已知等比数列｛an｝的前n项和为Sn,a1 ＝2.S1、2S2、3S3成等差数列,数列{bn-a

S1,2S2,3S3成等差数列,则有4S2=S1+3S3,3S3-3S2=S2-S1,3a3=3 a2 q=a2
a1=2,则可得q=1/3
等比数列an=2*(1/3)^(n-1)
bn-an=-6+(n-1)2=2n-8,则bn= 2n-8 + an
S`n= (-6+2n-8)n/2 + 2(1-1/3^n)/(1-1/3)= n^2-7n + 3-(1/3)^(n-1)