![](https://cos-cdn.shuashuati.com/pipixue-web/2020-1231-2005-12/ti_inject-812ce.png)
将数列{a n }中的所有项按每组比前一组项数多一项的规则分组如下:(a 1 ),(a 2 ,a 3 ),(a 4 ,a 5 ,a 6 ),(a 7 ,a 8 ,a 9 ,a 10 ),…每一组的第1个数a 1 ,a 2 ,a 4 ,a 7 ,…构成的数列为{b n },b 1 =a 1 =1,S n 为数列{b n }的前n项和,且满足S n+1 (S n +2)=S n (2-S n+1 ),n∈N * , (I)求证:数列{ 1 S n }成等差数列,并求出数列{b n }的通项公式; (Ⅱ)若从第2组起,每一组中的数自左向右均构成等比数列,且公比q为同一个正数,当a 18 =- 2 15 时,求公比q的值; (Ⅲ)在(Ⅱ)的条件下,记每组中最后一数a 1 ,a 3 ,a 6 ,a 10 ,…构成的数列为{c n },设d n =n 2 (n-1)?c n ,求数列{d n }的前n项和T n .