数列an的前n项和为Sn=n^2-10n-8,求{an绝对值}的前n项和Tn.

Question

如题所述

Answer 1

an=sn-s(n-1)
=n^2-10n-8-(n-1)^2+10(n-1)+8
=2n-1 -10
=2n-11
当n=6时 a6=1>0 当n=5时a5=-1 <0
而a1=-9 a2=-7 a3=-5 a4=-3
所以m=a1+a2+a3+a4+a5=-9-7-5-3-1=-25
sn=a6+a7+...an+ (a1+a2+a3+a4+a5)=(a6+a7+..+an) +m
所以Tn=la6l+la7l+...+lanl +la1l+la2l+la3l+la4l+la5l
=(a6+a7+..an) -m
=(a6+a7+..+an) +m -2m
=sn-2m
=n^2-10n-8+50
=n^2-10n+42本回答被提问者和网友采纳

Answer 2

首先，利用 an=Sn-S[n-1]，得到an的通项
an=Sn-S[n-1]
=(n^2-10n-8) - [(n-1)^2-10(n-1)-8]
= 2n - 11
所以 |an|=|2n-11|
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