(Ⅰ)当n=1时,a1=S1=2a1-1,解得a1=1. 又当n≥2时,an=Sn-Sn-1=2an-1-(2an-1-1)=2an-2an-1, ∴an=2an-1(n≥2). ∴数列{an}是首项为1,公比为2的等比数列. ∴an=1×2n-1=2n-1(n∈N*). 由bn-1-bn=bnbn-1,得-=1. 又b1=1,所以数列{}是首项为==1,公差为1的等差数列. ∴=1+(n-1)×1=n. ∴bn=. (Ⅱ)由(Ⅰ)可得:=n•2n-1, ∴Tn=1×20+2×21+…+n•2n-1, 2Tn=1×21+2×22+…+(n-1)•2n-1+n•2n,. 两式相减,得-Tn=1+21+22+…+2n-1-n•2n=-n•2n=2n-1-n•2n. ∴Tn=(n-1)•2n+1. |