Question

an的通项an=2n-8,an的前n项和Sn=n`2-7n-8,求/an/的前n项和Tn

Answer 1

an=2n-8
1<n<4
an<=0
|an|=-2n+8=6-2(n-1)
这是6为首项,-2为公差的等差数列
Tn=6n+n(n-1)/2*(-2)=-n^2+7n,1<=n<=4
n>=5
an>0
|an|=2n-8
此时an前4项是-6,-4,-2,0
所以|an|前4项和=12
bn=2n-8前n项和=-6n+n(n-2)/2*2=n^2-7n
前四项和=-6-4-2+0=-12
所以从第五项开始的和=n^2-7n-(-12)=n^2-7n+12
所以当n>=5时,|an|前n项和=n^2-7n+12+12=n^2-7n+24
所以1<=n<=4,则Tn=-n^2+7n
n>=5,Tn=n^2-7n+24</n<4