(1)∵Sn+1-Sn=2n+1,∴an+1=2n+1,∴an=2n,∴Sn= 2(1?2n) 1?2 =2n+1-2;(2)bn=2log2an+1=2n+1,∴ 1 bn?bn+1 = 1 2 ( 1 2n+1 - 1 2n+3 ),∴Tn= 1 2 ( 1 3 - 1 5 + 1 5 - 1 7 +…+ 1 2n+1 - 1 2n+3 )= 1 2 ( 1 3 - 1 2n+3 )= n 6n+9 .