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Question Number 118180 by mathocean1 last updated on 15/Oct/20
show that if n is odd , n(n^2 +3) is even.
$${show}\:{that}\:{if}\:{n}\:{is}\:{odd}\:,\:{n}\left({n}^{\mathrm{2}} +\mathrm{3}\right)\:{is}\:{even}. \\ $$
Answered by floor(10²Eta[1]) last updated on 15/Oct/20
if n is odd⇒n=2k+1, k∈Z  (2k+1)((2k+1)^2 +3)  =(2k+1)(4k^2 +4k+4)  =4(2k+1)(k^2 +k+1)=even because  it′s a multiple of 2
$$\mathrm{if}\:\mathrm{n}\:\mathrm{is}\:\mathrm{odd}\Rightarrow\mathrm{n}=\mathrm{2k}+\mathrm{1},\:\mathrm{k}\in\mathbb{Z} \\ $$$$\left(\mathrm{2k}+\mathrm{1}\right)\left(\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\right) \\ $$$$=\left(\mathrm{2k}+\mathrm{1}\right)\left(\mathrm{4k}^{\mathrm{2}} +\mathrm{4k}+\mathrm{4}\right) \\ $$$$=\mathrm{4}\left(\mathrm{2k}+\mathrm{1}\right)\left(\mathrm{k}^{\mathrm{2}} +\mathrm{k}+\mathrm{1}\right)=\mathrm{even}\:\mathrm{because} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{2} \\ $$$$ \\ $$
Commented by mathocean1 last updated on 15/Oct/20
thanks
$${thanks} \\ $$

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