Sn是等差数列前n项和,Sn=16n-n平方 Tn=a1绝对值+a2绝对值.an绝对值,则Tn

题目

Sn是等差数列前n项和,Sn=16n-n平方 Tn=a1绝对值+a2绝对值.an绝对值,则Tn

答案

Sn=16n-n²
an=Sn - S(n-1)=16n-n²-16(n-1)+(n-1)²=17-2n,n＞1
n＝1时,a1=S1=15,满足上式.
∴an=17-2n
an=17-2n＞0,n＜17/2,即n≤8
an=17-2n＜0,n＞17/2,即n≥9
①n≤8时,an＞0
Tn=|a1|+|a2|+|a3|+……+|an|
=a1+a2+a3+……+an
=Sn=16n-n²
②n≥9时,
Tn=|a1|+|a2|+|a3|+……+|a8|+|a9|+|a10|+……+|an|
= a1+ a2 +a3 +……+a8 - a9 - a10 - …… - an
= S8 - (a9+a10+……+an)
= S8 - [(a1+a2+……+an)-(a1+a2+……+a8)]
=S8 - (Sn-S8)
=2S8-Sn
=2×(16×8-8²) - (16n-n²)
= n²-16n+128