2006 AMC 12B Problem 1

Since $(-1)^k = -1$ for odd $k$ and $(-1)^k = 1$ for even $k,$ the terms alternate $-1, 1, -1, 1, \ldots$ There are $2006$ terms, forming $1003$ pairs, each equal to $(-1) + 1 = 0.$ The total is $0.$ Thus, the correct answer is C.