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== See Also ==
 
== See Also ==
 
*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems]]
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*[[2005 AMC 10A Problems/Problem 2|Next Problem]]

Revision as of 13:50, 2 August 2006

Problem

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

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

Solution

Since $-1$ raised to an odd power is $-1$ and $-1$ raised to an even power is $1$:

$(-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = 0 \Rightarrow C$

See Also

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