∵z=Z=1+ 2i 1?i =1+ 2i(1+i) (1?i)(1+i) ＝1+ ?2+2i 2 =1-1+i=i．∴1+Z+Z2+…+Z2014= 1?z2015 1?z = 1?i2015 1?i ＝ 1+i 1?i = (1+i)2 (1?i)(1+i) ＝ 2i 2 ＝i．故选：C．
