{k(k-1)}{2}\) is really just the sum of integers from 1 to k-1, and the cosine of some integer multiple \(\pi\) oscillates between -1 and 1 (if the integer is even, it's 1, and if it's odd, it's -1). Here, the sum alternates 2 negatives and 2 positives (starting with k=19)

This means that the first 4 terms of the sum are

\(-(1+2+3+\cdots+18)-(1+2+3+\cdots+19)+(1+2+3+\cdots+20)+(1+2+3+\cdots+21) \\=20+19+20+21 \\=4(20)\)

(notice a bunch of cancellation happens between the sums, as the 1+2+3+...18 is repeated throughout all the 4 terms)

So for every group of 4 terms, the sum of the group is the biggest number of the sum of the 3rd term of that group times 4.

Therefore, the sum is:

\(4(20+24+28+\cdots+96)\)

Use the sum formula to obtain your answer.

We want to compute \(\cos \frac{k(k - 1) \pi}{2}\) for various intergers k. The first values are as follows:

 \(\begin{array}{c|c} k & \cos \frac{k(k - 1) \pi}{2} \\ \hline 1 & 1 \\ 2 & -1 \\ 3 & -1 \\ 4 & 1 \\ 5 & 1 \\ 6 & -1 \\ 7 & -1 \\ 8 & 1 \end{array}\)

The values seem to be periodic with period 4. If we let \(f(k) = \frac{k(k - 1) \pi}{2},\) then 

\(f(k + 4) - f(k) = \frac{(k + 4)(k + 3) \pi}{2} - \frac{k(k - 1) \pi}{2} = (4k + 6) \pi \\ = (2k + 3) \cdot 2 \pi\)

Since f(k+4) and f(k) differ by a mutiple of \(2 \pi\), \(\cos f(k)\) is indeed period with period of 4. So for a positive integer m, 

\(A_{4m - 1} = \frac{(4m - 1)(4m - 2)}{2} \cos f(4m - 1) = -\frac{(4m - 1)(4m - 2)}{2}, \\ A_{4m} = \frac{(4m)(4m - 1)}{2} \cos f(4m) = \frac{(4m)(4m - 1)}{2}, \\ A_{4m + 1} = \frac{(4m + 1)(4m)}{2} \cos f(4m + 1) = \frac{(4m + 1)(4m)}{2}, \\ A_{4m + 2} = \frac{(4m + 2)(4m + 1)}{2} \cos f(4m + 2) = -\frac{(4m + 2)(4m + 1)}{2},\)

Therefore:

\(A_{4m - 1} + A_{4m} + A_{4m + 1} + A_{4m + 2} = -\frac{(4m - 1)(4m - 2)}{2} + \frac{(4m)(4m - 1)}{2} + \frac{(4m + 1)(4m)}{2} - \frac{(4m + 2)(4m + 1)}{2} = -2.\)

Summing over \(5 \le m \le 24,\) we find that \(A_{19} + A_{20} + \dots + A_{98} = 20(-2) = \boxed{-40}.\)