(Ⅰ)由 Sn=an-×2n+1+, n=1,2,3,… , ① 得 a1=S1= a1-×4+所以a1=2. 再由①有 Sn-1=an-1-×2n+, n=2,3,4,… 将①和②相减得: an=Sn-Sn-1= (an-an-1)-×(2n+1-2n),n="2,3," … 整理得: an+2n=4(an-1+2n-1),n="2,3," … , 因而数列{ an+2n}是首项为a1+2=4,公比为4的等比数列, 即 : an+2n=4×4n-1= 4n, n="1,2,3," …, 因而an=4n-2n, n="1,2,3," …, (Ⅱ)将an=4n-2n代入①得: Sn= ×(4n-2n)-×2n+1 += ×(2n+1-1)(2n+1-2) = ×(2n+1-1)(2n-1) Tn= = × = ×(- ) 所以, = - ) = ×(- ) < |