Question Number 152555 by Gbenga last updated on 29/Aug/21
$$\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\boldsymbol{\sum}}{k}}^{\mathrm{5}} =? \\ $$
Answered by EDWIN88 last updated on 29/Aug/21
$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}+\mathrm{1}\right)^{\mathrm{5}} −{k}^{\mathrm{5}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{5}{k}^{\mathrm{4}} +\mathrm{10}{k}^{\mathrm{3}} +\mathrm{10}{k}^{\mathrm{2}} +\mathrm{5}{k}+\mathrm{1}\right) \\ $$$$\left({n}+\mathrm{1}\right)^{\mathrm{5}} −\mathrm{1}=\mathrm{5}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{4}} +\mathrm{10}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} +\mathrm{10}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} +\mathrm{5}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}+{n} \\ $$
Commented by Gbenga last updated on 29/Aug/21
$${ok}\:{thanks} \\ $$
Answered by puissant last updated on 29/Aug/21
$${S}_{\mathrm{5}} =\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{12}}.. \\ $$
Commented by Tawa11 last updated on 02/Sep/21
$$\mathrm{Good} \\ $$