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近几年来,国内外还不断有人讨论自然数方幂和的求法(见[1]～[4]),本文研究等差数列的方幂和问题,得到一个简单的求和方法.并导出了一个自然数方幂和公式。首先,我们介绍“循环积和”的概念及其有关结果。设{a_n}为等差数列,记 a_ k~(r)=a_k a_(k+1)…a_(k+r-1),k=1,2,3,…我们把数列{a_k~(r)}叫做等差数列{a_n}的r阶循环积,而将和 S_n~(r)=a_1(r)+a_2(r)+…a_n~(r)叫做数列{a_n}的r阶循环积和,简称循环积和。我们有[5] S_n~(r)=1/(r+1)d[a_n(r+1)-a_0~(r+1)] (1)其中d为{a_n}的公差,a_0=a_1=d。特别,令a_n=n,则对自
In recent years, people at home and abroad have continued to discuss the method of sum of squares and exponents of natural numbers (see [1]–[4]). This paper studies the problem of the sum of squares of arithmetic sequences, and obtains a simple summation method. Natural number power sum formula. First, we introduce the concept of “circular product sum” and its related results. Let {a_n} be an arithmetic progression, note that a_k~(r)=a_k a_(k+1)...a_(k+r-1),k=1,2,3,... We set the sequence {a_k~( r)} is called the r-order circulant product of the arithmetic sequence {a_n}, and the sum S_n~(r)=a_1(r)+a_2(r)+...a_n~(r) is called the r-order loop of the sequence {a_n} Product sum, referred to as cyclic product sum. We have [5] S_n~(r)=1/(r+1)d[a_n(r+1)-a_0~(r+1)] (1) where d is the tolerance of {a_n}, a_0=a_1 =d. In particular, let a_n=n, then