Question Number 40876 by prof Abdo imad last updated on 28/Jul/18
$${prove}\:{by}\:{recurrence}\:{that}\: \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{4}} =\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{3}{n}−\mathrm{1}\right)}{\mathrm{30}} \\ $$