wallis为什么2k+1/2k等于多少度1

介绍Wallis公式及其应用_百度文库
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介绍Wallis公式及其应用
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>>>在数列{an}中,a1=0,且对任意k∈N*,a2k-1,a2k,a2k+1成等差数列..
在数列{an}中,a1=0,且对任意k∈N*,a2k-1,a2k,a2k+1成等差数列,其公差为2k, (Ⅰ)证明:a4,a5,a6成等比数列; (Ⅱ)求数列{an}的通项公式; (Ⅲ)记,证明。
题型:解答题难度:偏难来源:天津高考真题
解:(Ⅰ)证明:由题设可知,a2=a1+2=2,a3=a2+2=4,a4=a3+4=8,a5=a4+4=12,a6=a5+6=18,从而, 所以a4,a5,a6成等比数列. (Ⅱ)由题设,可得a2k+1-a2k-1=4k,k∈N*,所以a2k+1-a1=(a2k+1-a2k-1)+(a2k-1-a2k-3)+…+(a3-a1)=4k+4(k-1)+…+4×1=2k(k+1),k∈N*,由a1=0,得a2k+1=2k(k+l),从而a2k=a2k+1-2k=2k2,所以数列{an}的通项公式为或写为。 (Ⅲ)证明:由(Ⅱ)可知a2k+1=2k(k+1),a2k=2k2, 以下分两种情况进行讨论: (1)当n为偶数时,设n=2m(m∈N*),若m=1,则,若m≥2,则,所以,,从而,n=4,6,8,…… (2)当n为奇数时,设n=2m+1(m∈N*),,所以,从而,n=3,5,7,…… 综合(1)和(2)可知,对任意n≥2,n∈N*,有。
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据魔方格专家权威分析,试题“在数列{an}中,a1=0,且对任意k∈N*,a2k-1,a2k,a2k+1成等差数列..”主要考查你对&&反证法与放缩法,等比数列的定义及性质,一般数列的通项公式,数列求和的其他方法(倒序相加,错位相减,裂项相加等)&&等考点的理解。关于这些考点的“档案”如下:
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反证法与放缩法等比数列的定义及性质一般数列的通项公式数列求和的其他方法(倒序相加,错位相减,裂项相加等)
反证法的定义:
有些不等式无法利用题设的已知条件直接证明,我们可以用间接的方法——反证法去证明,即通过否定原结论——导出矛盾——从而达到肯定原结论的目的。
放缩法的定义:
把原不等式放大或缩小成一个恰好可以化简的形式,比较常用的方法是把分母或分子适当放大或缩小(减去或加上一个正数)使不等式简化易证。 反证法证题的步骤:
若A成立,求证B成立。共分三步:(1)提出与结论相反的假设;如负数的反面是非负数,正数的反面是非正数即0和负数;(2)从假设出发,经过推理,得出矛盾;(必须由假设出发进行推理否则不是反证法或证错);(3)由矛盾判定假设不正确,从而肯定命题的结论正确.矛盾:与定义、公理、定理、公式、性质等一切已有的结论矛盾甚至自相矛盾。反证法是一种间接证明命题的基本方法。在证明一个数学命题时,如果运用直接证明法比较困难或难以证明时,可运用反证法进行证明。
放缩法的意义:
放缩法理论依据是不等式的传递性:若,a&b,b&c,则a&c.
