(1)由Sn+2n=2an,知Sn=2an-2n.当n=1 时,S1=2a1-2,则a1=2,当n≥2时,Sn-1=2an-1-2(n-1),故an=2an-1+2,由此能够证明数列{an+2}是等比数列.并能求出数列{an}的通项公式an.
(2)由bn=log2(an+2)==n+1,得,故,由此利用错位相减法能够求出Tn,并证明.
证明:(1)由Sn+2n=2an得 Sn=2an-2n
当n∈N*时,Sn=2an-2n,①
当n=1 时,S1=2a1-2,则a1=2,
则当n≥2,n∈N*时,Sn-1=2an-1-2(n-1).②
①-②,得an=2an-2an-1-2,
即an=2an-1+2,
∴an+2=2(an-1+2)
∴,
∴{an+2}是以a1+2为首项,以2为公比的等比数列.
∴an+2=4•2n-1,
∴an=2n+1-2.
(2)证明:由bn=log2(an+2)==n+1,
得,
则,③
④
③-④,得
=
=
=,
所以 .