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

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(Problem)
 
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What is <math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} </math> ?
 
What is <math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} </math> ?
  
<math> \mathrm{(A) \ } -2006\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } 2006 </math>
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<math> \textbf{(A)} -2006\qquad \textbf{(B)} -1\qquad \textbf{(C) } 0\qquad \textbf{(D) } 1\qquad \textbf{(E) } 2006 </math>
  
 
== Solution ==
 
== Solution ==

Latest revision as of 10:32, 19 December 2021

Problem

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

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

Solution

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

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

See Also

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

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