Question Number 125526 by ka7th last updated on 11/Dec/20
Answered by Dwaipayan Shikari last updated on 11/Dec/20
$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{n}^{{m}} =\frac{\mathrm{1}}{{m}+\mathrm{1}}\left({n}^{{m}+\mathrm{1}} \begin{pmatrix}{{m}+\mathrm{1}}\\{\mathrm{0}}\end{pmatrix}\beta_{\mathrm{0}} +{n}^{{m}} \begin{pmatrix}{{m}+\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}\beta_{\mathrm{1}} +{n}^{{m}−\mathrm{1}} \begin{pmatrix}{{m}+\mathrm{1}}\\{\mathrm{2}}\end{pmatrix}\beta_{\mathrm{2}} +....\right) \\ $$$$\beta_{{n}} ={Bernoulli}\:{numbers} \\ $$
Commented by ka7th last updated on 11/Dec/20
this sis true for k=1, what about other k, I'm looking at a sum of sums so to speak not just sums to the mth power