2006 AMC 10B Problems/Problem 1

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

2006 AMC 10B (Problems • Answer Key • Resources)
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2006 AMC 10B Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

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