Question Number 62462 by naka3546 last updated on 21/Jun/19
Answered by MJS last updated on 21/Jun/19
$$\mathrm{let} \\ $$$${a}=\alpha−\sqrt{\beta} \\ $$$${b}=\alpha+\sqrt{\beta} \\ $$$${c}=\gamma−\sqrt{\delta} \\ $$$${d}=\gamma+\sqrt{\delta} \\ $$$$\Rightarrow \\ $$$${a}=\mathrm{1}−\mathrm{i} \\ $$$${b}=\mathrm{1}+\mathrm{i} \\ $$$${c}=−\mathrm{1} \\ $$$${d}=\mathrm{2} \\ $$$$\left(\mathrm{we}\:\mathrm{can}\:\mathrm{interchange}\:{a},\:{b},\:{c},\:{d}\right) \\ $$$$\Rightarrow \\ $$$$\sqrt[{{n}+\mathrm{1}}]{\frac{{a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} +{c}^{{n}+\mathrm{1}} +{d}^{{n}+\mathrm{1}} }{\sqrt[{\mathrm{3}}]{{a}^{{n}} +{b}^{{n}} +{c}^{{n}} +{d}^{{n}} }}}= \\ $$$$=\left(\frac{\mathrm{2}^{\frac{{n}}{\mathrm{2}}+\mathrm{1}} \left(\mathrm{2}^{\frac{{n}}{\mathrm{2}}} −\mathrm{1}\right)−\mathrm{1}}{\left(\mathrm{2}^{{n}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }\right)^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} ={f}\left({n}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left({n}\right)=\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\Rightarrow\:{f}\left(\mathrm{2018}\right)=\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} +\epsilon \\ $$$$\mathrm{if}\:\mathrm{you}\:\mathrm{have}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{value},\:\mathrm{post}\:\mathrm{it} \\ $$