等比数列前n项和Sn的一个性质的两个补充

对命题1，可以利用等比数列的性质和整体代换来判定真假．     

等差数列中Sm，Sn，Sm＋n之间的关系

在等差数列 {an}中 ,d为公差 ,Sn 为前n项和 ,则Sm,Sn,Sm +n有下列性质 .性质 1　在等差数列 {an}中 ,Sm +n=Sm+Sn+mnd(m ,n∈N ) .证明　Sm+n=a1+a2 +… +am+am +1+am +2 +… +am+n=Sm+ (a1+md) + (a2 +md) +… + (an+md) =Sm+Sn+mnd .性质 2　 Sm +nm +n=Sm-Snm -n (m ,n∈N ,且m≠n) .证明　∵Sm-Sn=ma1+ m(m - 1)d2 -na1-n(n - 1)d2=(m -n)a1+ (m +n - 1)d2=m -nm +n (m +n)a1+ (m +n) (m +n - 1)d2=m -nm +nSm +n,∴ Sm +nm +n=Sm-Snm -n .性质 3　若Sm =Sn,则Sm +n=0 (m ,n∈N ,且m≠n) .证明　由性质 2知 ,Sm +nm +n=Sm-…     

关于函数项级数∑(-1)n(x+n)n/nn+1在[0,1]上的一致收敛性判定

讨论了函数项级数∑(-1)n 1在[0,1]上的一致收敛性判定的两个方法,同时对[1]中的判别方法作了一些补充.     

分形迭代ak+1=ak+a2k/n第n项的估计

研究了分形迭代ak+1=ak+a2k/n第n项an的估计.利用迭代的等价形式给出了an的一个估计1-1/n＜an＜1.进一步地,将an展开成1/n的幂级数形式后改进了上述估计,最后an被控制在一个长度为4/n2的开区间内.     

m×n－1矩形的覆盖问题

ｍ×ｎ－１矩形的覆盖问题华中师大数学系徐高诚第３３届ＩＭＯ中国队选拔考试第５题为：给定（３ｎ＋１）×（３ｎ＋１）的方格纸，任剪一个方格后，余下的必可全部剪成形如的纸片．本文试将其推广到一般的ｍ×ｎ矩形的情况为方便起见，令为Ｌ形卡片，剪去一格简称为空格...     

分形迭代a_(k+1)=a_k+a_k~2/n第n项的估计

研究了分形迭代ak+1=ak+a2k/n第n项an的估计.利用迭代的等价形式给出了an的一个估计1-1n相似文献   

The [n 1,n 2, ...,n s]-th reduced KP hierarchy andW 1+∞ constraints

To every partition n=n₁+n₂+...+n_s one can associate a vertex operator realization of the Lie algebras a and gl_n. Using this construction, we obtain reductions of the s-component KP hierarchy, reductions which are related to these partitions. In this way we obtain matrix KdV-type equations. We show that the following two constraints on a KP -function are equivalent: (1) is a -function of the [n₁, n₂, ..., n_s]-th reduced KP hierarchy which satisfies the string equation L_–1=0; (2) satisfies the vacuum constraints of the W₁₊ algebra.Translated from Teoreticheskaya i Matematicheskaya Fizika, Vol. 104, No. 1, pp. 32–42, July, 1995.     

从(n 1)/n×(n 1)=(n 1)/n (n 1)谈起

一问题的提出本刊2003年第5期刊载了《运用发现法解题》(以下简称《解题》)一文,文章在谈到“归纳发现法”时,提到这样一个例子: 1.观察下列各式     

O n ( m n,n, m n,m) relative differe nce sets with $$\text {gcd}( m,n)=1$$gcd( m,n)=1

