已知等差数列{an}公差为d(d≠0),前n项和为Sn,Xn表示{an}前n项的平均数,且数列{Xn}补充如下:
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/13 23:38:32
已知等差数列{an}公差为d(d≠0),前n项和为Sn,Xn表示{an}前n项的平均数,且数列{Xn}补充如下:
已知等差数列{an}公差为d(d≠0),前n项和为Sn,Xn表示{an}前n项的平均数,且数列{Xn}的前n项和为Tn,且数列{1/(Sn+1-Tn+1)}的前n项和为An,则An=( )
已知等差数列{an}公差为d(d≠0),前n项和为Sn,Xn表示{an}前n项的平均数,且数列{Xn}的前n项和为Tn,且数列{1/(Sn+1-Tn+1)}的前n项和为An,则An=( )
等差数列{an}公差为d(d≠0),前n项和
Sn=na1+n(n-1)d/2
∴xn=Sn/n=a1+(n-1)d/2
∴{xn}为等差数列,首项为a1公差为d/2
∴{Xn}的前n项和
Tn=n[2a1+(n-1)d/2]/2=na1+n(n-1)d/4
Sn+1-Tn+1=(n+1)nd/2-(n+1)nd/4=(n+1)nd/4
∴1/[S(n+1)-T(n+1)]=4/d*1/[n(n+1)]
=4/d[1/n-1/(n+1)]
∴An=4/d[ (1/1-1/2)+(1/2-1/3)+.+(1/n)+1/(n+1)]
=4/d*[1-1/(n+1)]=4n/[(n+1)d]
Sn=na1+n(n-1)d/2
∴xn=Sn/n=a1+(n-1)d/2
∴{xn}为等差数列,首项为a1公差为d/2
∴{Xn}的前n项和
Tn=n[2a1+(n-1)d/2]/2=na1+n(n-1)d/4
Sn+1-Tn+1=(n+1)nd/2-(n+1)nd/4=(n+1)nd/4
∴1/[S(n+1)-T(n+1)]=4/d*1/[n(n+1)]
=4/d[1/n-1/(n+1)]
∴An=4/d[ (1/1-1/2)+(1/2-1/3)+.+(1/n)+1/(n+1)]
=4/d*[1-1/(n+1)]=4n/[(n+1)d]
已知等差数列{an}公差为d(d≠0),前n项和为Sn,Xn表示{an}前n项的平均数,且数列{Xn}补充如下:
{an}为等差数列,公差d>0,Sn是数列{an}前n项和,
数列an是公差d不等于0的等差数列,其前n项和为Sn,且
若等差数列{an}的首项为a1,公差为d,前n项的和为Sn,则数列(Sn/n)为等差数列,且通项
已知数列{an}是公差为d的等差数列,d≠0且a1=0,bn=2^(an)(n属于N*),Sn是{bn}的前n项和,Tn
已知等差数列{an}前n项和为Sn,公差d≠0,且S3+S5=50,a1,a4,a13,成等比数列 (2)若从数列
已知等差数列{an}的公差d不为零,首项a1=2且前n项和为sn
等差数列{an}中,公差为d,已知a4=84,前n项和为Sn,且S10>0,S11
已知数列{an}的前n项和为Sn,a1=3且an+1=2Sn+3,数列{bn}为等差数列,且公差d>0,b1+b2+b3
已知等差数列(an)中,公差d>0,其前n项和为Sn,且.
高中数学有关数列、极限的题:设等差数列{an}的公差d=2,前n项的和为sn,则lim(an^-n^0/sn=
已知数列An.是首项a1=1,公差d大于0的等差数列,且2a2,a10,5a5,成等差数列,数列An,前n项和为Sn 求