正整数的立方部分数列的求和

研究了数列u(n)和v(n)的求和问题,其中u(n)表示不超过n的最大立方部分,v(n)表示不小于n的最小立方部分,给出了关于这两个数列的两个求和公式。     

自然数方幂求和的比较法

自然数方幂求和的方法较多,为了使其方法更为初等化,本短文采用比较法作一尝试,简单介绍为下: 大家知道,若a/b=m(b≠0)则a=mb (1) 这是四则运算中一个基本公式。又自然数列求和公式1+2+3+……+n=1/2n(n+1) (2)是数列的一个最基本的求和计算公式。我们就从这两个基本公式出发,来探求自然数方幂的求和方法。把公式(2)前K(K=1,2,…)项和所组成的数列:     

数列求和的方法

<正>数列求和是数列的重要内容之一,是高考必考内容.除了等差数列和等比数列有求和公式外,大部分数列的求和都需要一定的技巧.下面就谈谈这类问题的解决方法和技巧.一、分组求和法如果数列的通项公式可分为几个等差、等比或常见的数列,这时就要分别求和,然后再相加.譬如数列{cn=an+bn},其中数列{an}、{bn}分别是等差、对比数列,前n项和Sn=(a1+b1)+(a1+b2)+…+(an+bn)=(a1+a2+…+an)+(b1+b2+…+bn).例1推测数列112,214,318,4116,…的前n项和Sn.解Sn=112+214+318+…+n+12()n=(1+2+3+…+n)+     

数列求和的方法

<正>数列求和是数列的重要内容之一,是高考必考内容.除了等差数列和等比数列有求和公式外,大部分数列的求和都需要一定的技巧.下面就谈谈这类问题的解决方法和技巧.一、分组求和法如果数列的通项公式可分为几个等差、等比或常见的数列,这时就要分别求和,然后再相加.譬如数列{cn=an+bn},其中数列{an}、{bn}分别是等差、对比数列,前n项和Sn=(a1+b1)+(a1+b2)+…+(an+bn)=(a1+a2+…+an)+(b1+b2+…+bn).例1推测数列112,214,318,4116,…的前n项和Sn.解Sn=112+214+318+…+n+12()n=(1+2+3+…+n)+     

拆项在数列中的应用

形如an=f(n)×qn(其中f(n)是关于n的多项式)的数列可用错位相减法求和,但f(n)的次数较高时用错位相减法比较麻烦.下面就来探讨拆项在相关数列问题中的应用. 一、拆项在数列求和中的应用 1.可行性分析 如果能找到一个数列{bn},使得an =bn+1-bn,那么数列{an}的前n项和Sn=a1 +a2+…+an=(b2-b1)+(b3-b2)+…+(bn+1-b1)一般地,当an=bn+k-bn或an=bn-bn+k(其中n∈N+,k∈N+,且k为常数)时,都可快速求和.     

关于正整数的立方部分数列

设n是正整数，u(n)表示不超过n的最大立方部分，v(n)表示不小于n的最小立方部分。本的主要目的是研究数列u(n)和v(n)的均值性质，并对罗马尼亚数论专家F.Smarandach教授在献[1仲提出的第41个问题做出实质性进展，利用解析分析方法给出了包含这两个数列及除数函数的几个有趣的渐近公式。     

中学数列求和探微

一、利用公式法求和 若数列的通项公式是an＋b（a，b为常数）的形式，则说明数列是等差数列，可直接用等差数列求和公式Sn=n（a1＋an）/2或Sn=na1＋n（n-1）/2d进行计算。     

数列求和的几种方法

数列这部分内容是重要的高考考点之一,而数列求和又是重中之重.除了等差数列和等比数列有求和公式外,大部分数列的求和都需要一定的技巧.下面,结合几道例题谈一谈高考中数列求和的几种重要方法和技巧,供同学们在学习时参考.一、裂项相消法这种方法是将数列的通项公式分成两个式子的代数和,即a=f(n)+1-f()n,然后累加抵消掉中间的许多项,     

合成数列的通项及其应用

设有两个数列{‘}及{右,}: al一a,一a3,.‘”口”, b:,西,,b3,…,b。,依次交错排列a:,西:(k=1,2,…)构成一个新的数列{x。}: a,,b:,a,,b:,…,a。,乙二,我们称上述数列{x。}为数列{么。}和{乙。}的合成数列。 本文讨论两个数列的合成数列的通项公式及其应用。 定理设数列{a’‘},{乙。}的通项分别为 a。=f(n),b矛==g(n),那么,数列{。、}与数列{阮、}的合成数列{x二}的通项为 解:将。,二f(。)=a,b。=g(:)二吞代入(1)得所求数列的通项为X”二例2合、一“,+合‘一‘,”+“口一的·求数歹l】{x。1:1,1,2,2,3,3,n,”,’..的通项.解:将a,=f(:)=n二…     

