首页 | 本学科首页   官方微博 | 高级检索  
相似文献
 共查询到20条相似文献,搜索用时 46 毫秒
1.
题147设数列{an}满足:当n=2k-1(k∈N*)时,an=n;当n=2k(k∈N*)时,an=ak.1)求a2 a4 a6 a8 a10 a12 a14 a16;2)若Sn=a1 a2 a3 … a2n-1 a2n,证明:Sn=4n-1 Sn-1(n≥2);3)证明:S11 S12 … S1n<1-41n.解1)原式=a1 a2 a3 a4 a5 a6 a7 a8=a1 a1 a3 a1 a5 a3 a7 a1=4a1 2a3 a5 a7=4×1 2  相似文献   

2.
与二项式系数有关的求和问题的解题策略   总被引:1,自引:0,他引:1  
1赋值求和例1设(2x-3)10=a10(x-1)10 a9(x-1)9 … a2(x-1)2 a1(x-1) a0,求a1 a2 a3 … a10的值.解令x=2,得a0 a1 a2 a3 … a10=1;令x=1,得a0=(-1)10=1,所以a1 a2 a3 … a10=1-1=0.例2设(1 x x2)n=a0 a1x a2x2 … a2nx2n,求a1 a3 a5 … a2n-1的值.解令x=1,得a0 a1 a2 … a2n=3n;令x=-1,得a0-a1 a2-…-a2n-1 a2n=1.两式相减得a1 a3 a5 … a2n-1=3n-12.2逆用定理例3已知等比数列{an}的首项为a1,公比为q,求和:a1C0n a2C1n a3C2n … an 1Cnn.解a1C0n a2C1n a3C2n … an 1Cnn=a1C0n a1qC1n a1q2C2n … a1qnCnn=a1(C0n qC1n q2C2n … qnCnn)…  相似文献   

3.
定理设{an}的各项全为正数,若a12a2+a22a3+…+an-12n=(a1+a2+…+an-1)2a2+a3+…+an,则a1,a2,…,an为等比数列.证令m=(a1a2,a2a3,…,an-1an).n=(a2,a3,…,an).由a12a2+a22a3+…+an-12an=(a1+a2+…+an-1)2a2+a3+…+an得a12a2+a22a3+…+an-12an·a2+a3+…+an=a1+a2+…+an-1.即|m||n|=m·n,所以m与n共线,故存在常数k,使得a2=ka1a2,a3=ka2a3,…,an=kan-1an,∴a2a1=a3a2=…=anan-1=k,从而{an}是等比数列.等比数列的一个判定条件@齐行超$单县二中!山东274300…  相似文献   

4.
题目 设a1 、a2 、m1 、m2 均为正实数 ,且m1 +m2 =1.求证 :m1 a1 +m2 a2 ≥m1 a1 +m2 a2 .证明 ∵a1 、a2 、m1 、m2 均为正实数 ,且m1 +m2 =1.要证 :  m1 a1 +m2 a2 ≥m1 a1 +m2 a2 m1 a1 +m2 a2 ≥m21 a1 +2m1 m2a1 a2 +m22 a2 m1 ( 1-m1 )a1 +m2 ( 1-m2 )a2≥ 2m1 m2 a1 a2 m1 m2 a1 +m2 m1 a2 ≥ 2m1 m2 a1 a2 m1 m2 (a1 -2a1 a2 +a2 )≥ 0 m1 m2 (a1 -a2 ) 2 ≥ 0 .上式显然成立 .∴m1 a1 +m2 a2 ≥m1 a1 +m2 a2 .思考设a1 、a2 、a3、m1 、m2 、m3均为正实数 ,且m1 +m2 +m3=1.则m1 a1 +m2 a2 +m3a3≥m1 a1 +m2 a2 +m3a3是否…  相似文献   

