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
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![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](/uploads/image/z/2779375-31-5.jpg?t=%E5%B7%B1%E7%9F%A5%E5%85%AC%E5%B7%AE%E5%A4%A7%E4%BA%8E%E9%9B%B6%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%2Ca2%EF%BC%8Ba3%EF%BC%8Ba4%EF%BC%9D9%2C%E4%B8%94a2%EF%BC%8B1%2Ca3%EF%BC%8B3%2Ca4%EF%BC%8B8%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97bn%E5%89%8D%E4%B8%89%E9%A1%B9%EF%BC%881%EF%BC%89%E6%B1%82an%2Cbn%E9%80%9A%E9%A1%B9%EF%BC%882%EF%BC%89%E8%AE%BEan%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%EF%BC%8C%E6%B1%821%2FS1%EF%BC%8B1%2FS2%EF%BC%8B1%2FSn)
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...
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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]=
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