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

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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>
 
<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==
 
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>:  
 
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>:  
  
 
<math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = \boxed{\textbf{(C) }0}. </math>
 
<math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = \boxed{\textbf{(C) }0}. </math>
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== Solution 2 ==
 +
The
 +
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:
 +
 +
<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>
  
 +
~anabel.disher
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2006|ab=B|before=First Problem|num-a=2}}
 
{{AMC10 box|year=2006|ab=B|before=First Problem|num-a=2}}

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

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