Difference between revisions of "2006 AMC 12B Problems/Problem 1"

Revision as of 09:05, 14 November 2007

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

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

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

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

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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All AMC 12 Problems and Solutions

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	== See also ==		== See also ==
−	* [[2006 ~~AMC 12B Problems]]~~	+	{{AMC12 box\|year=2006\|ab=B\|before=First Question\|num-a=2}}