放缩法的操作:
若求证P&Q,先证P&P1&P2&…&Pn,再证恰有Pn&Q.需注意:(1)只有同方向才可以放缩,反方向不可。(2)不能放(缩)得太大(小),否则不会有最后的Pn&Q.等比数列的定义:
一般地,如果一个数列从第2项起,每一项与它的前一项的比等于同一个常数,那么这个数列就叫做等比数列,这个常数叫做公比,公比通常用字母q表示(q≠0)。 等比数列的性质:
在等比数列{an}中,有 (1)若m+n=p+q,m,n,p,q∈N*,则aman=apaq;当m+n=2p时,aman=ap2; (2)若m,n∈N*,则am=anqm-n; (3)若公比为q,则{}是以为公比的等比数列; (4)下标成等差数列的项构成等比数列; (5)1)若a1>0,q>1,则{an}为递增数列; 2)a1<0,q>1, 则{an}为递减数列; 3)a1>0,0<q<1,则{an}为递减数列; 4)a1<0, 0<q<1, 则{an}为递增数列; 5)q<0,则{an}为摆动数列;若q=1,则{an}为常数列。
等差数列和等比数列的比较:
如何证明一个数列是等比数列:
证明一个数列是等比数列,只需证明是一个与n无关的常数即可(或an2=an-1an+1)。 一般数列的定义:
如果数列{an}的第n项an与序号n之间的关系可以用一个式子表示成an=f(n),那么这个公式叫做这个数列的通项公式。
&通项公式的求法:
(1)构造等比数列:凡是出现关于后项和前项的一次递推式都可以构造等比数列求通项公式; (2)构造等差数列:递推式不能构造等比数列时,构造等差数列; (3)递推:即按照后项和前项的对应规律,再往前项推写对应式。已知递推公式求通项常见方法:①已知a1=a,an+1=qan+b,求an时,利用待定系数法求解,其关键是确定待定系数λ,使an+1&+λ=q(an+λ)进而得到λ。②已知a1=a,an=an-1+f(n)(n≥2),求an时,利用累加法求解,即an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)的方法。③已知a1=a,an=f(n)an-1(n≥2),求an时,利用累乘法求解。数列求和的常用方法:
1.裂项相加法:数列中的项形如的形式,可以把表示为,累加时抵消中间的许多项,从而求得数列的和; 2、错位相减法:源于等比数列前n项和公式的推导,对于形如的数列,其中为等差数列,为等比数列,均可用此法; 3、倒序相加法:此方法源于等差数列前n项和公式的推导,目的在于利用与首末两项等距离的两项相加有公因式可提取,以便化简后求和。4、分组转化法:把数列的每一项分成两项,或把数列的项“集”在一块重新组合,或把整个数列分成两个部分,使其转化为等差或等比数列,这一求和方法称为分组转化法。5、公式法求和:所给数列的通项是关于n的多项式,此时求和可采用公式求和,常用的公式有:& 数列求和的方法多种多样,要视具体情形选用合适方法。 数列求和特别提醒:
(1)对通项公式含有的一类数列,在求时,要注意讨论n的奇偶性;(2)在用等比数列前n项和公式时,一定要分q=1和q≠1两种情况来讨论。
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与“在数列{an}中,a1=0,且对任意k∈N*,a2k-1,a2k,a2k+1成等差数列..”考查相似的试题有:
560568332891571587395108410982342628& 等差数列与等比数列的综合知识点 & “已知数列{an}中,...”习题详情
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已知数列{an}中,a1=1,对任意的k∈N*,a2k-1、a2k、a2k+1成等比数列,公比为qk;a2k、a2k+1、a2k+2成等差数列,公差为dk,且d1=2.(1)写出数列{an}的前四项;(2)设bk=1qk-1,求数列{bk}的通项公式;(3)求数列{dk}的前k项和Dk.
本题难度:较难
题型:解答题&|&来源:2014-上海模拟
分析与解答
习题“已知数列{an}中,a1=1,对任意的k∈N*,a2k-1、a2k、a2k+1成等比数列,公比为qk;a2k、a2k+1、a2k+2成等差数列,公差为dk,且d1=2.(1)写出数列&#12...”的分析与解答如下所示:
(1)由题意得{a22=a1a3a3=a2+2,求出a2=2或a2=-1,即可写出数列{an}的前四项;(2)由a2k,a2k+1,a2k+2成等差数列,其公差为dk,知2a2k+1=a2k+a2k+2,从而能够证明{bk}是等差数列,且公差为1.(3)由d1=2,得a3=a2+2,解得a2=2,或a2=-1.由此进行分类讨论,能够求出Dk.
解:(1)由题意得{a22=a1a3a3=a2+2,∴a22=a2+2,a2=2或a2=-1.…(2分)故数列{an}的前四项为1,2,4,6或1,-1,1,3.…(4分)(2)∵a2k-1,a2k,a2k+1成公比为qk的等比数列,a2k+1,a2k+2,a2k+3成公比为qk+1的等比数列,∴a2k+1=a2kqk,a2k+2=a2k+1qk+1又∵a2k,a2k+1,a2k+2成等差数列,∴2a2k+1=a2k+a2k+2.得2a2k+1=a2k+1qk+a2k+1qk+1,2=1qk+qk+1,…(6分)∴qk-1qk=qk+1-1,∴1qk+1-1=qkqk-1=1+1qk-1,1qk+1-1-1qk-1=1,即bk+1-bk=1.∴数列{bk}为公差d=1等差数列,且b1=1q1-1=1或b1=1q1-1=-12.…(8分)∴bk=b1+(k-1)o1=k或bk=k-32.…(10分)(3)当b1=1时,由(2)得bk=1qk-1=k,qk=k+1k.a2k+1a2k-1=(k+1k)2,a2k+1=a2k+1a2k-1oa2k-1a2k-3…a3a1oa1=(k+1k)2o(kk-1)2…(21)2o1=(k+1)2,a2k=a2k+1qk=k(k+1),dk=a2k+1-a2k=a2k+1qk=k+1,Dk=k(k+3)2.…(13分)当b1=-12时,同理可得dk=4k-2,Dk=2k2.…(16分)
本题考查数列的前n项和的计算,等差数列的证明,综合性强,难度大,是高考的重点.解题时要认真审题,仔细解答,注意计算能力的培养.
