Difference between revisions of "2006 AMC 12B Problems/Problem 1"
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\text {(A) } - 2006 \qquad \text {(B) } - 1 \qquad \text {(C) } 0 \qquad \text {(D) } 1 \qquad \text {(E) } 2006 | \text {(A) } - 2006 \qquad \text {(B) } - 1 \qquad \text {(C) } 0 \qquad \text {(D) } 1 \qquad \text {(E) } 2006 | ||
</math> | </math> | ||
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== Solution == | == Solution == | ||
<math>(-1)^n=1</math> if n is even and <math>-1</math> if n is odd. So we have | <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- | + | <math>-1+1-1+1-\cdots-1+1=0+0+\cdots+0+0=0 \Rightarrow \text{(C)}</math> |
== See also == | == See also == | ||
| − | + | {{AMC12 box|year=2006|ab=B|before=First Question|num-a=2}} | |
| + | {{MAA Notice}} | ||
Latest revision as of 09:40, 15 September 2017
Problem
What is 
?
Solution
if n is even and 
if n is odd. So we have
See also
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
