数列{An}中,Sn=9n-n^2,求Tn=a1的绝对值+a2的绝对值+.+an的绝对值

a1=S1=9-1=8.
an =Sn-S(n-1)= 9n-n^2-[9(n-1)-(n-1)^2]
=10-2n.
∴an=10-2n.(n∈N*)
所以该数列的前四项为正,第五项为0.以后各项为负.
N≤5时,Tn=|a1|+|a2|+……+|an|=Sn=9n-n^2.
N＞5时,Tn=S5+[-(Sn-S5)]=2 S5-Sn
=2×5(8+0)/2-(9n-n^2)= n^2-9n+40.