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

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== See Also ==
 
== See Also ==
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{{AMC10 box|year=2006|ab=B|num-b=First Problem|num-a=2}}
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*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems/Problem 2|Next Problem]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 22:49, 7 September 2011

Problem

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

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

Solution

Since $-1$ raised to an odd exponent is $-1$ and $-1$ raised to an even integer exponent is $1$:

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

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem First Problem
Followed by
Problem 2
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All AMC 10 Problems and Solutions