设数列{an}的前n项和Sn=n^2-4n+1,则/a1/+/a2/+/a3/+……+/a10/=

题目

设数列{an}的前n项和Sn=n^2-4n+1,则/a1/+/a2/+/a3/+……+/a10/=
需详细过程。

答案

a1=s1=1^2-4*1+1=-2
sn=n^2-4n+1
s(n-1)=(n-1)^2-4(n-1)+1
=n^2-2n+1-4n+4+1
=n^2-6n+6
an=sn-s(n-1)
=n^2-4n+1-(n^2-6n+6)
=n^2-4n+1-n^2+6n-6
=2n-5
an>0
2n>5
n>2.5
即当n=3时an>0
a2=2*2-5=-1
a3=2*3-1=1
a10=2*10-5=15
/a1/+/a2/+/a3/+……+/a10/
= /a1/+/a2/+a3+……+a10
= /-2/+/-1/+(1+15)*8/2
=2+1+64
=67