5.
设数列{an}是公差为d的等差数列,且对于n∈N,有an≠0,当d≠0时容易得到以下几个恒等式:1a1a2=1daa21-aa21=1d(a11-a12),1a1a2a3=21daa31a-2aa13.=21d(a11a2-a21a3)=21d[1d(a11-a12)-1d(a12-a13)]=21d2(a11-a22 a13).1a1a2a3a4=31daa1a4-2aa31a4=31d(a1a12a3-a2a13a4)=31d[21d2(a11-a22 a13)-21d2(a12-a23 a14)]=61d3(a11-a32 a33-a14).为了除去d≠0的限制,我们作出如下变形:1a1-a12=a1da2,1a1-a22 a13=a12ad22a3,1a1-a32 a33-a14=a1a62da33a4.显然d=0时,以上三式也是恒成立的,注意到系数与组合数之间的关系,因此以上三式可改写为:C10a1 (-a…  相似文献   

6.
张靖 《数学通讯》2003,(5):43-48
选择题(每小题6分,共60分) 1.设数列{an}是公比为2的等比数列,且a1·a2……a30=230,则a3·a6……a30等于 ( ) (A)210. (B)215. (C)216. (D)220. 解 令S1=a1·a4·a7……a28, S2=a2·a5·a8……a29, S3=a3·a6·a9……a30,  相似文献   

7.
解题新发现     
奥数课上,老师给我们出了这样一道题:证明:形如a4 4的数(a为任意整数,a≠±1)是一个合数.此题的证明是个因式分解问题.证明a4 4=a4 4a2 4-4a2 =(a2 2)2-4a2 =(a2 2a 2)(a2-2a 2).  相似文献   

8.
文[1]中给出了如下两个不等式及证明:1.设a1,a2,…,am均为正数,且a1 +a2+…+am=ms0,则(a1+1+a1)a+(a2+1/a2)a+…+(am+1/am)a≥m (s0+1/s0)a (m,a∈N*,m≥2)① 2.设a1,a2,…,am均为正数,且a1+a2+…+ am=ms0,若s0≤s≤1或1≤s≤s0,则(a1+1/a1)a+(a2+1/a2)a+…+(am+1/am)a≥m(s+1/s)a(m,a∈N*,m≥2) ②笔者认为当a是大于等于1的实数时,上述不等式也是成立的.  相似文献   

9.
设数列{an}是公差为d的等差数列,且对于n∈N^*,有an≠0,当d≠0时容易得到以下几个恒等式: 1/a1a2=1/d a2-a1/a1a2=1/d(1/a1-1/a2), 1/a1a2a3=1/2d a3-a1/a1a2a3。  相似文献   

10.
(a~(1/~a))2和a2~(1/~a2)是两个重要的根式,由于它们形相似,极易混淆.下面简析一下它们的异同. 一、区别 1. 写法不同(a~(1/a))2有括号,a2~(1/a2)没有括号. 2.读法不同(a~(1/a))2读作a的算术平方根的平方,a2~(1/a2)读作a的平方的算术平方根. 3.意义不同(a~(1/a))2表示非负数a的算术平方根的平方,a2~(1/a2)表示实数a的平方的算术平方根.  相似文献   

11.
在解析几何中有二次曲线与直线位置关系的讨论、二次曲面与直线位置关系的讨论,而二次曲面与平面相关位置关系的探讨较少.本文给出二次曲面a11x2+a22y2+a33z2+2a12xy+2a13xz+2a23yz+2a14x+2a24y+2a34z+a44=0(1)和平面Ax+By+Cz+D=0(2)的相对位置的判别式Δ=a11a12a13a14Aa21a22a23a24Ba31a32a33a34Ca41a42a43a44DA B C D0(aij=aji).(3)并证明了:若Δ>0,则二次曲面(1)与平面(2)相交;若Δ=0,则(1)和(2)相切;若Δ<0,则(1)和(2)相离.  相似文献   

12.
The main result is that a separable Banach space with the weak* unconditional tree property is isomorphic to a subspace as well as a quotient of a Banach space with a shrinking unconditional basis. A consequence of this is that a Banach space is isomorphic to a subspace of a space with a shrinking unconditional basis if and only if it is isomorphic to a quotient of a space with a shrinking unconditional basis, which solves a problem dating to the 1970s. The proof of the main result also yields that a uniformly convex space with the unconditional tree property is isomorphic to a subspace as well as a quotient of a uniformly convex space with an unconditional finite dimensional decomposition.  相似文献   

