已知数列 {an}\left\{ {a}_{n}\right\}{an} 的前 nnn 项和为 Sn,a1=12{S}_{n},{a}_{1} = \frac{1}{2}Sn,a1=21 , 且 anan+1=2an+1\frac{{a}_{n}}{{a}_{n} + 1} = 2{a}_{n + 1}an+1an=2an+1 ,记 {1an}\left\{ \frac{1}{{a}_{n}}\right\}{an1} 的前 nnn 项和为 Tn{T}_{n}Tn ,则( )A. a2=16{a}_{2} = \frac{1}{6}a2=61B. {1an}\left\{ \frac{1}{{a}_{n}}\right\}{an1} 是等比数列C. Tn=2n+2−2n−4{T}_{n} = {2}^{n + 2} - {2n} - 4Tn=2n+2−2n−4D. 12−12n+1<Sn≤1−12n\frac{1}{2} - \frac{1}{{2}^{n + 1}} < {S}_{n} \leq 1 - \frac{1}{{2}^{n}}21−2n+11<Sn≤1−2n1