若{sn/n}等差数列,证明an为等差数列 在(1)的条件下s1=2s2=6,求数列{1/SN}
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/01 12:00:25
若{sn/n}等差数列,证明an为等差数列 在(1)的条件下s1=2s2=6,求数列{1/SN}
证:
(1)
设公差为d.
Sn/n=S1/1 +(n-1)d=a1+(n-1)d
Sn=na1+n(n-1)d
S(n+1)=(n+1)a1+n(n+1)d
a(n+1)=S(n+1)-Sn=(n+1)a1+n(n+1)d-na1-n(n-1)d=a1+2nd
an=a1+2(n-1)d
a(n+1)-an=a1+2nd-a1-2(n-1)d=2d,为定值.
数列{an}是以2d为公差的等差数列.
(2)
S1=2 S2=6
d=S2/2 -S1/1=3 -2=1
Sn/n=S1/1 +(n-1)d=S1+(n-1)d=2+n-1=n+1
Sn=n(n+1)/2
1/Sn=2/[n(n+1)]=2[1/n-1/(n+1)]
前n项和Tn=2[1/1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)
(1)
设公差为d.
Sn/n=S1/1 +(n-1)d=a1+(n-1)d
Sn=na1+n(n-1)d
S(n+1)=(n+1)a1+n(n+1)d
a(n+1)=S(n+1)-Sn=(n+1)a1+n(n+1)d-na1-n(n-1)d=a1+2nd
an=a1+2(n-1)d
a(n+1)-an=a1+2nd-a1-2(n-1)d=2d,为定值.
数列{an}是以2d为公差的等差数列.
(2)
S1=2 S2=6
d=S2/2 -S1/1=3 -2=1
Sn/n=S1/1 +(n-1)d=S1+(n-1)d=2+n-1=n+1
Sn=n(n+1)/2
1/Sn=2/[n(n+1)]=2[1/n-1/(n+1)]
前n项和Tn=2[1/1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)
数列an是首项为3公差为2的等差数列其前n项和为Sn求An=1/S1+1/S2+1/S3+...+1/Sn
已知数列an的前项和为Sn,a1=1,nSn+1-(n+1)Sn=n^2+cn,S1,S2/2,S3/3成等差数列.(1
在数列{An}中,已知A1=1,An=2Sn^2/(2Sn-1),(n>=2),证明{1/Sn}是等差数列,并求Sn
已知数列{an}中的前几项和为Sn且满足a1=0.5,an=-2Sn*S(n-1).证明数列{1/Sn}为等差数列,求S
已知等差数列{an的公差为2,前n项和为Sn,且S1,S2,S3成等比数列.(1)求数列{an的通项公式
已知数列{an }是等差数列,其前n项和为Sn,a3=1/2,S3=6.求1/S1+1/S2+······+1/Sn的值
已知等比数列{an}的前n项和为sn,a1=2,s1,2s2,3s3成等差数列,1.求数列{an}的通项公式
已知数列{an}中,a1=1,且Sn,Sn+1,2S1成等差数列,用数学归纳法证明Sn=(2^n-1)/2^(n-1)
等比数列{an}的前n项和为Sn已知S1,S3,S2成等差数列,(1)求{an}的公比q(2)若a1-a3=3,求Sn
设数列an的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n,且a1,a2,a3成等差数列
已知公差不为0的等差数列{An}的首项A1=1,前n项和为Sn,若数列{Sn/An}是等差数列,求An?
设等差数列{An}的前n项和为Sn,S4=4S2,A2n=2An+1 ,(1)求数列{an}的通