Question

已知数列{an}的前n项和Sn=n*n-7n-8.（1)求{an}的通项公式（2）求{an的绝对值}的前n项和Tn

Answer 1

(1)

Sn=n^2-7n-8 (1)

S(n-1)=(n-1)^2-7(n-1)-8 (2)

(1)-(2)

an=2n-8

(2)

2n-8>=0

n>=4

bn = |an|

bn = -an ; n=1,2,3,4

= an ; n=5,6,.

Tn = b1+b2+..+bn

= -(a1+a2+a3+a4) + a5+a6+..+an

= (6+0)4/2 + (2+2n-8)(n-4)/2

= 12 + (n-3)(n-4)

= n^2-7n+24