己知公差大于零的等差数列an,a2＋a3＋a4＝9,且a2＋1,a3＋3,a4＋8为等比数列bn前三项（1）求an,bn通项（2）设an前n项和Sn，求1/S1＋1/S2＋1/Sn
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/20 02:39:34
己知公差大于零的等差数列an,a2＋a3＋a4＝9,且a2＋1,a3＋3,a4＋8为等比数列bn前三项（1）求an,bn通项（2）设an前n项和Sn，求1/S1＋1/S2＋1/Sn
己知公差大于零的等差数列an,a2＋a3＋a4＝9,且a2＋1,a3＋3,a4＋8为等比数列bn前三项（1）求an,bn通项
（2）设an前n项和Sn，求1/S1＋1/S2＋1/Sn
己知公差大于零的等差数列an,a2＋a3＋a4＝9,且a2＋1,a3＋3,a4＋8为等比数列bn前三项（1）求an,bn通项（2）设an前n项和Sn，求1/S1＋1/S2＋1/Sn
1)假设公差为t
则a3=a2+t
a4=a2+2t
==>3a2+3t=9==>a2+t=3==>a3=3
又a2＋1,a3＋3,a4＋8为等比数列
==》(3-t+1)(3+t+8)=36
==>t=1
==>an=n
==>bn=3*2^(n-1)
2)Sn=[(n+1)n]/2
==>1/S1＋1/S2＋...1/Sn=2/(1*2)+2/(2*3)+...2/[(n+1)n]
又1/[(n+1)n]=1/n-1/(n+1)
==>1/S1＋1/S2＋...1/Sn=2/(1*2)+2/(2*3)+...2/[(n+1)n]
=2[1-1/2+1/2-1/3+1/3-1/4.+n/1-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)
解设公差为d
(1)a2=a1+d,a3=a1+2d,a4=a1+3d
a2＋a3＋a4＝a1+d+a1+2d+a1+3d=9 ,得3a1+6d=9,得a1+2d=a3=3,a2=3-d,a4=3+d
(a2＋1):(a3＋3)=(a3＋3):(a4＋8) 得d=-8(不合题意舍去)或d=1
所以an=n
另：b1=a2+1=3,b2=a3+3=6,b3=a...
全部展开
解设公差为d
(1)a2=a1+d,a3=a1+2d,a4=a1+3d
a2＋a3＋a4＝a1+d+a1+2d+a1+3d=9 ,得3a1+6d=9,得a1+2d=a3=3,a2=3-d,a4=3+d
(a2＋1):(a3＋3)=(a3＋3):(a4＋8) 得d=-8(不合题意舍去)或d=1
所以an=n
另：b1=a2+1=3,b2=a3+3=6,b3=a4+8=12,所以q=2,bn=3×2^(n-1)
(2)Sn=[(n+1)n]/2
1/S1＋1/S2＋1/Sn=1+1/3+2/[(n+1)n]=
收起