If $\mathrm{\alpha }=\frac{5}{2!\times 3}+\frac{5\times 7}{3!\times 3^2}+\frac{5\times 7\times 9}{4!\times 3^3}+\dots \dots$ then ${\mathrm{\alpha }}^2+4\mathrm{\alpha }=$

$$\begin{gathered} \mathrm{\alpha }=\frac{5}{2!} \left(\frac{1}{3}\right)+\frac{5\times 7}{3!} {\left(\frac{1}{3}\right)}^2+.... \\ \mathrm{\alpha }=\frac{3.5}{2!} {\left(\frac{1}{3}\right)}^2+\frac{3.5.7}{3!} {\left(\frac{1}{3}\right)}^3+..... \\ \mathrm{p}=3, \mathrm{q}=2, \frac{\mathrm{x}}{\mathrm{q}}=\frac{1}{3} \Rightarrow \mathrm{x}=\frac{2}{3} \\ \mathrm{\alpha }={\left(1-\frac{2}{3}\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-1-\left(3\right) \left(\frac{1}{3}\right) \\ ={\left(\frac{1}{3}\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2 \\ \mathrm{\alpha }+2=3^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \Rightarrow {\mathrm{\alpha }}^2+4\mathrm{\alpha }+4=27 \\ {\mathrm{\alpha }}^2+4\mathrm{\alpha }=23 \end{gathered}$$