If $\quad \alpha=\frac{5}{2 ! 3}+\frac{5 \cdot 7}{3 ! 3^2}+\frac{5 \cdot 7 \cdot 9}{4 ! 3^3}+\ldots, \quad$ then $\alpha^2+4 \alpha$ is equal to

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), $$ \begin{aligned} & \left(1-\frac{2}{3}\right)^{-3 / 2}=1+1+\frac{5}{2 ! 3}+\frac{5 \cdot 7}{3 ! 3^2}+\ldots \\ & \Rightarrow \quad 3^{3 / 2}-2=\frac{5}{2 ! 3}+\frac{5 \cdot 7}{3 ! 3^2}+\ldots \\ & \Rightarrow \quad \alpha=3^{3 / 2}-2 \quad \text { [from Eq. (i)] } \\ & \text { Now, } \alpha^2+4 \alpha=\left(3^{3 / 2}-2\right)^2+4\left(3^{3 / 2}-2\right) \\ & =27+4-4 \cdot 3^{3 / 2}+4 \cdot 3^{3 / 2}-8 \\ & =23 \\ & \end{aligned} $$