$1+\frac{2}{4}+\frac{2.5}{4.8}+\frac{\mathrm{2.5.8}}{\mathrm{4.8.12}}+\frac{\mathrm{2.5.8.11}}{\mathrm{4.8.12.16}}+\mathrm{...}$ is equal to

Let, $S=1+\frac{2}{4}+\frac{2.5}{4.8}+\frac{\mathrm{2.5.8}}{\mathrm{4.8.12}}+\mathrm{..}$
On comparing with ${\left(1+x\right)}^n=1+nx+\frac{n\left(n-1\right)}{2!}{x}^2+\mathrm{..}$
We get, $nx=\frac{2}{4}\to \left(1\right)$        
And, $\frac{n\left(n-1\right)}{2!}{x}^2=\frac{2.5}{4.8}\to \left(2\right)$          
From Eqs.(i) and (ii) we get
$\frac{\frac{n\left(n-1\right)}{2!}{x}^2}{n^2{x}^2}=\frac{\frac{2.5}{4.8}}{\frac{2}{4}.\frac{2}{4}}$ 
$\Rightarrow \frac{n-1}{n}=\frac{5}{2}$ 
$\Rightarrow 2n-2=5n$ 
$\Rightarrow n=-\frac{2}{3}$ 
On putting the value of n in (1), $x=\frac{-3}{4}$
$s={\left(1+x\right)}^n={\left(1-\frac{3}{4}\right)}^{\frac{-2}{3}}=\sqrt[3]{16}$