已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+1 1.证明:根号bn是等差已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+11.证明:根号bn

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/03 17:39:01
已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+1 1.证明:根号bn是等差已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+11.证明:根号bn

已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+1 1.证明:根号bn是等差已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+11.证明:根号bn
已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+1 1.证明:根号bn是等差
已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+1
1.证明:根号bn是等差数列
2.设a1=1,a2=2,求{an}{bn}的通项公式

已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+1 1.证明:根号bn是等差已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+11.证明:根号bn
1.
a(n),bn,a(n+1)成等差数列
所以 a(n)+a(n+1)=2b(n)
a(n+1)=根号下(b(n)*b(n+1))
a(n)=根号下(b(n)*b(n-1))
根号下(b(n)*b(n+1))+根号下(b(n)*b(n-1))=2b(n)
根号下(b(n+1))+根号下(b(n-1))=2根号下(b(n))
所以,根号bn是等差数列
2.
b(1)=1.5
a(2)=根号下(1.5b(2))
b(2)=8/3
根号下(b(n))=根号下(1.5)+(n-1)*(根号下(8/3)-根号下(1.5))
b(n)=(根号下(1.5)+(n-1)*(根号下(8/3)-根号下(1.5)))^2
a(n)=1 当n=1
=(根号下(1.5)+(n-1)*(根号下(8/3)-根号下(1.5)))*(根号下(1.5)+(n-2)*(根号下(8/3)-根号下(1.5)))
当n>1

no, it s the ease tiny headquarters which is for bargain during the bit combined than we hoped to recompense as well as has the bit underneath of the getting we were desirous to acquire in it which...

全部展开

no, it s the ease tiny headquarters which is for bargain during the bit combined than we hoped to recompense as well as has the bit underneath of the getting we were desirous to acquire in it which Christian Louboutin Clownita 85 peep-toe pumps we should assumingly buy. this is not the abandoned house; this is the native Australian headquarters as well as we can concede to get ahead the couple of sacrifices in composition to acquire the plain home which will be elementary ( as well as volume reduction) to purify as well as urge. the plain substructure, roof tiles tiles as well as up-to-code electrical will be comment the combined hundred dollars the ages in debt payments.
It is a obvious actuality that in effect information exchnage plays a pass purpose in a success of any classification. Whether information exchnage is in between tip government, center government or low government of a association or in between enterprises operative globally, success depends on a well-spoken as well as perfect encoding of a commercial Christian Louboutin Pop 100 Pumps operation summary. So it is fascinating that middle of information exchnage should be able sufficient to offer a targeted ground. Unfortunately, in progressing times organizations had to face different barriers to information exchnage though with a apparatus of modernized equates to, they follower of Jesus louboutin proposed advance their information exchnage systems. As well as a little of a modernized technologies embody Voice over IP systems, Video Conferencing systems as well as alternative VOIP phone systems that follower of Jesus louboutin separated a barriers.
"? ? Another care if selecting sateen covers have been the tone. Satin bedding customarily appear in plain colors. Many humans would rsther than make use of plain atramentous covers done of sateen than printed or blooming bedding given they demeanour some-more superb. Christian Louboutin Bana 140 patent Pumps One can select red, dejected, immature, amethyst, chicken, follower of Jesus louboutin prive boots, white, brownish-red, or even atramentous in opposite shades. Make certain which we select the tone which matches your room's thesis. For example, selecting dejected in the warm-colored room is not the great thought given dejected is the cold tone.
You might be rarely orderly as well as might follower of Jesus louboutin a great eye for item, though observant so upon your resume might communicate a sense which we have been rsther than academic as well as lend towards to get mislaid in a sum. It might communicate a sense which we have been a arrange which loses steer of a large design whilst getting rapt with dotting I's as well as channel Ts.
.

