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

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What is <math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} </math> ?
 
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>
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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>
  
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== Solution ==
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== Solution 1==
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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>:  
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Since <math>-1</math> raised to an odd integer is <math>-1</math> and <math>-1</math> raised to an even integer exponent is <math>1</math>:  
  
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<math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = 0 \Rightarrow C </math>
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<math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = \boxed{\textbf{(C) }0}. </math>
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== Solution 2 ==
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The
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Using the formula for the first <math>n</math> terms of a geometric series, with <math>n = 2006</math>, <math>a = (-1)^{1} = -1</math>, and <math>r = -1</math>, the sum can also be obtained:
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<math>\frac{a(1-r^n)}{1-r} = \frac{-1(1 - (-1)^{2006})}{1 + 1} = \frac{-1(1 - 1)}{2} = \frac{0}{2} = \boxed{\textbf{(C) }0}</math>
  
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~anabel.disher
 
== See Also ==
 
== See Also ==
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*[[2006 AMC 10B Problems]]
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{{AMC10 box|year=2006|ab=B|before=First Problem|num-a=2}}
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*[[2006 AMC 10B Problems/Problem 2|Next Problem]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 15:52, 10 April 2025

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 1

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

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

Solution 2

The Using the formula for the first $n$ terms of a geometric series, with $n = 2006$, $a = (-1)^{1} = -1$, and $r = -1$, the sum can also be obtained:

$\frac{a(1-r^n)}{1-r} = \frac{-1(1 - (-1)^{2006})}{1 + 1} = \frac{-1(1 - 1)}{2} = \frac{0}{2} = \boxed{\textbf{(C) }0}$

~anabel.disher

See Also

2006 AMC 10B (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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