Compute 1+i+i^2+i^3+i^4+....i^2009

Everything cancels out except 2009 positive "1s" PLUS the last term of: i^2009 =i. Therefore the answer is: =2,009 + i

Compute 1+i+i^2+i^3+i^4+....i^2009

This is the sume of a  GP

a=1, r=i, n=2010

$\boxed{s_n=\frac{a(1-r^n)}{1-r}}\\ S_{2010}=\frac{1(1-i^{2010})}{1-i}\\ S_{2010}=\frac{1-i^{2010}}{1-i}\\$

$i^1=i \\ i^2=-1\\ i^3=-i\\ i^4=+1\\ ...For \;\;k\in Z \\i^{4k+1}=i\\ i^{4k+2}=-1\\ i^{4k+3}=-i\\ i^{4k}=1\\ so\\ i^{2010}=i^{4*{502}+2}=-1$

$S_{2010}=\frac{1-i^{2010}}{1-i}\\ S_{2010}=\frac{1-\color{red}{-1}}{1-i}\qquad \text {Error corrected here}\\ S_{2010}=\frac{2}{1-i}\\ S_{2010}=\frac{2}{1-i}\times\frac{1+i}{1+i} \\ S_{2010}=\frac{2(1+i)}{1-i^2}\\ S_{2010}=\frac{2(1+i)}{1--1}\\ S_{2010}=\frac{2(1+i)}{2}\\ S_{2010}=1+i$

$1+i+i^2+i^3+i^4+....i^{2009}=1+i$

Melody: Can you please explain this "goofy" answer by W/A? Thanks.

∑[1 + i^n], n=1 to 2009 =2,009 + i

OK. I think I found the mistake !! The first 1 should be outside the "Sigma" sign, but it still gives a different answer: 1 + ∑{ i^n}, n=1 to 2009 = 1 + i  ?????.