等差数列An=2n+1 Bn=1/An^2-1 求Bn前N项和Tn

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/06 07:52:33
等差数列An=2n+1 Bn=1/An^2-1 求Bn前N项和Tn

等差数列An=2n+1 Bn=1/An^2-1 求Bn前N项和Tn
等差数列An=2n+1 Bn=1/An^2-1 求Bn前N项和Tn

等差数列An=2n+1 Bn=1/An^2-1 求Bn前N项和Tn
Bn=1/An^2-1=1/[(2n+1)^2-1]=1/(4n^2+4n)=1/4*1/[n*(n+1)]
又∵1/[n*(n+1)]=1/n-1/(n+1)]
∴Tn=1/4*[1/1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)]
=1/4*[1/1-1/(n+1)]
=1/4*[(n+1)/(n+1)-1/(n+1)]
=1/4*[(n+1-1)/(n+1)]
=n/(4n+4)
有什么不懂的请追问,我会为您详细解答,

等差数列{an},{bn}的前n项和分别为An,Bn,切An/Bn=2n/3n+1,求lim(n→∞)an/bn 计算等差数列{an}{bn}的前n项和分别为An.Bn,且An/Bn=2n/(n+1)求limn→∞(an/bn) 已知数列{an}是等差数列,且bn=an+a(n-1),求证bn也是等差数列 等差数列求和已知{an}=1+(n-1)/2求{bn}=1/(n×an) 设{an}是等差数列,an=2n-1,{bn}是等比数列,bn=2^(n-1)求{an/bn}前n项和Sn {an}{bn}都是等差数列,已知An/Bn(各自前n项和)=(5n+3)/(2n-1)则an/bn=? 已知等差数列{an}和{bn},他们的前n项之和为An和Bn,若An/Bn=(5n+3)/(2n-1)A9/B9 已知{an},{bn}均为等差数列,前n项的和为An,Bn,且An/Bn=2n/(3n+1),求a10/b10的值 等差数列{an}的前n项和为An,等差数列{bn}的前n项和为Bn若Bn/An=n/3n+1 1.求a5/b5 2 an/bn 等差数列An=2n+1 Bn=1/An^2-1 求Bn前N项和Tn 等差数列bn=(a1+a2+3a….an)/n(1)bn=n^2,求{an}(2){bn}为等差数列,求证{an}也为等差数列错了 是 (a1+a2-+a3...+an)/n an+1^2-a^2=bn an是等差数列 求证bn是等差数列 在数列{an}和{bn}中,an>0,bn>0,且an,bn,a(n+1)成等差数列,bn,a(n+1),b(n+1)成等比数列,a1=1,b1=2,求an/bn. an是等差数列,an=13+(n-1)2.an=log2bn.证明bn是等比数列 数列{an}与{bn}满足an=1/n(b1+b2+…+bn)(n∈N).求证:数列{bn}为等差数列的充要条件是数列{an}为等差数列 数列an,bn各项均为正数,a1=1,b1=2,a2=3,对任意n,an,bn,an+1成等差数列,bn,an+1,bn+1成等比数列,求an,bn的通项公式 若两个等差数列{an}和{bn}的前n项An和Bn,满足关系式An/Bn=2n+1/4n+27(n∈N*),求an/bn. 高二数列练习题 数列{an}中,a1=4,an=4-4/a(n-1),数列{bn},bn=1/an-2,求:(1){bn}为等差数列; (2){an}数列{an},a1=4,an=4-4/a(n-1),数列{bn},bn=1/an-2,求:(1){bn}为等差数列;(2){an}的通项公式.