Infinite sum

Let $\displaystyle{\sum_{n=0}^{\infty}{\dfrac{4}{5^n}} = a \implies a = \dfrac{4}{5^0}+\dfrac{4}{5^1}+\dfrac{4}{5^2}+\dfrac{4}{5^3}+\cdots \\ \text{So } 5a = 20+\dfrac{4}{5^0}+\dfrac{4}{5^1}+\dfrac{4}{5^2}+\dfrac{4}{5^3}+\cdots \implies 5a = 20+a \implies a = 5 \\ \therefore 25a = 25(5) = \boxed{125}}$ .

Relevant wiki: Geometric Progression Sum

$\begin{aligned} S & = 25 \sum_{n=0}^\infty \frac 4{5^n} \\ & = 100 {\color{#3D99F6} \sum_{n=0}^\infty \left(\frac 15 \right)^n} & \small \color{#3D99F6} \text{Infinite GP: } \sum_{n=0}^\infty r^n = \frac 1{1-r}, \ |r| < 1 \\ & = 100 {\color{#3D99F6} \left(\frac 1{1-\frac 15}\right)} \\ & = \boxed{125} \end{aligned}$