What is (-1)^1 + (-1)^2 + … + (-1)^2006?

2006 AMC 10B Problem 1

Attempt Problem 1 of the 2006 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10B solutions, or check the answer key.

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What is

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

-2006

-1

0

1

2006

Group the terms in consecutive pairs

In each pair, compare the signs of the odd-power and even-power terms

There are

2006

terms. Pairing consecutive terms gives

(-1+1)+(-1+1)+\cdots.

Since

2006

is even, every term pairs off and the sum is

0.

Thus, the correct answer is C.