(Ⅰ)∵a1=2,Sn=an+1-1(n∈N*), ∴当n=1时,S1=a2-1=a1=2, 解得a2=6. 当n=2时,S2=a3-1=2+6=8, 解得a3=18. (Ⅱ)∵a1=2,Sn=an+1-1(n∈N*), ∴当n≥2时,Sn=an+1-1,Sn-1=an-1, ∴an=Sn-Sn-1=an+1-an, 即an+1=3an. 对于a2=3a1也满足上式, ∴数列{an}是首项为2,公比为3的等比数列, ∴an=2•3n-1(n∈N*). ( III)∵an=2•3n-1(n∈N*), ∴nan=2n•3n-1, ∴Tn=2•1+4•3+6•32+8•33+…+2n•3n-1, 3Tn=2•3+4•32+6•33+8•34+…+2n•3n, 相减得,-2Tn=2(1+3+32+33+…+3n-1)-2n•3n =2•-2n•3n =3n-1-2n•3n, ∴Tn=. |