设等差数列{an}的前n项和为Sn,且S3=2S2+4,a5=36(1)求an,Sn(2)设bn=Sn-1(n∈N﹢),Tn=1/b1+1/b2+1/b3+·····+1/bn,求Tn

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/27 07:22:58
设等差数列{an}的前n项和为Sn,且S3=2S2+4,a5=36(1)求an,Sn(2)设bn=Sn-1(n∈N﹢),Tn=1/b1+1/b2+1/b3+·····+1/bn,求Tn
xRN0`NQnURe>U0"RQT LTtjh28;b@l߻wOڵh=wh}&NEڽYz[M_CpW~(ݨ~>Ox>!`(0D<%$q IKRDX%'v-~Y,/] eĩ$Cb& BM&~\ZnQNE\ nW˂j!i Omн ^!}CNá0 ٚb@

设等差数列{an}的前n项和为Sn,且S3=2S2+4,a5=36(1)求an,Sn(2)设bn=Sn-1(n∈N﹢),Tn=1/b1+1/b2+1/b3+·····+1/bn,求Tn
设等差数列{an}的前n项和为Sn,且S3=2S2+4,a5=36
(1)求an,Sn
(2)设bn=Sn-1(n∈N﹢),Tn=1/b1+1/b2+1/b3+·····+1/bn,求Tn

设等差数列{an}的前n项和为Sn,且S3=2S2+4,a5=36(1)求an,Sn(2)设bn=Sn-1(n∈N﹢),Tn=1/b1+1/b2+1/b3+·····+1/bn,求Tn
1、由S3=2S2+4 a5=36得:
3a1+3*(3-1)*d/2=2*[2a1+2*(2-1)*d/2]+4
a1+4d=36
化简得:
d-a1=4
a1+4d=36
解之得:a1=4 d=8
则:an=4+(n-1)*8=8n-4
Sn=(4+8n-4)*n/2=4n^2
2、bn=Sn-1=4n^2-1=(2n-1)*(2n+1) 则:
Tn=1/(1*3)+1/(3*5)+.+1/[(2n-1)(2n+1)]
=1/2*[1-1/3+1/3-1/5+.+1/(2n-1)-1/(2n+1)]
=1/2*[1-1/(2n+1)]
=n/(2n+1)

设Sn为等差数列an的前n项和.求证Sn/n为等差数列 设Sn为等差数列{An}的前n项和,求证:{Sn/n}是等差数列 设等差数列{an}的前n项和为Sn,且满足S15>0,S16 设等差数列{an}的前n项和为Sn,且S6 设等差数列{an}前n项和为Sn,且a1>0,S13=S19,求Sn的最大值 设等差数列An的前n项和为Sn,且S4=4S2,A2n=2An+1 设数列{an}为正项数列,前n项的和为Sn,且an,Sn,an^2成等差数列,求an通项公式 设数列An的前n项和为Sn,已知A1=1.A2=6,A3=11,且(5n-8)S(n+1)-(5n+2)Sn=-20n-8,求证An为等差数列 设等差数列{an}的前n项和为Sn,且S5=-5,S10=15,求数列{Sn/n}的前n项和Tn 若等差数列{an}的前n项和为Sn,且满足Sn/S2n为常数,则称该数列为S数列 若首项为a1的各项为正数的等差数列{an}是S数列,设n+h=2008,(n,h为正数) 求1/Sn+1/Sh的最小值 Sn、Sh分别是数列的前n项和和 设等差数列{an}的前n项和为Sn 若a1=Sn> 设等差数列an前n项和为sn,且s10=s20.则s30=? 设等比数列an的公比为q,前n项和为sn,若s(n+1),sn,s(n+2)成等差数列,求q的值 设等比数列[an]的公比为q,前n项和为Sn,若S(n+1),Sn,S(n+2)成等差数列,则q的值? 设等比数列 {an} 的公比为q,前n项和为Sn,若S(n+1),Sn,S(n+2)成等差数列,则q= 设等差数列{an}的前n项和为Sn,且a4-a2=8,S10=190,(1)求等差数列{an}的通项公式an 设Sn为等差数列{an}的前n项和,已知s6=36,Sn=324 ,S(n-6)=144 ,(n>6) ,求n的值 设等差数列{an}的前n项和为Sn ,且S15>0,a8+a90,a8+a9