The value of $(0.2{)}^{{\log }_{\sqrt{5}}\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots \mathrm{\infty }\right)}$ is 

We have,$\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots \mathrm{\infty }=\frac{\frac{1}{4}}{1-\frac{1}{2}}=\frac{1}{2}=2^{-1}$

$\therefore (0.2{)}^{{\log }_{\sqrt{5}}\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots \mathrm{\infty }\right)}={\left(\frac{1}{5}\right)}^{{\log }_{\sqrt{5}}{2}^{-1}}$

$\begin{array}{l}={\left(5^{-1}\right)}^{\frac{-1}{1/2}{\log }_{\mathrm{s}}2}\\ =5^{2{\log }_{\mathrm{s}}2}=5^{{\log }_5(2{)}^2}=2^2=4\end{array}$