(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=2, 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=-+n2n+1=2+(n-1)2n+1, ∴Bn=Tn-(1+2+3++n)=2+(n-1)2n+1-(12分) |