$1+\frac{2}{4}+\frac{2.5}{4.8}+\frac{\mathrm{2.5.8}}{4.8\cdot 12}+\frac{\mathrm{2.5.8.11}}{4\cdot 8\cdot 12.16}+\dots \text{ is equal to }$

$$\begin{gathered} \text{ Let }\mathrm{S}=1+\frac{2}{4}+\frac{2.4}{4.8}+\frac{2.5.8}{4.8.12}+.. \\ \mathrm{On} \mathrm{comparing} \mathrm{with} (1+\mathrm{x}{)}^{\mathrm{n}}=1+\mathrm{nx}+\frac{\mathrm{n}(\mathrm{n}-1)}{2!}{\mathrm{x}}^2+. \\ \mathrm{We} \mathrm{get} \mathrm{nx}=\frac{2}{4}\to (1) \\ \text{ And }\frac{\mathrm{n}(\mathrm{n}-1)}{2!}{\mathrm{x}}^2=\frac{2.5}{4.8}\to (2) \\ \mathrm{From} \mathrm{Eqs}.(\mathrm{i}) \mathrm{and} (\mathrm{ii}) \mathrm{we} \mathrm{get} \\ \begin{array}{l}\frac{\frac{\mathrm{n}(\mathrm{n}-1)}{2!}}{{\mathrm{n}}^2{\mathrm{x}}^2}=\frac{\frac{2.5}{4.8}}{\frac{2}{4}\cdot \frac{2}{4}}\\ \Rightarrow \frac{\mathrm{n}-1}{\mathrm{n}}=\frac{5}{2}\\ \Rightarrow 2\mathrm{n}-2=5\mathrm{n}\\ \Rightarrow \mathrm{n}=-\frac{2}{3}\\ \text{ On putting the value of }\mathrm{n}\text{ in (1) }\mathrm{x}=\frac{-3}{4}\\ \mathrm{s}=(1+\mathrm{x}{)}^{\mathrm{n}}={\left(1-\frac{3}{4}\right)}^{\frac{-2}{3}}=\sqrt[3]{16}\end{array} \end{gathered}$$