If [tex]\tilde n = 2n(n+1)[/tex], then
[tex]\tilde 8 = 2\times8\times9 = 144[/tex]
[tex]\tilde 2 = 2\times2\times3 = 12[/tex]
so that
[tex]\dfrac{\tilde8}{\tilde2} = \dfrac{144}{12} = \boxed{12}[/tex]
We are given that [tex]{\bf{\tilde{n}=2n(n+1)\:\: \forall \:\: n\in \mathbb{N}}}[/tex] and need to find the value of [tex]{\bf{\dfrac{\tilde{8}}{\tilde{2}}}}[/tex] . Now , let's find it ;
[tex]{:\implies \quad \sf \dfrac{\tilde{8}}{\tilde{2}}=\dfrac{2\times 8(8+1)}{2\times 2(2+1)}}[/tex]
[tex]{:\implies \quad \sf \dfrac{\tilde{8}}{\tilde{2}}=\dfrac{\cancel{2}\times 8(9)}{\cancel{2}\times 2(3)}}[/tex]
[tex]{:\implies \quad \sf \dfrac{\tilde{8}}{\tilde{2}}=\dfrac{8(9)}{2(3)}}[/tex]
[tex]{:\implies \quad \sf \dfrac{\tilde{8}}{\tilde{2}}=\dfrac{2\times 4\times 3\times 3}{2\times 3}}[/tex]
Cancelling 2 and 3 from denominator
[tex]{:\implies \quad \sf \dfrac{\tilde{8}}{\tilde{2}}=4\times 3}[/tex]
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{\dfrac{\tilde{8}}{\tilde{2}}=12}}}[/tex]
