(1)令n=1,由Sn=2an-2n可得a1=2,再由sn+1=2an+1-2(n+1),相减后化简可得 an+1+2=2(an +2 ),可得数列{an+2}是以4为首项,以2为公比的等比数列.
(2)由(1)知,an +2=4×2n-1,由此求得bn=n+1,故 ==,再用错位相减法求出数列的前n项和Tn的值,从而得出结论.
【解析】
(1)令n=1,由Sn=2an-2n可得a1=2.
再由Sn=2an-2n(n∈N+),可得 sn+1=2an+1-2(n+1),
∴sn+1-Sn =2an+1-2an-2,即 an+1=2an +2,故有 an+1+2=2(an +2 ),
故数列{an+2}是以4为首项,以2为公比的等比数列.
(2)由(1)知,an +2=4×2n-1,∴bn=log2(an+2)=log2(4×2n-1 )=n+1,
∴==.
∴数列的前n项和Tn=+++…+ ①,
∴ Tn=+++…+ ②,
①-②可得 Tn=+++…+-=+-
=+,
故Tn=+,显然满足 .