Given that, $\alpha=\frac{5}{2 ! 3}+\frac{5 \cdot 7}{3 ! 3^2}+\frac{5 \cdot 7 \cdot 9}{4 ! 3^3}+\ldots$
We know that,
$$
\begin{aligned}
(1+x)^n= & 1+\frac{n x}{1 !}+\frac{n(n-1)}{2 !} x^2 \\
& +\frac{n(n-1)(n-2)}{3 !} x^3+\ldots
\end{aligned}
$$
On comparing Eqs. (i) and (ii), with respect to factorial
$$
\begin{aligned}
n(n-1) x^2 & =\frac{5}{3} \\
n(n-1)(n-2) x^3 & =\frac{5 \cdot 7}{3^2}
\end{aligned}
$$
and
$$
n(n-1)(n-2)(n-3) x^4=\frac{5 \cdot 7 \cdot 9}{3^3}
$$
On dividing Eq. (iv) by (iii) and Eq. (v) by (iv), we get
$$
(n-2) x=\frac{7}{3}
$$
and
$$
(n-3) x=3
$$
Again, dividing Eq. (vi) by (vii), we get
$$
\begin{array}{rlrl}
& & \frac{n-2}{n-3} & =\frac{7}{9} \\
\Rightarrow & & 9 n-18 & =7 n-21 \\
\Rightarrow & & 2 n & =-3 \\
\Rightarrow & n & =-\frac{3}{2}
\end{array}
$$
On putting the value of $n$ in Eq. (vi), we get
$$
\left(-\frac{3}{2}-2\right) x=\frac{7}{3} \Rightarrow x=-\frac{2}{3}
$$
$\therefore$ From Eq. (ii),
$$
\begin{aligned}
& \left(1-\frac{2}{3}\right)^{-3 / 2}=1+1+\frac{5}{2 ! 3}+\frac{5 \cdot 7}{3 ! 3^2}+\ldots \\
& \Rightarrow \quad 3^{3 / 2}-2=\frac{5}{2 ! 3}+\frac{5 \cdot 7}{3 ! 3^2}+\ldots \\
& \Rightarrow \quad \alpha=3^{3 / 2}-2 \quad \text { [from Eq. (i)] } \\
& \text { Now, } \alpha^2+4 \alpha=\left(3^{3 / 2}-2\right)^2+4\left(3^{3 / 2}-2\right) \\
& =27+4-4 \cdot 3^{3 / 2}+4 \cdot 3^{3 / 2}-8 \\
& =23 \\
&
\end{aligned}
$$