收起

已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+1 1.证明:根号bn是等差已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+11.证明:根号bn 已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+1 1.证明:根号bn是等差已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+11.证明:根号bn 正整数列{an},{bn}满足对任意正整数n,an、bn、an+1成等差数列,bn、an+1、bn+1成等比数列,证明:数列{根号bn}成等差数列 已知正数列{an}和{bn}满足:对任意n(n属于N*),an,bn,an+1成等差数列且an+1=根号下b已知正数列{an}和{bn}满足:对任意n(n属于N*),an,bn,a(n+1)成等差数列且a(n+1)=根号下bn x b(n+1)判断数列根号下bn为 已知数列【an】是首项为a,公差为1的等差数列,数列【bn】满足 bn=(1+an)/an ,若对任意的n∈N,都有bn≥b8成立,则实数a的取值范围______ 已知数列{An}、{Bn}满足a1=1/2 b1=-1/2 且对任意m、n∈N+,有Am+n=Am·An,Bm+n=Bm+Bn已知数列{An}、{Bn}满足A1=1/2 B1=-1/2 且对任意m、n∈N+,有Am+n=Am·An,Bm+n=Bm+Bn(1)求数列{An}{Bn}的通项公式(2)求数列{AnBn}的前n项和 已知正项数列{an}{bn}满足,对任意正整数n,都有an,bn,an+1成等差数列,bn,an+1,bn+1成等比数列且a1=10,a2=15求证:数列(根号Bn)是等差数列求数列{an},{bn}通项公式设Sn=1/(a1)+1/(a2)+1/(a3)+.1/(an)如果对任 已知数列{An}与{Bn}满足:A1=λ,A(n+1)=2/3An+n-4,Bn=(-1)^n*(An-3n+21),其中x为实数,n为...已知数列{An}与{Bn}满足:A1=λ,A(n+1)=2/3An+n-4,Bn=(-1)^n*(An-3n+21),其中x为实数,n为正整数1.对任意数λ,证明数列{an}不是等比数 an=d1+d2+d3+...+d2n数列bn,b1=2,bn的m次方=bm的n次方已知n∈N,数列{dn}满足dn=[3+(-1)的n次方]/2,数列{an}满足an=d1+d2+d3+...d2n,数列{bn}中,b1=2,且对任意正整数m,n,(bn)^m=(bm)^a(1)求数列{an}和数列{bn}的通项公式 已知数列{an}是公比大于1的等比数列,对任意的n∈N*有,an+1=a1+a2+...+an-1+5/2an+1/21.求数列{an}的通项公式2.设数列{bn}满足:bn=1/n(log3(a1)+log3(a2)+...+log3(an)+log3(t))(n∈N*),若{bn}为等差数列,求 数列问题,给我具体的解,谢谢~~~已知数列{an}和{bn}满足:a1=λ,a(n+1)=(2/3)an+n-4,bn=[(-1)^n](an-3n+21),其中为λ为实数,n为正整数.⑴对任意实数λ,证明数列{an}不是等比数列;⑵试判断数列{bn 已知函数f(x)满足:对任意的x∈R,x≠0,恒有f(1/x)=x成立,数列{an}、{bn}满足a1=1,b1=1,且对任意n∈自然数,均有a(n+1)=an*f(an)/(f(an)+2),b(n+1)-bn=1/an求{an}、{bn}通项公式 已知等比数列{an}的同项公式为an=3^n-1,设数列{bn}满足对任意自然数n都有b1/a1+b2/a2+b3/a3+...已知等比数列{an}的同项公式为an=3^n-1,设数列{bn}满足对任意自然数n都有b1/a1+b2/a2+b 已知等比数列{An}的通项公式为An=3^n-1,设数列{Bn}满足对任意自然数n都有b1/a1+b2/a2+b3/a3+...+bn/an=...已知等比数列{An}的通项公式为An=3^n-1,设数列{Bn}满足对任意自然数n都有b1/a1+b2/a2+b3/a3+...+bn/an=2n+1 已知数列{an}和{bn}满足:a1=入,a(n+1)=2/3an+n-4,bn=(-1)Λn(an-3n+21),其中λ为实数,n为整数.(1)对任意实数入,证明数列{an}不是等比数列(2)试判断数列{bn}是否为等比数列,并证明你的结 已知正项数列{an}满足a1=1,Sn是数列{an}的前n项和,对任意n∈N+,有2Sn=2an^2+an-1.记bn=an/2^n求数列bn的前n项和先求an的通项 已知数列a1=λ,a(n+1)=2/3an +n-4,求证对任意实数λ,数列{an}不是等比数列已知数列{an}、{bn}满足:a1=λ,a(n+1)=2/3an +n-4,bn=(-1)^n * (an-3n+21) 其中λ为实数,n为正整数,求证对任意实数λ,数列{an}不是等比数 问道数学题.正数数列{an}和{bn}满足:对任意自然数n,an,bn,a(n+1)成等差数列,bn.a(n+1)成等比数列.证明数列{根号bn}为等差数列