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

2006 AMC 12B Problem 1

Attempt Problem 1 of the 2006 AMC 12B 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 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

What is

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

-2006

-1

0

1

2006

(-1)^k

equals

1

when

k

is even and

-1

when

k

is odd

Group the

2006

terms into consecutive pairs

(-1)+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.