已知复数等比数列an以w=-1/2+(根号3)/2为公比,且a1=1,它的前n 项和为Sn

(1) w=(cos120+sin120*i )w^3=cos360+sin360*i=cos0=1(2)sn=a1*(1-w^n)/(1-w)=(1-w^n)/(1-w) 讨论w^n: n=1w1^1=w1n=2w2^2=-1/2-(根号3)/2n=3w3=1n=4w4=w1n=5w5=w2n=6w6=w3n=7w7=w1n=8w8=w2n=9w9=w...