设等差数列 {an}\left\{ {a}_{n}\right\}{an} 的公差为 ddd ,且 d>1d > 1d>1 . 令 bn=n2+nan{b}_{n} = \frac{{n}^{2} + n}{{a}_{n}}bn=ann2+n ,记 Sn,Tn{S}_{n},{T}_{n}Sn,Tn 分别为数列 {an}\left\{ {a}_{n}\right\}{an} , {bn}\left\{ {b}_{n}\right\}{bn} 的前 nnn 项和.(1)若 3a2=3a1+a3,S3+T3=213{a}_{2} = 3{a}_{1} + {a}_{3},{S}_{3} + {T}_{3} = {21}3a2=3a1+a3,S3+T3=21 ,求 {an}\left\{ {a}_{n}\right\}{an} 的通项公式;(2)若 {bn}\left\{ {b}_{n}\right\}{bn} 为等差数列,且 S99−T99=99{S}_{99} - {T}_{99} = {99}S99−T99=99 ,求 ddd .