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== Problem ==
What is <math>( - 1)^1 + ( - 1)^2 + \cdots + ( - 1)^{2006}</math>?

<math>
\text {(A) } - 2006 \qquad \text {(B) } - 1 \qquad \text {(C) } 0 \qquad \text {(D) } 1 \qquad \text {(E) } 2006
</math>
== Solution ==
<math>(-1)^n=1</math> if n is even and <math>-1</math> if n is odd. So we have

<math>-1+1-1+1-\cdots-1+1=0+0+\cdots+0+0=0 \Rightarrow \text{(C)}</math>

== See also ==
{{AMC12 box|year=2006|ab=B|before=First Question|num-a=2}}