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2006 AMC 12B Problems/Problem 1

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Problem

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

\text {(A) } - 2006 \qquad \text {(B) } - 1 \qquad \text {(C) } 0 \qquad \text {(D) } 1 \qquad \text {(E) } 2006

Solution

(-1)^n=1 if n is even and -1 if n is odd. So we have

-1+1-1+1-\cdots-1+1=0+0+\cdots+0+0=0 \Rightarrow \text{(C)}

See also

2006 AMC 12B (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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