Problem:
What is (β1)1+(β1)2+β―+(β1)2006 ?
Answer Choices:
A. β2006
B. β1
C. 0
D. 1
E. 2006
Solution:
Because
(β1)k={1, if k is even β1, if k is odd β
the sum can be written as
(β1+1)+(β1+1)+β―+(β1+1)=0+0+β―+0=0β
The problems on this page are the property of the MAA's American Mathematics Competitions