(1)由Tn=1-an,,n≥2,知,从而=1,由此能证明数列{bn}是等差数列.
(2)由(1)知bn=2+(n-1)=n+1,从而cn=(n+1)•2n,故Sn=2•2+3•22+…+(n+1)•2n,利用错位相减法能求出数列{cn}的前n项和Sn.
(3)由,知n≥2时,,由,知,由此利用放缩法能够证明.
【解析】
(1)∵Tn=1-an,,n≥2,
∴,从而=1,(n≥2)
∴bn-bn-1=1,(n≥2)
∵T1=a1=1-a1,
∴,,
∴{bn}是以2为首项,1为公差的等差数列.
(2)由(1)知bn=2+(n-1)=n+1,从而cn=(n+1)•2n,
∴Sn=2•2+3•22+…+(n+1)•2n,
2Sn=2•22+3•23+…+n•2n+(n+1)•2n+1,
两式相减,得-Sn=4+(22+23+…+2n)-(n+1)•2n+1
=4+-(n+1)•2n+1
=-n•2n+1,
∴Sn=n•2n+1.
(3)∵,
∴n≥2时,,
∵,∴,
=
>
=++…+
=
=,
∴,
又∵当n≥2时,
=
=+
=
==,
.