Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 125526 by ka7th last updated on 11/Dec/20

Answered by Dwaipayan Shikari last updated on 11/Dec/20

$$\underset{n=1}{\sum ^n}{n}^m=\frac{1}{m+1}(n^{m+1}\left(\begin{array}{c}m+1\\ 0\end{array}\right){\beta }_0+n^m\left(\begin{array}{c}m+1\\ 1\end{array}\right){\beta }_1+n^{m-1}\left(\begin{array}{c}m+1\\ 2\end{array}\right){\beta }_2+....)$$$${\beta }_n=Bernoullinumbers$$

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com