Difference between revisions of "2006 AMC 12B Problems/Problem 1"

(Solution)
Line 1: Line 1:
 
== Problem ==
 
== 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 ==
 
== 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 ==
 
== See also ==
 
* [[2006 AMC 12B Problems]]
 
* [[2006 AMC 12B Problems]]

Revision as of 13:54, 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

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

See also

Invalid username
Login to AoPS