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

Latest revision as of 09:32, 19 December 2021

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

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

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

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

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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 == Problem ==
+What is <math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} </math> ?
+<math> \textbf{(A)} -2006\qquad \textbf{(B)} -1\qquad \textbf{(C) } 0\qquad \textbf{(D) } 1\qquad \textbf{(E) } 2006 </math>
 == Solution ==
+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>
 == See Also ==
-*[[2006 AMC 10B Problems]]
+{{AMC10 box|year=2006|ab=B|before=First Problem|num-a=2}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}