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

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== Problem ==
 
== Problem ==
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What is <math>( - 1)^1 + ( - 1)^2 + \cdots + ( - 1)^{2006}</math>?
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<math>
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\text {(A) } - 2006 \qquad \text {(B) } - 1 \qquad \text {(C) } 0 \qquad \text {(D) } 1 \qquad \text {(E) } 2006
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</math>
  
 
== Solution ==
 
== Solution ==
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<math>(-1)^n=1</math> if n is even and <math>-1</math> if n is odd. So we have
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<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]]
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{{AMC12 box|year=2006|ab=B|before=First Question|num-a=2}}
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{{MAA Notice}}

Latest revision as of 10:40, 15 September 2017

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

2006 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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