正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式;令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/18 19:19:37
正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式;令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn正数数列an的前n项和为

正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式;令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn
正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式;
令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn<9/2

正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式;令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn
(1)
2Sn+nan=6n+3
n=1
3a1=9
a1=3
n=2
2(a1+a2)+2a2=15
4a2=9
a2=9/4
n=3
2(a1+a2+a3)+3a3=21
5a3=21/2
a3=21/10

2Sn+nan=6n+3 (1)
2S(n-1)+(n-1)a(n-1)=6(n-1)+3 (2)
(1)-(2)
2an+nan-(n-1)a(n-1)=6
(2+n)[an -2] = (n-1)( a(n-1) - 2)
(an-2)/(a(n-1)-2) = (n-1)/(n+2)
(an-2)/(a1 -2 ) = 6/[(n+2)(n+1)n]
an = 2+ 6/[(n+2)(n+1)n]

(2)
bn = (an-2)(n+1)
= 6/[n(n+2)]
= 3[ 1/n - 1/(n+2) ]
Tn =b1+b2+..+bn
= 3[ 1+1/2 - 1/(n+1)-1/(n+2) ]
< 3(3/2)
=9/2

(1)
2Sn+nan=6n+3
n=1
3a1=9
a1=3
n=2

已知正数数列{an}的前n项和为Sn,且对于任意正整数n满足2根号Sn=an+1 求an通项 已知正数数列{an}的前n项和为Sn,且对于任意正整数n满足2根号Sn=an+1 求an通项 已知正数数列{an}的前n项和为Sn,且对于任意正整数n满足2根号Sn=an+1 求an通项 已知正数数列{an}的前n项和为Sn,且对任意的正整数n满足 2倍的根号下Sn等于an+1,求数列{an}的通项公式? 已知数列{an}的前n项和为Sn,且对任意n属于N ,有n,an,Sn成等差数列.(1).求数列{an}的通项公式;(2)求数列{nan}的前n项和Tn. 在数列{an}中前n项和为Sn,且对任意正整数n,an+sn=20481.求数列{an}的通项公式2.设数列{log2 an}的前n项和为Tn 求Tn 设数列{an}的前n项和为Sn,且对任意正整数n,an+Sn=4096(2)设数列{log an}的前n项和为Tn,对数列{Tn},从第几项起Tn 设数列{an}的前n项和为Sn,且对任意正整数n,an+Sn=4096,)求{an}的通项公式 数列{an}的各项均为正数,Sn为其前n项和,对于任意n∈N+,总有an,Sn,an 设数列{an}的前n项和为Sn,且对任意正整数n,an+Sn=4096,)求{an}的通项公式设数列{log(2)A(n)},前n项和是Tn(n),(2)是下角标 设数列{an}的前n项和为Sn,且对任意正整数n,an+Sn=4096 (2)设数列{log an}的前n项和为Tn,当n=多少时,Tn=0 设数列{an}的前n项和为Sn,且对任意正整数n,an+Sn=4096若数列{log2底an}的前n项和记为f(n),求函数最大 设数列{an}是正数组成的数列,其前n项和Sn,且对任意n属于N*,an与2的等差中项等于Sn与2的等比中项,求 数列{an}的各项均为正数,Sn为其前n项和,对于任意n∈N+,总有an,Sn,an^2成等差数列,设数列{bn}的前n项和为Tn,且bn=(lnx)^n /an^2,则对任意实数x∈(1,e]和任意正数n,Tn小于的最小正整数是多少A1 B2 C3 D4 设数列(an )的前n 项和为S ,且对任意正整数n ,an +Sn =4096 求数列的通项公式 数列an的各项均为正数,sn为其前n项和,对于任意的n∈N*,总有an,sn,an^2成等差数列.(1)求数列an的通项公式.(2)设数列bn的前n项和为Tn,且bn=lnx/an^2,求证:对任意的实数x∈(1,e]和任意的正整数n,总 首项为正数的等比数列{an}的前n项和为Sn,数列{an^2/a(n+1)}的前n项和为Tn,且对一切正整数n都有Sn 正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式;令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn