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

 
(Added problem and solution)
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== Problem ==
 
== Problem ==
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What is <math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} </math> ?
 +
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<math> \mathrm{(A) \ } -2006\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } 2006 </math>
 +
 
== Solution ==
 
== 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>
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== See Also ==
 
== See Also ==
 
*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems]]

Revision as of 17:50, 13 July 2006

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 power is $-1$ and $-1$ raised to an even power is $1$: $(-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + (-1) + (1) + ... + (-1)+(1) = 0 \Rightarrow C$

See Also

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