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

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<math>(-1)^n=1</math> if n is even and <math>-1</math> if n is odd. So we have
<math>(-1)^n=1</math> if n is even and <math>-1</math> if n is odd. So we have


<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>
<math>-1+1-1+1-\cdots-1+1=0+0+\cdots+0+0=0 \Rightarrow \text{(C)}</math>


== See also ==
== See also ==
* [[2006 AMC 12B Problems]]
* [[2006 AMC 12B Problems]]

Revision as of 12:55, 13 November 2007

Problem

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

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

Solution

$(-1)^n=1$ if n is even and $-1$ if n is odd. So we have

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

See also

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