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)求{an}的通项公式
an=Sn-Sn-1=n*n-7n-8-[（n-1）²-7（n-1）-8]=2n-8

（2）求{an的绝对值}的前n项和Tn
{an的绝对值}={6,4,2,0,2,4,6,.....2n-8}
当n<=4时，Tn=[6+(8-2n)]*n/2=7n-n²
当n>4时,，Tn=(6+0)*4/2+[2+(2n-8)]*(n-4)/2=n²-7n+24

Answer 2

(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

Answer 3

Sn=n*n-7n-8
S(n-1)=(n-1)*(n-1)-7(n-1)-8
相减得an=2n-8
当n<=4时,Tn=6n-n(n-1)
当n>4时,Tn=2(n-1)+(n-4)(n-5)+12