1-2-3-4-n-n-n-1-2-1-2-2-2-3-2-n-2-n-n-1-2n-1-6-1-x-2-x-3-x-n-x-Formula-x-R-

Question Number 9232 by geovane10math last updated on 24/Nov/16

$$\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+…+\:{n}\:=\:\frac{{n}\left({n}\:+\:\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \:+\mathrm{3}^{\mathrm{2}} \:+…+\:{n}^{\mathrm{2}} \:=\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\mathrm{1}^{{x}} \:+\:\mathrm{2}^{{x}} \:+\:\mathrm{3}^{{x}} \:+…+\:=\:{n}^{{x}} \:=\:\mathrm{Formula}? \\ $$$${x}\:\in\:\mathbb{R} \\ $$

Commented by sou1618 last updated on 25/Nov/16