已知数列an的前n项和Sn,且an>0,n属于N*,an,Sn,an^2成等差数列已知数列an的前n项和Sn,且an>0,n属于N*,,,成等差数列求an,若Tn=1/S1+1/S2+1/S3+...+1/Sn,求Tn

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/18 16:50:56
已知数列an的前n项和Sn,且an>0,n属于N*,an,Sn,an^2成等差数列已知数列an的前n项和Sn,且an>0,n属于N*,,,成等差数列求an,若Tn=1/S1+1/S2+1/S3+

已知数列an的前n项和Sn,且an>0,n属于N*,an,Sn,an^2成等差数列已知数列an的前n项和Sn,且an>0,n属于N*,,,成等差数列求an,若Tn=1/S1+1/S2+1/S3+...+1/Sn,求Tn
已知数列an的前n项和Sn,且an>0,n属于N*,an,Sn,an^2成等差数列
已知数列an的前n项和Sn,且an>0,n属于N*,,,成等差数列
求an,

若Tn=1/S1+1/S2+1/S3+...+1/Sn,求Tn

已知数列an的前n项和Sn,且an>0,n属于N*,an,Sn,an^2成等差数列已知数列an的前n项和Sn,且an>0,n属于N*,,,成等差数列求an,若Tn=1/S1+1/S2+1/S3+...+1/Sn,求Tn
an + (an)^2 = 2Sn
n=1
(a1)^2 -a1=0
a1= 1
an + (an)^2 = 2Sn
Sn = (1/2) {an + (an)^2}
an = Sn -S(n-1)
=(1/2) {an + (an)^2} - (1/2) {a(n-1) + [a(n-1)]^2}
[ (an)^2 - an ] -[a(n-1))^2 + a(n-1) ]=0
{(an)^2 - [a(n-1)]^2} - [an+a(n-1)]=0
[an+a(n-1)][ an- a(n-1) -1 ] =0
an- a(n-1) -1 =0
an- a(n-1) =1
an -a1 = n-1
an =n
Sn = n(n+1)/2
1/Sn = 2[1/n -1/(n+1)]
Tn =1/S1+1/S2 +...+1/Sn
= 2[ 1 - 1/(n+1) ]
= 2n/(n+1)

已知数列{an}中,an>0,Sn为{an}的前n项和,且an+1/an=2Sn,求an. 已知数列{an}中,an>0,Sn为{an}的前n项和,且an+1/an=2Sn,求an. “已知数列{an}中,an>0,Sn是数列{An}中的前n项和,且An+1/An=2Sn”An>0,求An 已知数列{an}中a1=1,且满足an+an-1不等于0,Sn=1/6*(an+1)(an+2).(1)求通项an,并说明{an}是什么数列(2)求数列{an}的前n项和Sn 已知数列{an}的前n项和为sn,且满足sn=n 已知{an}的前n项和为Sn,且an+Sn=4求证:数列{an}是等比数列 已知数列an的前n项和为Sn,且An=3^n+2n,则Sn等于 已知数列{an}的前n项和Sn,且(1-k)Sn=1-kan求an、sn 1.已知数列an的前n项和为Sn,且Sn=2^n,求通项an;2.已知数列an的前n项和为Sn,且Sn=n^2+3n,求通项an; 已知数列{an}的前n项和为Sn,且Sn=lgn 求通项公式 已知数列an是等差数列,且a1不等于0,Sn为这个数列的前n项和,求limnan/Sn.limSn+Sn-1/Sn+Sn-1 已知数列{an}的前n项和为Sn,且满足Sn=2an-1(n属于正整数),求数列{an}的通项公式an 三校生数学,an和sn的关系.已知数列{an}中,an>0,前n项之和为An,且满足An三校生数学,an和sn的关系.已知数列{an}中,an>0,前n项之和为An,且满足An=1/8(an+2)²,求数列{an} 已知数列{an}的前n项和为Sn,且an=n2的n次方,则Sn= 已知数列{an}的前n项和为Sn,且满足an+2Sn*Sn-1=0,a1=1/2.求证:{1/Sn}是等差数列 已知数列{an}的前n项和为Sn,且Sn=2an-n(n∈N*),求数列{an}的通项公式. 已知数列an的前n项和为sn,且sn+an=n^2+3n+5/2,证明数列{an-n}是等比数列 已知数列an的前n项和Sn,且an>0,n属于N*,an,Sn,an^2成等差数列已知数列an的前n项和Sn,且an>0,n属于N*,,,成等差数列求an,若Tn=1/S1+1/S2+1/S3+...+1/Sn,求Tn