jmc

The powers of $i$ cycle through $i^0=1,$ $i^1=i,$ $i^2=-1,$ and $i^3=-i,$ and the sum of any four consecutive powers of $i$ is $$1+i+(-1)+(-i)=0.$$Thus, the sum reduces to $i^{2008}+i^{2009}=\boxed{1+i}.$