2006 AMC 12B Problem 1

What is

(-1)^1 + (-1)^2 + \cdots + (-1)^{2006}?

Show solution

Solution

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.

Concepts: exponent · pairing and grouping

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.