(Ⅰ)∵Sn=an+1-2n+1+1(n∈N*),且a1=1, ∴S1=a2-22+1,a1=a2-22+1,∴a2=4, S2=a3-23+1,a1+a2=a3-23+1,∴a3=12; (Ⅱ)由Sn=an+1-2n+1+1,(n∈N*)①, 得Sn-1=an-2n+1,(n∈N*,n≥2)②, ①-②得:an=an+1-an-2n,即an+1=2an+2n(n≥2), 检验知a1=1,a2=4满足an+1=2an+2n(n≥2). ∴an+1=2an+2n(n≥1). 变形可得=+1(n≥1). ∵==1, ∴数列{}是以1为首项,1为公差的等差数列. ∴=1+(n-1)×1=n, 则an=n•2n-1(n≥1); (Ⅲ)证明:由(Ⅱ)知an=n•2n-1(n≥1),代入bn= 得bn==1-, ∵22n+1-(n+1)•2n-2=(2n+1-n-1-)2n>0, ∴(n+1)•2n+2<22n+1 又∵2n+1<(n+1)•2n+2, ∴2n+1<(n+1)•2n+2<22n+1, 则<< ∴1-<1-<1- ∴1-<bn<1- ∴n-(++…+)<Tn<n-(++…+) 即n-3×<Tn<n-3× ∴n-[1-()n]<Tn<n-[1-()n] ∴n-<Tn<n-<n-. |