已知数列{an}满足3(an+1)+an=4(n>=1),且a1=9,其前n项和为Sn,则满足不等式{Sn-n-6}

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|