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

Revision as of 18:40, 13 July 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 $1$ :

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

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 == Solution ==
 Since <math>-1</math> raised to an odd power is <math>-1</math> and <math>-1</math> raised to an even power is <math>1</math>:
-<math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + (-1) + (1) + ... + (-1)+(1) = 0 \Rightarrow C </math>
+<math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = 0 \Rightarrow C </math>
 == See Also ==
 *[[2006 AMC 10B Problems]]