已知数列{an}满足3an+1+an=4,a1=9,前n项和为sn,则满足不等式/sn-n-6/

已知数列{an}满足3an+1+an=4,a1=9,前n项和为sn,则满足不等式/sn-n-6/<1/125的最小正整数为

对3a(n+1)+an=4 变形得：
3[a(n+1)-1]=-(an-1)
a(n+1)/an=-1/3
an=8*(-1/3)^(n-1)+1
Sn=8{1+(-1/3)+(-1/3)^2+……+(-1/3)^(n-1)]+n
=6-6*(-1/3)^n+n
|Sn-n-6|=|-6*(-1/3)^n|