Question 1199960
<br>
sum k = 0 to 3 (i^k) = 1 + i - 1 - i = 0<br>
Every four consecutive integer powers of i have a sum of 0.<br>
So<br>
sum k = 0 to 2007 = 0<br>
So<br>
sum k = 0 to 2009 = 0 + i^2008 + i^2009 = 0 + 1 + i = 1 + i<br>
ANSWER: 1 + i