In this paper, by using superposition method, we aim to show that ∑^n i=1 （2/- 1）^2k-1 is the product of n2 and a rational polynomial in n2 with degree k- 1, and that ∑^ni=1 （2i - 1）^2k is the product of n（2n ...In this paper, by using superposition method, we aim to show that ∑^n i=1 （2/- 1）^2k-1 is the product of n2 and a rational polynomial in n2 with degree k- 1, and that ∑^ni=1 （2i - 1）^2k is the product of n（2n - 1）（2n ＋ 1） and a rational polynomial in （2n - 1）（2n ＋ 1） with degree k - 1. Moreover, recurrence formulas to compute the coefficients of the corresponding rational polynomials are also obtained.展开更多
This paper presents the way to make expansion for the next form function: to the numerical series. The most widely used methods to solve this problem are Newtons Binomial Theorem and Fundamental Theorem of Calculus (t...This paper presents the way to make expansion for the next form function: to the numerical series. The most widely used methods to solve this problem are Newtons Binomial Theorem and Fundamental Theorem of Calculus (that is, derivative and integral are inverse operators). The paper provides the other kind of solution, except above described theorems.展开更多
基金Supported by the Natural Science Foundation of Hainan Province (Grant No.111004)
文摘In this paper, by using superposition method, we aim to show that ∑^n i=1 （2/- 1）^2k-1 is the product of n2 and a rational polynomial in n2 with degree k- 1, and that ∑^ni=1 （2i - 1）^2k is the product of n（2n - 1）（2n ＋ 1） and a rational polynomial in （2n - 1）（2n ＋ 1） with degree k - 1. Moreover, recurrence formulas to compute the coefficients of the corresponding rational polynomials are also obtained.
文摘This paper presents the way to make expansion for the next form function: to the numerical series. The most widely used methods to solve this problem are Newtons Binomial Theorem and Fundamental Theorem of Calculus (that is, derivative and integral are inverse operators). The paper provides the other kind of solution, except above described theorems.
