Question Number 74923 by aliesam last updated on 04/Dec/19
![Σ_(k=1) ^n k^3 =[((n(n+1) )/2)]^2](https://www.tinkutara.com/question/Q74923.png)
$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} =\left[\frac{{n}\left({n}+\mathrm{1}\right)\:}{\mathrm{2}}\right]^{\mathrm{2}} \\ $$
Commented by aliesam last updated on 04/Dec/19
$${god}\:{bless}\:{you} \\ $$
Commented by mathmax by abdo last updated on 04/Dec/19
$${let}\:{P}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} \:\:\:{we}\:{have} \\ $$$${P}\left({x}+\mathrm{1}\right)−{P}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{2}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\left\{\:\:\left({x}+\mathrm{2}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \right\}\:=\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\left\{\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}−{x}^{\mathrm{2}} \right\} \\ $$$$=\left({x}+\mathrm{1}\right)^{\mathrm{3}} \:\Rightarrow{P}\left({x}\right)−{P}\left({x}−\mathrm{1}\right)\:={x}^{\mathrm{3}} \:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{3}} =\sum_{{k}=\mathrm{1}} ^{{n}} \left({P}\left({k}\right)−{P}\left({k}−\mathrm{1}\right)\right) \\ $$$$={P}\left(\mathrm{1}\right)−{P}\left(\mathrm{0}\right)+{P}\left(\mathrm{2}\right)−{P}\left(\mathrm{1}\right)+….+{P}\left({n}\right)−{P}\left({n}−\mathrm{1}\right) \\ $$$$={P}\left({n}\right)−{P}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{4}}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{0}\:=\frac{\left({n}\left({n}+\mathrm{1}\right)\right)^{\mathrm{2}} }{\mathrm{4}}\:\:{so}\:{the}\:{equaly}\:{is} \\ $$$${proved}. \\ $$
Commented by mathmax by abdo last updated on 04/Dec/19
$${you}\:{are}\:{welcome}. \\ $$