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已知数列{an}中,a1=1,对任意的k∈N*,a2k-1、a2k、a2k+1成等比数列,公比为qk;a2k、a2k+1、a2k+2成等差数列,公差为dk,且d1=2.(1)写出数...
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经过分析,习题“已知数列{an}中,a1=1,对任意的k∈N*,a2k-1、a2k、a2k+1成等比数列,公比为qk;a2k、a2k+1、a2k+2成等差数列,公差为dk,且d1=2.(1)写出数列&#12...”主要考察你对“等差数列与等比数列的综合”
等考点的理解。
因为篇幅有限,只列出部分考点,详细请访问。
等差数列与等比数列的综合
等差数列与等比数列的综合.
与“已知数列{an}中,a1=1,对任意的k∈N*,a2k-1、a2k、a2k+1成等比数列,公比为qk;a2k、a2k+1、a2k+2成等差数列,公差为dk,且d1=2.(1)写出数列&#12...”相似的题目:
已知{an}是公差不为零的等差数列,a1=1,且a1,a3,a9成等比数列.(Ⅰ)求数列{an}的通项;(Ⅱ)求数列{2an}的前n项和Sn.&&&&
已知-9,a1,a2,-1四个实数成等差数列,-9,b1,b2,b3,-1五个实数成等比数列,则b2(a2-a1)=&&&&A.8B.-8C.&8D.&&&&
在等差数列是{an}中,已知a4与a2与a8的等比中项,a3+2是a2与a6的等差中项,Sn是前n项和,则满足的所有n值的和为&&&&.
“已知数列{an}中,...”的最新评论
该知识点好题
1已知{an}是公差为d的等差数列,{bn}是公比为q的等比数列.(1)若an=3n+1,是否存在m、k∈N*,有am+am+1=ak?说明理由;(2)找出所有数列{an}和{bn},使对一切n∈N*,an+1an=bn,并说明理由;(3)若a1=5,d=4,b1=q=3,试确定所有的p,使数列{an}中存在某个连续p项的和是数列{bn}中的一项,请证明.
2已知等比数列{an}的前n项和为Sn=ao2n+b,且a1=3.(1)求a、b的值及数列{an}的通项公式;(2)设bn=nan,求数列{bn}的前n项和Tn.
3已知等差数列{an}的公差为-1,且a2+a7+a12=-6,(1)求数列{an}的通项公式an与前n项和Sn;(2)将数列{an}的前4项抽去其中一项后,剩下三项按原来顺序恰为等比数列{bn}的前3项,记{bn}的前n项和为Tn,若存在m∈N*,使对任意n∈N*总有Sn<Tm+λ恒成立,求实数λ的取值范围.
该知识点易错题
1已知{an}是公差为d的等差数列,{bn}是公比为q的等比数列.(1)若an=3n+1,是否存在m、k∈N*,有am+am+1=ak?说明理由;(2)找出所有数列{an}和{bn},使对一切n∈N*,an+1an=bn,并说明理由;(3)若a1=5,d=4,b1=q=3,试确定所有的p,使数列{an}中存在某个连续p项的和是数列{bn}中的一项,请证明.
2已知等比数列{an}的前n项和为Sn=ao2n+b,且a1=3.(1)求a、b的值及数列{an}的通项公式;(2)设bn=nan,求数列{bn}的前n项和Tn.
3已知等差数列{an}的公差为-1,且a2+a7+a12=-6,(1)求数列{an}的通项公式an与前n项和Sn;(2)将数列{an}的前4项抽去其中一项后,剩下三项按原来顺序恰为等比数列{bn}的前3项,记{bn}的前n项和为Tn,若存在m∈N*,使对任意n∈N*总有Sn<Tm+λ恒成立,求实数λ的取值范围.
欢迎来到乐乐题库,查看习题“已知数列{an}中,a1=1,对任意的k∈N*,a2k-1、a2k、a2k+1成等比数列,公比为qk;a2k、a2k+1、a2k+2成等差数列,公差为dk,且d1=2.(1)写出数列{an}的前四项;(2)设bk=1/qk-1,求数列{bk}的通项公式;(3)求数列{dk}的前k项和Dk.”的答案、考点梳理,并查找与习题“已知数列{an}中,a1=1,对任意的k∈N*,a2k-1、a2k、a2k+1成等比数列,公比为qk;a2k、a2k+1、a2k+2成等差数列,公差为dk,且d1=2.(1)写出数列{an}的前四项;(2)设bk=1/qk-1,求数列{bk}的通项公式;(3)求数列{dk}的前k项和Dk.”相似的习题。Wallis Jordan
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