Difference between revisions of "2006 AMC 10B Problems/Problem 1"
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== Solution == | == Solution == | ||
− | Since <math>-1</math> raised to an | + | Since <math>-1</math> raised to an odd integer 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) = | + | <math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = \boxed{\textbf{(C) }0}. </math> |
== See Also == | == See Also == |
Revision as of 10:31, 19 December 2021
Problem
What is ?
Solution
Since raised to an odd integer 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 |
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