求sum from i=1 to n i^k的简便递推公式

A Simple Deductive Formula to Solve sum from i=1 to n(i^k)

寻找求sum from i=1 to n i^k值的方法,研究得不浅[1-9]都有介绍。这里仅用微积分的最基本知识推出较简便的自然数幂之和的求值递推公式:S_n^(k+1)=(k+1)[integral from n=0 to n(S^k(x)dx)-n integral from n=-1 to 0 (S^k(x)ds)。其中S^k(x)是S_n^k=sum from i=1 to i^k的派生函数。

Much has been said alrealy about the ways to seek the value of sum from i=1 to n(i^k). Here we offer a smple deductiveformula to seek the sum of nature and power by using rudimentary differential and integral equation: S_n^(k+1)=(k+1)[integral from n=0 to n(S^k(x)dx)-n(integral from n=-1 to 0(S^k(x)dx)]of which S^k(x) is the derivative function S_n^k=sum from i=1 to n(i^k).