伯努利数

[S_p(n)=Sigma_{k=0}^nk^p\ G(z,n)=Sigma_{p=0}^{infty}Sp(n)/p!z^p=Sigma_{k=0}^nSigma_{p=0}^{infty}(kz)^p/p!\ G(z,n)=Sigma_{k=0}^ne^{kz}=(e^{(n+1)z}-1)/(e^z-1)\ z/(e^z-1)=Sigma_{k=0}^{infty}B_k/k!*z^k\ (e^{(n+1)z}-1)/z=(n+1)Sigma_{k=0}^{infty}((n+1)z)^k/(k+1)/k!\ G(z,n)=Sigma_{k=0}^{infty}z^k/k!(n+1)Sigma_{j=0}^{k}C_k^jBj*(n+1)^{k-j}/(k-j+1)\ Sp(n) = (n+1)Sigma_{j=0}^pC_p^jB_j*(n+1)^{p-j}/(p-j+1)\ B_0=1, B_1=-1/2, B_2=1/6, B_3=0, B_4=-1/30\ S_2(n)=(n+1)^2/3-(n+1)/2+1/6=(n+1)^3/3-(n+1)^2/2+(n+1)/6\ S_1(n)=(n+1)^2/2-(n+1)/2=(n+1)*n/2 ]

原文地址:https://www.cnblogs.com/Merodach/p/7450755.html