等差数列an,前n项和为sn,a3=6,s3=12求证:1/s1+1/s2+...+1/sn
易知:d=2,sn=1/n(n+1)= 1/n - 1/(n+1) 所以:1/s1+1/s2+...+1/sn=1/1x2 + 1/2x3+.1/nx(n+1) =1/1-1/2+1/2-1/3+.+1/n-1/(n+1)=1-1/(n+1)
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扫描下载二维码等比数列{an}的前n项和为Sn,已知S1.S3.S2.成等差数列,(2)若a1-a3=-3/2,求{n.an}的前n项和_百度知道
等比数列{an}的前n项和为Sn,已知S1.S3.S2.成等差数列,(2)若a1-a3=-3/2,求{n.an}的前n项和
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[1-(-1/2)Tn=(3/解;)a1=-3/2a1=(-3/2)^(n-2)nan=n×(-1/2)^(n-1)=[(4+6n)/2)+;3Tn=[(8+12n)×(-1/]/)=(-3/,则2S3=S1+S22(a1+a1q+a1q²2(1-q²2)^0+;2)/2)^(n-1)=(-2)×[1-(-1/.;2)Tn=1×(-1/2)^(n-1)=(-1/.;2)^(n-2)(-1/2)] -n×(-1/(1-q²:设公比为qS1;2a1-a1q²2)Tn=(-1/.、S3...+n×(-1/[1-(-1/.+(-1/2)ⁿ.+nan=1×(-1/2)^(n-1)Tn-(-1/2)/2)ⁿ -4/2)^(-1)+(-1/2)^0+3×(-1/.,得2q²2)ⁿ2)+;2)^(n-2)Tn=a1+2a2+;3](-1/2)^(-1) +2×(-1/.+(n-1)×(-1/]=-2an=a1q^(n-1)=(-2)×(-1/2)^(n-2) -n×(-1/ -8]/+q=0q(2q+1)=0q=0(舍去)或q=-1/)=a1+a1+a1q整理.;2)².;=-3/2)^0+2×(-1/2a1-a3=-3/、S2成等差数列;2)^(n-2)+n×(-1/
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出门在外也不愁(1/2)已知等差数列an的前n项和为sn.a3=6.s3=12.(1)求数列an的前n项和
(2).求1/s+1/s2+1/s3+.._百度知道
(1/2)已知等差数列an的前n项和为sn.a3=6.s3=12.(1)求数列an的前n项和
(2).求1/s+1/s2+1/s3+..
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s2+1/s+1/2-1/s3=12=3a2=a2=4d=a3-a2=2a1=2;3+;s3+;2+1/.=1-1/,Sn=n(n+1)1/....=1-1/
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+1/(2×3)+..+1/..+1/S3=3a1+3d=3a2=12 a2=4d=a3-a2=6-4=2a1=a2-d=4-2=2 an=2+2(n-1)=2nSn=2(1+2+..;(n+1)=n/S1+1/2-1/n-1/S2+;(1×2)+1/.;[n(n+1)]=1-1/3+;Sn=1/..;2+1/..+n)=n(n+1)=n²+n1/
1.S3=3a3-3d=12,a3-d=4d=2a2=4,a1=2an=2n2.Sn=2n+n(n-1)=n2+n=n(n+1)1/Sn=1/n-1/(n+1)1/S1+1/S2+······+1/Sn=1-1/2+1/2-1/3+1/3-1/4+······+1/n-1/(n+1)=1-1/(n+1)=n/(n+1)
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出门在外也不愁设{an}是等差数列,其前n项和是Sn,a3=6,S3=12.(1)求数列{an}的通项公式;(2)求1S1+1S2+…+1Sn的值_百度知道
设{an}是等差数列,其前n项和是Sn,a3=6,S3=12.(1)求数列{an}的通项公式;(2)求1S1+1S2+…+1Sn的值
1wordWwordWrap:normal:1px solid black">1S<span style="vertical-align设{an}是等差数列:1px solid black">12+…+S<span style="vertical-wordSpacing,S3=12.(1)求数列{an}的通项公式;(2)求+
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overflow:normal">1n+1:1wordSpacing:6wordWrap: initial initial?.5px:9font-font-size:normal:normal">S<td style="font-size?1)2×2=n2+n,a3=6:1px:sub.jpg') no-repeat:90%">n=S<span style="vertical-align?13+…+<table cellpadding="-1" cellspacing="-1" style="/zhidao/pic/item/50da81cb39dbb6fd0ba3af920a24abd:1px solid black,∴an=2+(n-1)×2=2n.(2)∵a1=2://hiphotos:1px">n=1-+3d=12<td style="font-size,S3=12; width: overflow-x; margin- width:nowrap:normal"><table cellpadding="-1" cellspacing="-1" style="margin-font-overflow?<table cellpadding="-1" cellspacing="-1" style="margin-overflow,∴+2d=6<td style="border-bottom: 13,∴++n==; height: no-repeat repeat:hidden">2+…+a1n+1=1-1<td style="padding- height: url('http: left. background-origin
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出门在外也不愁已知等差数列,an=2n+1,令Tn=1/S1+1/S2+1/S3+……+1/Sn,求证:Tn<3/4
由等差数列,an=2n+1得a1=3所以sn=n(3+2n+1)/2=n(n+2)所以1/sn=1/[n(n+2)]=1/2*[1/n-1/(n+2)]因此Tn=1/S1+1/S2+1/S3+……+1/Sn=1/2*[1-1/3+1/2-1/4+1/3-1/5+.+1/(n-1)-1/(n+1)+1/n-1/(n+2)]=1/2*[1+1/2-1/(n+1)-1/(n+2)]=1/2*[3/2-1/(n+1)(n+2)]=3/4-1/[2(n+1)(n+2)]<3/4所以Tn<3/4
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an=2n+1Sn=a1+a2+...+n=2(1+2+...+n)+n=2n(n+1)/2 +n=n²+2n=n(n+2)1/Sn=1/[n(n+2)]=(1/2)[1/n -1/(n+2)]Tn=1/S1+1/S2+...+1/Sn=(1/2)[1-1/3+1/2-1/4+...+1/n-1/(n+2)]=(1/2)[(1+1/2+1...
an=2n+1则a1=3Sn=n(a1+an)/2=n(3+2n+1)/2=n^2+2n1/Sn=1/(n^2+2n)=1/[n(n+2)]=1/2[1/n -1/(n+2)]∴Tn=1/S1+1/S2+1/S3+……+1/Sn
=1/2{ 1-1/3 +1/2 -1/4 +1/3-1/5+……+1/n-1/(n+2)}
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