Question Number 2157 by Yozzi last updated on 05/Nov/15
$${Evaluate}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{1}!}+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{2}!}+\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$${by}\:{considering}\:{the}\:{series}\:{expansion} \\ $$$${of}\:{an}\:{expression}\:{of}\:{the}\:{form}\:{P}\left({x}\right){e}^{{x}} \\ $$$${where}\:{P}\left({x}\right)\:{is}\:{a}\:{suitably}\:{chosen} \\ $$$${polynomial}\:{in}\:{x}.\: \\ $$$$ \\ $$$$ \\ $$
Commented by RasheedAhmad last updated on 15/Nov/15
$$\mathrm{1}+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{1}!}+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{2}!}+\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$=\frac{\mathrm{1}^{\mathrm{3}} }{\mathrm{0}!}+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{1}!}+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{2}!}+\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$=\frac{\mathrm{1}^{\mathrm{3}} .\mathrm{1}}{\mathrm{0}!.\mathrm{1}}+\frac{\mathrm{2}^{\mathrm{3}} .\mathrm{2}}{\mathrm{1}!.\mathrm{2}}+\frac{\mathrm{3}^{\mathrm{3}} .\mathrm{3}}{\mathrm{2}!.\mathrm{3}}+\frac{\mathrm{4}^{\mathrm{3}} .\mathrm{4}}{\mathrm{3}!.\mathrm{4}}+… \\ $$$$=\frac{\mathrm{1}^{\mathrm{4}} }{\mathrm{1}!}+\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{2}!}+\frac{\mathrm{3}^{\mathrm{4}} }{\mathrm{3}!}+\frac{\mathrm{4}^{\mathrm{4}} }{\mathrm{4}!}+…+\frac{{n}^{\mathrm{4}} }{{n}!}+… \\ $$$$ \\ $$