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

Revision as of 21:50, 7 September 2011

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

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

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

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

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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 == See Also ==
-{{AMC10 box|year=2006|ab=B|num-b=First Problem|num-a=2}}
+{{AMC10 box|year=2006|ab=B|before=First Problem|num-a=2}}
 *[[2006 AMC 10B Problems]]
 [[Category:Introductory Algebra Problems]]