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Question Number 144482 by ArielVyny last updated on 25/Jun/21

show that ∀n∈Z E(((n−1)/2))+E(((n+2)/4))+E(((n+4)/4))=n

$${show}\:{that}\:\forall{n}\in\mathbb{Z}\: \\ $$$${E}\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)+{E}\left(\frac{{n}+\mathrm{2}}{\mathrm{4}}\right)+{E}\left(\frac{{n}+\mathrm{4}}{\mathrm{4}}\right)={n} \\ $$

Answered by Olaf_Thorendsen last updated on 25/Jun/21

x = E(((n−1)/2))+E(((n+2)/4))+E(((n+4)/4)) 1st case : n = 4p x = E(((4p−1)/2))+E(((4p+2)/4))+E(((4p+4)/4)) x = E(2p−(1/2))+E(p+(1/2))+E(p+1) x = (2p−1)+(p)+(p+1) x = 4p = n (1) 2nd case : n = 4p+1 x = E(((4p)/2))+E(((4p+3)/4))+E(((4p+5)/4)) x = E(2p)+E(p+(3/4))+E(p+(5/4)) x = (2p)+(p)+(p+1) x = 4p+1 = n (2) 3rd case : n = 4p+2 x = E(((4p+1)/2))+E(((4p+4)/4))+E(((4p+6)/4)) x = E(2p+(1/2))+E(p+1)+E(p+(3/2)) x = (2p)+(p+1)+(p+1) x = 4p+2 = n (3) 4th case : n = 4p+3 x = E(((4p+2)/2))+E(((4p+5)/4))+E(((4p+7)/4)) x = E(2p+1)+E(p+(5/4))+E(p+(7/4)) x = (2p+1)+(p+1)+(p+1) x = 4p+3 = n (4) (1), (2), (3), (4) : In all cases possible x = n

$${x}\:=\:\mathrm{E}\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)+\mathrm{E}\left(\frac{{n}+\mathrm{2}}{\mathrm{4}}\right)+\mathrm{E}\left(\frac{{n}+\mathrm{4}}{\mathrm{4}}\right) \\ $$$$\mathrm{1st}\:\mathrm{case}\::\:{n}\:=\:\mathrm{4}{p} \\ $$$${x}\:=\:\mathrm{E}\left(\frac{\mathrm{4}{p}−\mathrm{1}}{\mathrm{2}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{2}}{\mathrm{4}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{4}}{\mathrm{4}}\right) \\ $$$${x}\:=\:\mathrm{E}\left(\mathrm{2}{p}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{E}\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{E}\left({p}+\mathrm{1}\right) \\ $$$${x}\:=\:\left(\mathrm{2}{p}−\mathrm{1}\right)+\left({p}\right)+\left({p}+\mathrm{1}\right) \\ $$$${x}\:=\:\mathrm{4}{p}\:=\:{n}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{2nd}\:\mathrm{case}\::\:{n}\:=\:\mathrm{4}{p}+\mathrm{1} \\ $$$${x}\:=\:\mathrm{E}\left(\frac{\mathrm{4}{p}}{\mathrm{2}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{3}}{\mathrm{4}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{5}}{\mathrm{4}}\right) \\ $$$${x}\:=\:\mathrm{E}\left(\mathrm{2}{p}\right)+\mathrm{E}\left({p}+\frac{\mathrm{3}}{\mathrm{4}}\right)+\mathrm{E}\left({p}+\frac{\mathrm{5}}{\mathrm{4}}\right) \\ $$$${x}\:=\:\left(\mathrm{2}{p}\right)+\left({p}\right)+\left({p}+\mathrm{1}\right) \\ $$$${x}\:=\:\mathrm{4}{p}+\mathrm{1}\:=\:{n}\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\mathrm{3rd}\:\mathrm{case}\::\:{n}\:=\:\mathrm{4}{p}+\mathrm{2} \\ $$$${x}\:=\:\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{1}}{\mathrm{2}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{4}}{\mathrm{4}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{6}}{\mathrm{4}}\right) \\ $$$${x}\:=\:\mathrm{E}\left(\mathrm{2}{p}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{E}\left({p}+\mathrm{1}\right)+\mathrm{E}\left({p}+\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${x}\:=\:\left(\mathrm{2}{p}\right)+\left({p}+\mathrm{1}\right)+\left({p}+\mathrm{1}\right) \\ $$$${x}\:=\:\mathrm{4}{p}+\mathrm{2}\:=\:{n}\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\mathrm{4th}\:\mathrm{case}\::\:{n}\:=\:\mathrm{4}{p}+\mathrm{3} \\ $$$${x}\:=\:\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{2}}{\mathrm{2}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{5}}{\mathrm{4}}\right)+\mathrm{E}\left(\frac{\mathrm{4}{p}+\mathrm{7}}{\mathrm{4}}\right) \\ $$$${x}\:=\:\mathrm{E}\left(\mathrm{2}{p}+\mathrm{1}\right)+\mathrm{E}\left({p}+\frac{\mathrm{5}}{\mathrm{4}}\right)+\mathrm{E}\left({p}+\frac{\mathrm{7}}{\mathrm{4}}\right) \\ $$$${x}\:=\:\left(\mathrm{2}{p}+\mathrm{1}\right)+\left({p}+\mathrm{1}\right)+\left({p}+\mathrm{1}\right) \\ $$$${x}\:=\:\mathrm{4}{p}+\mathrm{3}\:=\:{n}\:\:\:\:\left(\mathrm{4}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right),\:\left(\mathrm{2}\right),\:\left(\mathrm{3}\right),\:\left(\mathrm{4}\right)\:: \\ $$$$\mathrm{In}\:\mathrm{all}\:\mathrm{cases}\:\mathrm{possible}\:{x}\:=\:{n} \\ $$

Commented by ArielVyny last updated on 25/Jun/21

thank Mr olaf

$${thank}\:{Mr}\:{olaf} \\ $$

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