2006 AMC 12B Problems/Problem 1: Difference between revisions

Latest revision as of 09:40, 15 September 2017

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

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

$(-1)^n=1$ if n is even and $-1$ if n is odd. So we have

$-1+1-1+1-\cdots-1+1=0+0+\cdots+0+0=0 \Rightarrow \text{(C)}$

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by First Question	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 12 Problems and Solutions

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 == 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 ==
+<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-\cdots-1+1=0+0+\cdots+0+0=0 \Rightarrow \text{(C)}</math>
 == See also ==
-* [[2006 AMC 12B Problems]]
+{{AMC12 box|year=2006|ab=B|before=First Question|num-a=2}}
+{{MAA Notice}}