数列(an),a1=1,当n≥2,其前n项和Sn满足Sn^2=an(Sn-1)证(1/Sn)是等差数列.设bn=log以2为底Sn/S(n+2),bn的前n项和Tn,求满足Tn≥6的最小正整数n

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/23 04:34:57
数列(an),a1=1,当n≥2,其前n项和Sn满足Sn^2=an(Sn-1)证(1/Sn)是等差数列.设bn=log以2为底Sn/S(n+2),bn的前n项和Tn,求满足Tn≥6的最小正整数n数列(

数列(an),a1=1,当n≥2,其前n项和Sn满足Sn^2=an(Sn-1)证(1/Sn)是等差数列.设bn=log以2为底Sn/S(n+2),bn的前n项和Tn,求满足Tn≥6的最小正整数n
数列(an),a1=1,当n≥2,其前n项和Sn满足Sn^2=an(Sn-1)证(1/Sn)是等差数列.设bn=log以2为底Sn/S(n+2),bn的前n项和Tn,求满足Tn≥6的最小正整数n

数列(an),a1=1,当n≥2,其前n项和Sn满足Sn^2=an(Sn-1)证(1/Sn)是等差数列.设bn=log以2为底Sn/S(n+2),bn的前n项和Tn,求满足Tn≥6的最小正整数n
n>1时
an=Sn-S(n-1)
(Sn)^2=[Sn-S(n-1)](Sn-1)
=(Sn)^2-Sn+S(n-1)-SnS(n-1)
因此S(n-1)-Sn=SnS(n-1)
由题知道Sn不等于0
两边同时除以SnS(n-1)得
1/Sn-1/S(n-1)=1
因此(1/Sn)是等差数列
首项为1/S1=1/a1=1,公差为1
因此1/Sn=1+(n-1)*1=n
因此Sn=1/n
S(n+2)=1/(n+2)
Sn/S(n+2)=(n+2)/n
bn=log(2)[(n+2)/n]=log(2)(n+2)-log(2)n
当n为奇数时
Tn=1og(2)3-log(2)1+log(2)4-log(2)2+…+log(2)(n+1)-log(2)(n-1)+log(2)(n+2)-log(2)n
=log(2)(n+2)-log(2)1+log(2)(n+1)-log(2)2
=log(2)(n+2)+log(2)(n+1)-1
当n为偶数时
Tn=1og(2)3-log(2)1+log(2)4-log(2)2+…+log(2)(n+1)-log(2)(n-1)+log(2)(n+2)-log(2)n
=log(2)(n+2)-log(2)2+log(2)(n+1)-log(2)1
=log(2)(n+2)+log(2)(n+1)-1
综上所述
Tn
=log(2)(n+2)+log(2)(n+1)-1
=log(2)(n+2)+log(2)(n+1)-log(2)2
=log(2)[(n+2)(n+1)/2]
Tn≥6则
log(2)[(n+2)(n+1)/2]≥6
即log(2)[(n+2)(n+1)/2]≥log(2)(2^6)
又f(x)=log(2)x是增函数,2^6=64
因此
(n+2)(n+1)/2≥64
化简得到
n^2+3n-126≥0
由n>0解得
n≥(-3+√516)/2
又22

已知数列an中,a1=1,当n≥2时,其前n项和Sn平方=an(Sn-1/2) 求Sn表达式. 数列{an}中,a1=1,当n大于等于2时,其前n项的和Sn,满足Sn的平方=an(Sn-1) 数列{an}中,a1=1.Sn是其前n项和,当n≥2时,an=3Sn,则lim(n→∞0)Sn+1/S(n+1)-3为多少? 谢谢)数列{An}中,A1=8,A4=2,且满足A(n+2)=2A(n+1)-An,n属于N*数列{An}中,A1=8,A4=2,且满足A(n+2)=2A(n+1)-An,n属于N* (1)求数列{An}的通项公式;(2)当n为何值时,其前n项和Sn最大?求出最大值;———————— 数列(an)中,a1=1,当n≥2时,其前n项的和Sn满足Sn平方=an(Sn-1).证明:数列1/Sn为等差数列 数列an中,a1=1,当n大于等于2时,其前n项和满足sn^2=an(sn-1) 证明:数列{1/sn}是等差数列 数列{an}中,a1,=1,Sn为其前n项和,当t>0时,有3tSn-(2t+3) Sn-1 =3t(n∈N*,n≥2)(1) 求证:数列{an}是等比数列;(2)设数列{an}的公比为f(t),作数列{bn},使b1=1,bn=f( )(n∈N*,n≥2),求数 设数列{an}中,Sn是其前n项和,若首项a1=1,且满足2Sn^2=an(2Sn-1),(n为下标),(n∈N*,n≥2),求通项. 设数列{an}中,Sn是其前n项和,若首项a1=1,且满足2Sn^2=an(2Sn-1),(n为下标),(n∈N*,n≥2),求通项.不要跳步, 数列〔an〕满足an+1+an=4n-3,当a1=2时,求数列〔an〕前n项和 已知正项数列{an}中,a1=3,前n项和为Sn(n是N*)当n≥2时,有√Sn-√S(n-1)已知正项数列{an}中,a1=3,前n项和为Sn(n是N*)当n≥2时,有√Sn-√S(n-1)=√3 求通项 1:在数列{an}中,a1=1,当n>=2时,其前n项和sn满足an+2sn*s(n-1)=0(1) 求sn表达式(2)设 bn=sn/2n+1,求数列{bn}的前n项和2:已知数列{an}的通向公式为an=(n-√70)/(n-√71),求数列的最大项和最小项 数列(an),a1=1,当n≥2,其前n项和Sn满足Sn^2=an(Sn-1)证(1/Sn)是等差数列.设bn=log以2为底Sn/S(n+2),bn的前n项和Tn,求满足Tn≥6的最小正整数n 设Sn是数列{an}的前n项和,a1=a,且Sn^2=3n^2an+S(n-1)^2,证明数列{a(n+2)-an}是常数数列设Sn是数列{an}的前n项和,a1=a,且Sn^2=3n^2an+S(n-1)^2,an≠0,n=2,3,4……证明数列{a(n+2)-an}(n≥2)是常数数列 等差数列、等比数列1、数列{a n}中,a1=1,当n≥2,其前n项和S n满足(S n)^2=a n (S n -1/2),求数列{a n}2、已知数列{a n}满足a1=1/2,a1+a2+a3+……+a n=n^2 a,求数列{a n}的通项公式2、已知数列{a n}满足a1=1/2,a1+a2+ 已知数列{an}中,a1=3,a2=5,其前n项和Sn满足Sn+S(n-2)=2S(n-1)+2^ n-1(n≥3),(2)令bn=1/an*a(n+1),Tn是数列bn前n项和,证明Tnn0时,(2)中的Tn>n成立. 已知数列{an}的前n项和为Sn,a1=1,当n≥2时,an+2S(n-1)=n,则S2011= 在数列{an}中,a1=1,Sn是{an}前n项的和,若当n≥2时,an,Sn,S(n-1/2)成等比数列,1.在数列{an}中,a1=1,Sn是{an}前n项的和,若当n≥2时,an,Sn,S(n-1/2)成等比数列,求a2,a3,a4的值,并由此猜想{an}的通项公式,并证