an=n+2 Sn=1/a1a2+1/a2a3+.+1/ana(n+1),求Sn
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/27 02:41:36
an=n+2Sn=1/a1a2+1/a2a3+.+1/ana(n+1),求Snan=n+2Sn=1/a1a2+1/a2a3+.+1/ana(n+1),求Snan=n+2Sn=1/a1a2+1/a2a3
an=n+2 Sn=1/a1a2+1/a2a3+.+1/ana(n+1),求Sn
an=n+2 Sn=1/a1a2+1/a2a3+.+1/ana(n+1),求Sn
an=n+2 Sn=1/a1a2+1/a2a3+.+1/ana(n+1),求Sn
首先lz要知道1/[n*(n+1)]=1/n-1/(n+1)
所以 Sn=1/a1a2+1/a2a3+.+1/ana(n+1)
=1/(3*4)+1/(4*5)+.+1/[(n+2)*(n+3)]
=1/3-1/4+1/4-1/5+.+1/(n+2)-1/(n+3)
=1/3-1/(n+3)
=n/(3n+9)
an=n+2 Sn=1/a1a2+1/a2a3+.+1/ana(n+1),求Sn
已知数列an的前n项和是Sn=n^2,则1/a1a2+1/a2a3+...+1/a(n-1)an=?
已知数列an的前n项和是Sn=n^2,则1/a1a2+1/a2a3+...+1/a(n-1)an=?
已知{an}是等比数列,an>0,sn=a1+a2+.an,Tn=1/a1+1/a2+.1/an,求证a1a2.an=(sn/Tn)^n/2
已知数列an满足a1=1,a(n+1)=an/{3(an)+1} Sn=a1a2+a2a3+.+an(an+1),求Sn已知数列an满足a1=1,a(n+1)=an/{3(an)+1}Sn=a1a2+a2a3+......+an(an+1),求Sn
等差数列an中,a(n+1)=2n+1,则Sn=(1/a1a2)+(1/a2a3)…(1/a99a100)=
已知数列AN满足an=1 且an=2A(n-1)+2的N次 1求a1a2 (2)证明数列AN/2n次是等差 (30)前N项和SN
已知{an} a1=-3.4Sn=(2n+3)an+1 (1)求an(2)求和1/a1a2+1/a2a3+```+1/an(an+1)为多少
已知数列前n项和为sn=1/3(an-1) 求a1a2
已知函数F(x)=2x/(x+2)数列An满足A1=4/3,A(n+1)=F(An)记Sn=A1A2+A2A3+.+AnA(n+1),求证Sn
求自然数a1a2.an,使得12×2 a1a2.an 1=21×1 a1a2.an 2
已知数列an的前n项和Sn=n^2+2n,求通项an 求和1/a1a2+1/a2a3+1/a3a4…1/anan+1
设数列{an}的前n项和为sn,且a1=1,sn=nan-2n(n-1)(n∈正整数)证明,证明1/a1a2+1/a2a3+.+1/an-an+1
已知数列an的前n项和Sn=2n^2+n,则lim[1/a1a2+1/a2a3+1/a3a4+...+1/anan+1]的值为
已知等差数列an前n项和为Sn,Sn=n^2,求和1/(a1a2)+1/(a2a3)+.+1/[(an-1an] (n≥2 )老师说,用裂项相消法,求完整过程,
已知数列{an}前n项和Sn=n^2+2n (1)求数列的通项公式an (2)设Tn=1/a1a2已知数列{an}前n项和Sn=n^2+2n(1)求数列的通项公式an(2)设Tn=1/a1a2+1/a2a3+…+1/(anan+1)求Tn
已知数列{an}是一个各项为正数的等比数列,Sn为它的前n项和,Bn=1/a1+1/a2+...+1/an,Pn=a1a2...an求证:Pn=(Sn/Bn)^(n/2)
已知数列{An}的前n项和为Sn,A1=A2=1,bn=nSn+(n+2)An,数列{bn}是公差为d的等差数列,证(A1A2.An)(S1.Sn)