2006 AMC 12B Problems/Problem 1: Difference between revisions

Revision as of 12:54, 13 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

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

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 == Problem ==
+What is <math>( - 1)^1 + ( - 1)^2 + \cdots + ( - 1)^{2006}</math>?
+<math>
+\text {(A) } - 2006 \qquad \text {(B) } - 1 \qquad \text {(C) } 0 \qquad \text {(D) } 1 \qquad \text {(E) } 2006
+</math>
 == Solution ==
-The answer was C, 0.
+<math>(-1)^n=1</math> if n is even and <math>-1</math> if n is odd. So we have
-For a full list of answers, go to this site.
+<math>-1+1-1+1-1+1-1+1\cdots-1+1-1+1-1+1=0+0+0+0+\cdots+0+0+0+0+0=0 \Rightarrow \text{(C)}</math>
-[[http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2006-12a/06AMC12AB-answers.html]]
 == See also ==
 * [[2006 AMC 12B Problems]]