S2/S1=(a1+a2)/a1=4,a1=1
所以a2=3
公差d=a2-a1=2
an=a1+(n-1)d=2n-1
所以bn=(2n-1)*2^(n-1)
设bn的前n项和是Tn
则Tn=1*2^0+3*2^1+5*2^2+……+(2n-1)*2^(n-1)
2*Tn=1*2^1+3*2^2+5*2^3+……+(2n-3)*2^(n-1)+(2n-1)*2^n
两式相减得-Tn=2^0+2*2^1+2*2^2+……+2*2^(n-1)-(2n-1)*2^n
=2^1+2^2+2^3+2^4+……+2^n-(2n-1)*2^n-1
=2(1-2^n)/(1-2)-(2n-1)*2^n-1
=2*2^n-2-(2n-1)*2^n-1=(3-2n)*2^n-3
所以数列｛bn｝的前n项和Tn=3+(2n-3)*2^n