(I)由已知得Sn=2an-3n,Sn+1=2an+1-3(n+1),所以an+1=2an+3,3+an+1=2(3+an),由此能求出an.
(II)bn=n(2n-1)=n2n-n,设Tn=1×2+2×22+3×23++n×2n(1),2Tn=1×22+2×23++(n-1)2n+n×2n+1,Tn=-(2+22+23+…+2n)+n2n+1=,由此能求出数列{bn}的前n项和Bn.
【解析】
(I)由已知得Sn=2an-3n,
Sn+1=2an+1-3(n+1),两式相减并整理得:an+1=2an+3(2分)
所以3+an+1=2(3+an),又a1=S1=2a1-3,a1=3可知3+a1=6≠0,
进而可知an+3≠0
所以,
故数列{3+an}是首相为6,公比为2的等比数列,
所以3+an=6•2n-1,即an=3(2n-1)(6分)
(II)bn=n(2n-1)=n2n-n
设Tn=1×2+2×22+3×23++n×2n(1),
2Tn=1×22+2×23++(n-1)2n+n×2n+1(2)
由(2)-(1)得Tn=-(2+22+23+…+2n)+n2n+1=,
∴(12分)