伯努利数

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

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转载自www.cnblogs.com/Merodach/p/7450755.html