Question Number 144482 by ArielVyny last updated on 25/Jun/21
$${show}\:{that}\:\forall{n}\in\mathbb{Z}\: \\ $$$${E}\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)+{E}\left(\frac{{n}+\mathrm{2}}{\mathrm{4}}\right)+{E}\left(\frac{{n}+\mathrm{4}}{\mathrm{4}}\right)={n} \\ $$
Answered by Olaf_Thorendsen last updated on 25/Jun/21
$${x}\:=\:\mathrm{E}\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)+\mathrm{E}\left(\frac{{n}+\mathrm{2}}{\mathrm{4}}\right)+\mathrm{E}\left(\frac{{n}+\mathrm{4}}{\mathrm{4}}\right) \\ $$$$\mathrm{1st}\:\mathrm{case}\::\:{n}\:=\:\mathrm{4}{p} \\ $$$${x}\:=\:\mathrm{E}\left(\frac{\mathrm{4}{p}−\mathrm{1}}{\mathrm{2}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{2}}{\mathrm{4}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{4}}{\mathrm{4}}\right) \\ $$$${x}\:=\:\mathrm{E}\left(\mathrm{2}{p}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{E}\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{E}\left({p}+\mathrm{1}\right) \\ $$$${x}\:=\:\left(\mathrm{2}{p}−\mathrm{1}\right)+\left({p}\right)+\left({p}+\mathrm{1}\right) \\ $$$${x}\:=\:\mathrm{4}{p}\:=\:{n}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{2nd}\:\mathrm{case}\::\:{n}\:=\:\mathrm{4}{p}+\mathrm{1} \\ $$$${x}\:=\:\mathrm{E}\left(\frac{\mathrm{4}{p}}{\mathrm{2}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{3}}{\mathrm{4}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{5}}{\mathrm{4}}\right) \\ $$$${x}\:=\:\mathrm{E}\left(\mathrm{2}{p}\right)+\mathrm{E}\left({p}+\frac{\mathrm{3}}{\mathrm{4}}\right)+\mathrm{E}\left({p}+\frac{\mathrm{5}}{\mathrm{4}}\right) \\ $$$${x}\:=\:\left(\mathrm{2}{p}\right)+\left({p}\right)+\left({p}+\mathrm{1}\right) \\ $$$${x}\:=\:\mathrm{4}{p}+\mathrm{1}\:=\:{n}\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\mathrm{3rd}\:\mathrm{case}\::\:{n}\:=\:\mathrm{4}{p}+\mathrm{2} \\ $$$${x}\:=\:\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{1}}{\mathrm{2}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{4}}{\mathrm{4}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{6}}{\mathrm{4}}\right) \\ $$$${x}\:=\:\mathrm{E}\left(\mathrm{2}{p}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{E}\left({p}+\mathrm{1}\right)+\mathrm{E}\left({p}+\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${x}\:=\:\left(\mathrm{2}{p}\right)+\left({p}+\mathrm{1}\right)+\left({p}+\mathrm{1}\right) \\ $$$${x}\:=\:\mathrm{4}{p}+\mathrm{2}\:=\:{n}\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\mathrm{4th}\:\mathrm{case}\::\:{n}\:=\:\mathrm{4}{p}+\mathrm{3} \\ $$$${x}\:=\:\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{2}}{\mathrm{2}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{5}}{\mathrm{4}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{7}}{\mathrm{4}}\right) \\ $$$${x}\:=\:\mathrm{E}\left(\mathrm{2}{p}+\mathrm{1}\right)+\mathrm{E}\left({p}+\frac{\mathrm{5}}{\mathrm{4}}\right)+\mathrm{E}\left({p}+\frac{\mathrm{7}}{\mathrm{4}}\right) \\ $$$${x}\:=\:\left(\mathrm{2}{p}+\mathrm{1}\right)+\left({p}+\mathrm{1}\right)+\left({p}+\mathrm{1}\right) \\ $$$${x}\:=\:\mathrm{4}{p}+\mathrm{3}\:=\:{n}\:\:\:\:\left(\mathrm{4}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right),\:\left(\mathrm{2}\right),\:\left(\mathrm{3}\right),\:\left(\mathrm{4}\right)\:: \\ $$$$\mathrm{In}\:\mathrm{all}\:\mathrm{cases}\:\mathrm{possible}\:{x}\:=\:{n} \\ $$
Commented by ArielVyny last updated on 25/Jun/21
$${thank}\:{Mr}\:{olaf} \\ $$