0,前n项和Sn=【(an+1)/2】^2,bn=(-1)^n*Sn,求数列{bn}的前n项和Tn

网友回答

由 S1 = a1 = [(a1 + 1) / 2]^2 ,
得 a1 = 1 ,
所以 S2 = 1 + a2 = [(a2 + 1) / 2]^2 ,
得 a2 = 3 或 -1 ,
因为数列{an}是等差数列,公差d>0,所以 a2 = 3 ,
所以 d = 2 ,
所以 an = 2 n - 1 ,
所以 Sn = n^2 ,
所以 Tn = - 1 + 2^2 - 3^2 + 4^2 - 5^2 + 6^2 - …
若 n 为偶数,则
Tn = (2+1)(2-1)+(4+3)(4-3)+(6+5)(6-5)+ … +(n + n-1)[n - (n-1)]
= 1 + 2 + 3 + 4 + … + n-1 + n
= n (n + 1) / 2 ,
若 n 为奇数,则
Tn = (n - 1) n / 2 - n^2
= - n (n + 1) / 2 ,
综上,Tn = (-1)^n * n * (n + 1) / 2