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

Revision as of 14:02, 9 November 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 exponent is $-1$ and $-1$ raised to an even integer exponent is $1$ :

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

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 == Solution ==
-Since <math>-1</math> raised to an [[odd integer | odd]] [[exponent]] is <math>-1</math> and <math>-1</math> raised to an [[even integer]] exponent is <math>1</math>:
+Since <math>-1</math> raised to an [[odd integer | odd]] [[exponentiation | exponent]] is <math>-1</math> and <math>-1</math> raised to an [[even integer]] exponent is <math>1</math>:
 <math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = 0 \Longrightarrow C </math>