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/

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/04 04:35:03
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/
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/
3a(n+1)+an=4
3a(n+1)=-an+4
3a(n+1)-3=-an+4-3=-an+1=-(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为公比的等比数列
an -1=8×(-1/3)^(n-1)
an=1+ 8×(-1/3)^(n-1)
Sn=a1+a2+...+an
=n+8×[1+(-1/3)+...+(-1/3)^(n-1)]
=n+8×1×[1-(1/3)^n]/[1-(-1/3)]
=n+6- 6×(-1/3)^n
|Sn-n-6|