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Question Number 62462 by naka3546 last updated on 21/Jun/19

Answered by MJS last updated on 21/Jun/19

$$\mathrm{let}$$$$a=\alpha -\sqrt{\beta }$$$$b=\alpha +\sqrt{\beta }$$$$c=\gamma -\sqrt{\delta }$$$$d=\gamma +\sqrt{\delta }$$$$\Rightarrow$$$$a=1-\mathrm{i}$$$$b=1+\mathrm{i}$$$$c=-1$$$$d=2$$$$(\mathrm{we}\mathrm{can}\mathrm{interchange}a,b,c,d)$$$$\Rightarrow$$$$\sqrt[n+1]{\frac{a^{n+1}+b^{n+1}+c^{n+1}+d^{n+1}}{\sqrt[3]{a^n+b^n+c^n+d^n}}}=$$$$={\left(\frac{2^{\frac{n}{2}+1}(2^{\frac{n}{2}}-1)-1}{{(2^n+1)}^{\frac{1}{3}}}\right)}^{\frac{1}{n+1}}=f\left(n\right)$$$$\underset{n\to \mathrm{\infty }}{\lim }f\left(n\right)=2^{\frac{2}{3}}$$$$\Rightarrow f\left(2018\right)=2^{\frac{2}{3}}+ϵ$$$$\mathrm{if}\mathrm{you}\mathrm{have}\mathrm{an}\mathrm{exact}\mathrm{value},\mathrm{post}\mathrm{it}$$

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