SOLUTION: sum k = 0 to 2009 i ^ k

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Question 1199960: sum k = 0 to 2009 i ^ k
Answer by greenestamps(13334) About Me  (Show Source):
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sum k = 0 to 3 (i^k) = 1 + i - 1 - i = 0

Every four consecutive integer powers of i have a sum of 0.

So

sum k = 0 to 2007 = 0

So

sum k = 0 to 2009 = 0 + i^2008 + i^2009 = 0 + 1 + i = 1 + i

ANSWER: 1 + i