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}
问题描述：
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}
答
3a(n+1)+an=43a(n+1)=-an+43a(n+1)-3=-an+1[a(n+1)-1]/(an -1)=-1/3,为定值.a1-1=9-1=8,数列{an -1}是以8为首项,-1/3为公比的等比树列.an=8×(-1/3)^(n-1) +1Sn=8×[1-(-1/3)ⁿ]/[1-(-1/3)]+n=n+6-6×(-1/3)...