13.
Scalarization of Henig Proper Efficient Points in a Normed Space   总被引:1,自引:0,他引:1  
In a general normed space equipped with the order induced by a closed convex cone with a base, using a family of continuous monotone Minkowski functionals and a family of continuous norms, we obtain scalar characterizations of Henig proper efficient points of a general set and a bounded set, respectively. Moreover, we give a scalar characterization of a superefficient point of a set in a normed space equipped with the order induced by a closed convex cone with a bounded base.  相似文献   

14.
给出了最佳参数α_1,α_2,α_3,β_1,β_2,β_3∈R,使得双向不等式α_1Q(a,b)+(1-α_1)G(a,b)0且a≠b成立.其中A(a,b)=(a+b)/2,H(a,b)=2ab/(a+b),G(a,b)=(ab)~(1/2),Q(a,b)=((a~2+b~2)/2)~(1/2),C(a,b)=(a~2+b~2)/(a+b),T(a,b)=2/π∫_0~(π/2)(a~2cos~2t+b~2sin~2)~(1/2)tdt分别是两个正数a和b的算术平均,调和平均,几何平均,二次平均,反调和平均和Toader平均.  相似文献   

15.
We study a decomposition of a diffusion process in a manifold that is invariant under the action of a Lie group. As applications, we consider a diffusion process in a Euclidean space that is invariant under translations in a subspace, and a skew-product in a general setting.  相似文献   

16.
We define new parameters, a zero interval and a dual zero interval, of subsets in P- or Q-polynomial association schemes. A zero interval of a subset in a P-polynomial association scheme is a successive interval index for which the inner distribution vanishes, and a dual zero interval of a subset in a Q-polynomial association scheme is a successive interval index for which the dual inner distribution vanishes. We derive bounds of the lengths of a zero interval and a dual zero interval using the degree and dual degree respectively, and show that a subset in a P-polynomial association scheme (resp. a Q-polynomial association scheme) having a large length of a zero interval (resp. a dual zero interval) induces a completely regular code (resp. a Q-polynomial association scheme). Moreover, we consider the spherical analogue of a dual zero interval.  相似文献   

17.
James G. Oxley 《Combinatorica》1984,4(2-3):187-195
Seymour has shown that a matroid has a triad, that is, a 3-element set which is the intersection of a circuit and a cocircuit, if and only if it is non-binary. In this paper we determine precisely when a matroidM has a quad, a 4-element set which is the intersection of a circuit and a cocircuit. We also show that this will occur ifM has a circuit and a cocircuit meeting in more than four elements. In addition, we prove that if a 3-connected matroid has a quad, then every pair of elements is in a quad. The corresponding result for triads was proved by Seymour.  相似文献   

18.
Very recently, Takahashi and Takahashi [S. Takahashi, W. Takahashi, Strong convergence theorem for a generalized equilibrium problem and a nonexpansive mapping in a Hilbert space, Nonlinear Anal. 69 (2008) 1025–1033] suggested and analyzed an iterative method for finding a common solution of a generalized equilibrium problem and a fixed point problem of a nonexpansive mapping in a Hilbert space. In this paper, based on Takahashi–Takahashi’s iterative method and well-known extragradient method we introduce a relaxed extragradient-like method for finding a common solution of a generalized mixed equilibrium problem, a general system of generalized equilibria and a fixed point problem of a strictly pseudocontractive mapping in a Hilbert space and then obtain a strong convergence theorem. Utilizing this theorem, we establish some new strong convergence results in fixed point problems, variational inequalities, mixed equilibrium problems and systems of generalized equilibria.  相似文献   

19.
It was proved in [4] that every group ring of a torsion abelian group over a commutative local ring is a semi-clean ring. It was asked in [4] whether every group ring of a torsion abelian group over a commutative clean ring is a semi-clean ring and whether every group ring of a torsion abelian group over a commutative semi-clean ring is a semi-clean ring. In this paper, we give a positive answer to question 1 and a negative answer to question 2.  相似文献   

20.
图的分数k-因子   总被引:13,自引:0,他引:13  
给定图G=(V,E).设a和b是两个非负整数.fE→[0,1]是一个函数.如果  相似文献   

设为首页 | 免责声明 | 关于勤云 | 加入收藏

Copyright©北京勤云科技发展有限公司  京ICP备09084417号