# Difference between revisions of "2006 AMC 12B Problems/Problem 1"

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== Problem == | == Problem == | ||

+ | What is <math>( - 1)^1 + ( - 1)^2 + \cdots + ( - 1)^{2006}</math>? | ||

+ | <math> | ||

+ | \text {(A) } - 2006 \qquad \text {(B) } - 1 \qquad \text {(C) } 0 \qquad \text {(D) } 1 \qquad \text {(E) } 2006 | ||

+ | </math> | ||

== Solution == | == Solution == | ||

− | + | <math>(-1)^n=1</math> if n is even and <math>-1</math> if n is odd. So we have | |

− | + | <math>-1+1-1+1-1+1-1+1\cdots-1+1-1+1-1+1=0+0+0+0+\cdots+0+0+0+0+0=0 \Rightarrow \text{(C)}</math> | |

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== See also == | == See also == | ||

* [[2006 AMC 12B Problems]] | * [[2006 AMC 12B Problems]] |