Difference between revisions of "2006 AMC 10B Problems/Problem 1"
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Since <math>-1</math> raised to an [[odd integer | odd]] [[exponentiation | exponent]] is <math>-1</math> and <math>-1</math> raised to an [[even integer]] exponent is <math>1</math>: | Since <math>-1</math> raised to an [[odd integer | odd]] [[exponentiation | exponent]] is <math>-1</math> and <math>-1</math> raised to an [[even integer]] exponent is <math>1</math>: | ||
| − | <math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = 0 \Longrightarrow C </math> | + | <math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = 0 \Longrightarrow \boxed{C} </math> |
== See Also == | == See Also == | ||
| − | {{AMC10 box|year=2006|ab=B| | + | {{AMC10 box|year=2006|ab=B|before=First Problem|num-a=2}} |
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 18:14, 13 February 2016
Problem
What is 
?
Solution
Since 
raised to an odd exponent is and raised to an even integer exponent is 
:
See Also
| 2006 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
