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2006 AMC 10B Problems/Problem 1

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Problem

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

Solution

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

See Also

Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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