综合题新编选登

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     

与二项式系数有关的求和问题的解题策略

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)…     

一个不等式的证明与推广

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…     

等比数列的一个判定条件

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…     

等差数列的一个性质

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…     

2002年全国高中数学联赛山东赛区预赛题解

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,     

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).     

一道不等式的推广及证明

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.     

简析a~(1/2)与(a~2)~（1/2）的异同

(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.     

一道北大自主招生试题的另证

这是2008年北京大学自主招生数学试题第三题: 　　题目已知a1,a2,a3,b1,b2,b3,是非负数,且满足a1+a2+a3=b1+b2+b3,a1a2+a2·a1a3=b1b2+b2b3+b1b3,若min{a1,a2,a3}≤min{b1,b2,b3}. 　　求证:max{a1,a2,a3}≤max{b1,b2,b3}.……     

二次曲面和平面位置关系的判式

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.     

Subspaces and quotients of Banach spaces with shrinking unconditional bases

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.     

Scalarization of Henig Proper Efficient Points in a Normed Space

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.     

Two questions on semi-clean group rings

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.     

A generalized duality and applications

The aim of this paper is to present a nonconvex duality with a zero gap and its connection with convex duality. Since a convex program can be regarded as a particular case of convex maximization over a convex set, a nonconvex duality can be regarded as a generalization of convex duality. The generalized duality can be obtained on the basis of convex duality and minimax theorems. The duality with a zero gap can be extended to a more general nonconvex problems such as a quasiconvex maximization over a general nonconvex set or a general minimization over the complement of a convex set. Several applications are given.On leave from the Institute of Mathematics, Hanoi, Vietnam.     

图的分数k-因子

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     

Inventory replenishment model: lot sizing versus just-in-time delivery

Motivated by a practical industrial problem where a manufacturer stipulates a minimum order from each buyer but where a local dealer promises the buyer a just-in-time delivery with a slightly higher unit cost, this paper uses a dynamic lot-sizing model with a stepwise cargo cost function and a minimum order amount constraint to help the buyer select the supplier with minimum total cost.     

A recognition algorithm for the intersection graphs of directed paths in directed trees

Consider a finite family of non-empty sets. The intersection graph of this family is obtained by representing each set by a vertex, two vertices being connected by an edge if and only if the corresponding sets intersect. The intersection graph of a family of directed paths in a directed tree is called a directed path graph. In this paper we present an efficient algorithm which constructs to a given graph a representation by a family of directed paths on a directed tree, if one exists. Also, we prove that a graph is a proper directed path graph if and only if it is a directed path graph.     

Topological representation of posets

There is a canonical imbedding of a poset into a complete Boolean lattice and hence into a Boolean lattice. This gives it a representation as a collection of clopen sets of a Boolean space. There are reflective functions from a category of distributive posets to the subcategories of distributive and Boolean lattices and consequently a topological dual equivalence that extends the Stone duality of Boolean lattices.Presented by B. Jonsson.     

A splitting theorem for groups acting on quasi-trees

It is well known that a group is free if and only if it acts freely without inversions on a tree. We prove a generalisation of this fact by defining a quasi-tree to be a graph with a bound on the size of its simple loops. It is shown that a finitely generated group acting freely on such a graph is isomorphic to a free product of free groups and finite groups.