(Ⅰ)因为Sn =(an -1),n∈N+,所以Sn+1 =(an+1 -1). 两式相减,得Sn+1 -Sn =(an+1 -an );,即an+1 =(an+1 -an ) ∴an+1=3an,n∈N+. 又s1 =(a1 -1);,即a1 =(a1 -1);,所以a1=3. ∴{an}是首项为3,公比为3的等比数列.从而{an}的通项公式是{an=3n,n∈N+; (Ⅱ)由(Ⅰ)知bn=log3an=n,设数列{anbn}的前n项和为Tn, 则Tn=1×3+2×32+3×33++n•3n,3Tn =1×32+2×33+3×34++(n-1)•3n+n•3n+1, 两式相减得-2Tn=1×3+1×32+1×33++1×3n-n•3n+1 =(3n-1)-n•3n+1, 所以Tn=•3n+1+. |