令S=i+2i^2+3i^3+4i^4+…+2010i^2010
则is= i^2+2i^3+3i^4+…+2009i^2010+2010i^2011
得S-iS=i+i^2+i^3+i^4+…+i^2010-2010i^2011
即(1-i)S=i(1-i^2010)/(1-i)-2010i^2011
(1-i)S=0-2010i
得S=-2010i/(1-i)=1005-1005i