If $\mathrm{x}=\frac{2.5}{3.6}-\frac{2.5.8}{3.6.9}\left(\frac{2}{5}\right)+\frac{2.5.8.11}{3.6.9.12}{\left(\frac{2}{5}\right)}^2-\dots \dots \infty ,$ then $7^2{\left(12\mathrm{x}+55\right)}^3=$

$\begin{array}{rcl}\mathrm{x}& =& \frac{2.5}{2!}{\left(\frac{1}{3}\right)}^2-\frac{2.5.8}{3!}\left(\frac{2}{5}\right) {\left(\frac{1}{3}\right)}^3+.....\\ {\left(\frac{2}{5}\right)}^2 \mathrm{x}& =& \frac{2.5}{2!} {\left(\frac{2}{15}\right)}^2-\frac{2.5.8}{3!} {\left(\frac{2}{15}\right)}^3+.....\\ & =& {(1+\mathrm{X})}^{\raisebox{1ex}{$-\mathrm{p}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{q}$}\right.}-1+\frac{2}{1!} \left(\frac{2}{15}\right)\end{array}$

Where  $$\begin{gathered} \mathrm{p}=2, \mathrm{q}=3, \frac{\mathrm{X}}{\mathrm{q}}=\frac{2}{15} \\ \mathrm{x}=\frac{2}{5} \end{gathered}$$

$\begin{array}{rcl}\frac{4}{25}\mathrm{x}& =& {\left(1+\frac{2}{5}\right)}^{\frac{-2}{3}}-1+\frac{4}{15}\\ & =& {\left(\frac{7}{5}\right)}^{\frac{-2}{3}}-\frac{11}{15}\\ \frac{4}{25} \mathrm{x}+\frac{11}{15}& =& {\left(\frac{5}{7}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ \frac{12\mathrm{x}+55}{75}& =& {\left(\frac{25}{49}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ 7^2 {(12\mathrm{x}+55)}^3& =& {75}^3\times 25=3^3\times 5^8\end{array}$