Difference between revisions of "2006 AMC 10B Problems/Problem 1"

Revision as of 13:50, 2 August 2006

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

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

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

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

Revision as of 13:50, 2 August 2006 (view source) Xantos C. Guin (talk \| contribs) m (added link to next problem) ← Older edit		Revision as of 13:50, 2 August 2006 (view source) Xantos C. Guin (talk \| contribs) m (corrected typo in link) Newer edit →
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	*[[2006 AMC 10B Problems]]		*[[2006 AMC 10B Problems]]

−	*[[~~2005~~ AMC ~~10A~~ Problems/Problem 2\|Next Problem]]	+	*[[2006 AMC 10B Problems/Problem 2\|Next Problem]]