There has been much research on \((p^{a},p^{b},p^{a},p^{a-b})\) relative difference sets with p a prime, while there are only a few results on (mn, n, mn, m) relative difference sets with \(\text {gcd}(m,n)=1\). The non-existence results on (mn, n, mn, m) relative difference sets with \(\text {gcd}(m,n)=1\) have only been obtained for the following five cases: (1) \(m=p,\ n=q,\ p>q\); (2) \(m=pq,\ n=3,\ p,q>3\); (3) \(m=4,\ n=p\); (4) \(m=2\) and (5) \(n=p\), where p, q are distinct odd primes. For the existence results, there are only four constructions of semi-regular relative difference sets in groups of size not a prime power with the forbidden subgroup having size larger than 2. In this paper, we present some more non-existence results on (mn, n, mn, m) relative difference sets with \(\text {gcd}(m,n)=1\). In particular, our result is a generalization of the main result of Hiramine’s work (J Comb Theory Ser A 117(7):996–1003, 2010). Meanwhile, we give a construction of non-abelian (16q, q, 16q, 16) relative difference sets, where q is a prime power with \(q\equiv 1\pmod {4}\) and \(q>4.2\times 10^{8}\). This is the third known infinite classes of non-abelian semi-regular relative difference sets.     

带有中止检查规则（n－i）的CSP－1方案

本文引人转移概率母函数及幂级数等方法讨论带有中止规则（ｎ＊－ｉ）的连续抽样方案ＣＳＰ－１的中止概率Ｐ＊得到了Ｐ＊作为ＣＳＰ－１方案的度量的理论特性，证明了Ｄｏｄｇｅ型ＣＳＰ－１方案与带有中止规则（ｎ＊－ｉ）的ＣＳＰ－１方案具有相同的ＡＯＱ，ＡＦＩ，ＯＣ函数．文末还给出了一些数值结果．     

a_n={S_n-S_(n-1),n≥2,S_1,n=1的应用

设数列 {ａｎ}的前ｎ项和为Ｓｎ,则ａｎ 与Ｓｎ 间有 :ａｎ=Ｓｎ-Ｓｎ - 1 ,Ｓ1 ,ｎ≥ 2 ,ｎ=1.它是ａｎ 与Ｓｎ 之间相互转化的重要工具 ,应用它能较简洁地解出许多条件中含有Ｓｎ 的问题 .下面举例说明它的应用 .1　消去Ｓｎ,转化为ａｎ 来解例 1　设 {ａｎ}是正数组成的数列 ,其前ｎ项和为Ｓｎ.并且对所有自然数ｎ,ａｎ 与 2的等差中项等于Ｓｎ 与2的等比中项 .求数列 {ａｎ}的通项公式 .解　由题意有ａｎ 22 =2Ｓｎ (ｎ∈Ｎ ) ,整理得Ｓｎ=18(ａｎ 2 ) 2 ,由此得Ｓｎ 1 =18(ａｎ 1 2 ) 2 ,∴ａｎ 1=Ｓｎ 1 -Ｓｎ=18[(ａｎ 1 2 ) …     

欧拉等式n=1∑∞1/n^=π^2/6的一个证法

基于函数项级数一致收敛性概念，给出了欧拉等式（Basel问题）的一个证明．     

唇齿相依的an与Sn

从所周知,如何由数列|an|的递推关系式求数列的通项公式an,如何由数列的通项公式an求数列的前n项和公式Sn是<数列>这一章中我们要解决的两大基本问题,当已知数列的前n项和公式Sn时,则通过an={S1=(n=1),Sn-Sn-1(n≥2),很容易求得数列通项公式an,an与Sn可谓"唇齿相依";在高考与竞赛中还时常出现由an与Sn的关系式,求数列的通项公式的问题,本文通过一个典型题目的多种解法,介绍解决这类问题的几何常用策略,供大家参考.……     

Global Behavior of Solutions of x n + 1 = max {x n , A}/x n x n – 1

We investigate the asymptotic behavior, the oscillatory character, and theperiodic nature of solutions of the difference equation

On permutations of $$\{1,\ldots,n\}$$ { 1 , … , n } and related topics

Journal of Algebraic Combinatorics - In this paper, we study combinatorial aspects of permutations of $$\{1,\ldots ,n\}$$ and related topics. In particular, we prove that there is a unique...