求一些特殊数列前n项和的方法

对于等差、等比数列的前n项求和问题,一般只要根据已知条件,灵活应用公式,不难求出.而对一些特殊数列的求和问题,学生时常感到束手无策,无从下手.实际上,我们只要把这些特殊数列的求和稍加巧妙变化,转化为基本类型或熟知的数列求和问题,从而简捷地解答此类问题.现将解决这些特殊数列前n项和的方法归纳如下.1分项求和法所谓“分项求和法”,就是把一个数列分解为几个基本数列后再求和.例1求和S=1·n 2(n-1) 3(n-2) … n·1.分析这是一个数列求和问题,考察其通项k(n-k 1)=k(n 1)-k2,则可将其分解成两个数列的求和问题求解.解S=1·n 2(n-1) 3(n…     

关于类似广义Dedekind和S_2(h,m,n,k)的几个恒等式

利用初等方法研究了类似广义Dedekind和S2(h,m,n,,k)的算术性质.借助Bernoulli多项式及三角恒等式,探究了S2(qh,m,n,qk)与S2(h,m,n,k)的关系,以及当P为奇素数时∑S2(h+bk,m,n,pk)与S2(h,m,n,k)和S2(ph,m,n,k)的关系,提出并证明了两个恒等式,推广了有关文献的结论.     

基于最小二乘法的SOR法松弛因子自动生成算法

基于使每一迭代步线性方程组中各个方程的余量平方和最小的原理,给出了松弛因子的自动生成方法,从而有效地避免了最佳松弛因子的选取带来的不便.并通过数值实验说明此算法切实可行,只要系数矩阵满足严格对角占优或弱对角占优且不可约条件就能够取得很好的收敛效果.     

关于Lucas数任意次幂的恒等变换与同余式

研究了Lucas数的和式a b=n↑∑Vak↑m Vak↑m/a! b!,得出了一些关于Lucas数的恒等变换和一些有趣的同余式。     

Understanding the degrees of freedom of sample variance by using Microsoft Excel

In this article, the degrees of freedom of the sample variance are simulated by using the Visual Basic for Applications of Microsoft Excel 2010. The simulation file dynamically displays why the sample variance should be calculated by dividing the sum of squared deviations by n‐1 rather than n, which is helpful for students to grasp the meaning of degrees of freedom.     

关于Lucas数方幂和的一般形式

设U_(en)和V_(en)是广Lucas数,用发生函数的方法得到方幂和sum from k=1 to n(U～R_(ek)和sum from k=1 to n(U～_(-ek)),以及正负相间方幂和sum from k=1 to n((-1)～kU～r_(ek))和sum from k=1 to n((-1)～kU～r_(-ek))的计算公式.     

"人本主义"价值概念的超越

人本主义价值观把人视为价值的唯一主体。生态学等现代科学揭示出生态系统机能的相关性和整体性特性 ,这种机能整体对维护人类和其他生物的生存及整个生物圈的稳定、完整和美丽都是有价值的。哲学作为时代精神的精华必须对科学提供的新成果作出总结 ,建立一种以生态系统为主体的新的价值体系。这是价值哲学发展的趋势。     

Inequalities for inscribed simplexes

The problem on the geometrc inequalities involving an n-dimensional simplex and its inscribed simplex is studied. Aninequality is established, which reveals that the difference between the squared circumradius of the n-dimensional simplex andthe squared distance between its circumcenter and barycenter times the squared circumradius of its inscribed simplex is not lessthan the 2(n-1)th power of n times its squared inradius, and is equal to when the simplex is regular and its inscribed siplex is atangent point one. Deduction from this inequality reaches a generalization of n-dimensional Euler inequality indicating that thecircumradius of the simplex is not less than the n-fold inradius. Another inequality is derived to present the relationship betweenthe circumradius of the n-dimensional simplex and the circumradius and inradius of its pedal simplex.     

Creating IRT-Based Parallel Test Forms Using the Genetic Algorithm Method

In educational measurement, the construction of parallel test forms is often a combinatorial optimization problem that involves the time-consuming selection of items to construct tests having approximately the same test information functions (TIFs) and constraints. This article proposes a novel method, genetic algorithm (GA), to construct parallel test forms effectively. The sum of squared errors of the generated TIFs produced by GA were compared with those of the Swanson and Stocking method, and the Wang and Ackerman method. Experimental results show that tests constructed using GA yielded lower error, and an average improvement ratio above 90%.     

锦屏一级水电站建库前的浮游动物调查

2002年4、5月和9、10月分两次调查了雅砻江锦屏一级水电站及其影响区的浮游动物多样性概况.结果表明:雅砻江中、下游浮游动物由3门、4纲、11目、23科、28属、46种组成.其中原生动物31种,占总种数的67.39%;轮虫12种,占总种数的26.09%;节肢动物3种,占总种数的6.52%.     

Invariance of Item Characteristic Functions With Variations in Instructional Coverage

An assumption of item response theory is that a person's score is a function of the item response parameters and the person's ability. In this paper, the effect of variations in instructional coverage on item characteristic functions is examined. Using data from the Second International Mathematics Study (1985), curriculum clusters were formed based on teachers' ratings of their students' opportunities to learn the items on a test. After forming curriculum clusters, item response curves were compared using signed and unsigned sum of squared differences. Some of the differences in the item response curves between curriculum clusters were found to be large, but better performance was not necessarily related to greater opportunity to learn. The item response curve differences were much larger than differences reported in prior studies based on comparisons of black and white students. Implications of the findings for applications of item response theory to educational achievement test data are discussed