(1){bn}的通项公式;
(2){bn}的前n项和Tn.
(1){bn}的通项公式;
(2){bn}的前n项和Tn.
解:(1)当n≥2时,an=Sn-Sn-1=2an-4-2an-1+4,
即得an=2an-1,
当n=1时,a1=S1=2a1-4=4,∴an=2n+1.
∴bn+1=2n+1+2bn.∴
∴{
∴
(2)Tn=1·2+2·22+…+n·2n, ①
2Tn=1·22+2·23+…+(n-1)·2n+n·2n+1, ②
①-②,得-Tn=2+22+23+…+2n-n·2n+1=
∴Tn=(n-1)·2n+1+2.