设数列an的通项公式an=(2n-1)*a^(n-1)(a≠0),求前n项和

Sn=1+(4-1)*a^1+.+(2n-3)*a^(n-2)+(2n-1)*a^(n-1).(1)
若a=1则Sn=1+3+.+(2n-1)=n^2
a≠1时aSn= (2-1)*a^1+(4-1)*a^2+.+(2n-3)*a^(n-1)+(2n-1)*a^n.(2)
(2)-(1)有 (a-1)Sn=-1-2[a^1+a^2+.+a^(n-1)]+(2n-1)*a^n=(2n-1)*a^n-(a^n-a)/(a-1)-1
所以Sn=[(2n-1)*a^n-(a^n-a)/(a-1)-1]/(a-1)