Explanation

(b)
$Given,\ \ \ \alpha = \frac{5}{2!3} + \frac{5 \bullet 7}{3!3^{2}} + \frac{5 \bullet 7 \bullet 9}{{4!3}^{2}} + \ldots\ \ \ \ \ \ \ldots(i)$
On comparing
$(1 + x)^{n} = 1 + \frac{nx}{1!} + \frac{n(n - 1)}{2!}x^{2} + \frac{n(n - 1)(n - 2)}{3!}x^{3} + \ldots$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
With respect to factorial, we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n({n - 1)x}^{2} = \frac{5}{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(iii)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n(n - 1)(n - 2{)x}^{3} = \frac{5 \bullet 7}{3^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ldots(iv)$
$and\ n(n - 1)(n - 2)(n - 3)x^{4} = \frac{5 \bullet 7 \bullet 9}{3^{3}}\ \ ..(v)$
on dividing Eq. (iv) by (iii) and Eq. (v) by Eq. (iv), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (n - 2)x = \frac{7}{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(vi)$
$and\ \ (n - 3)x = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ $\ldots$(vii)
Again, dividing Eq. (vi) by Eq. (vii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n - 2}{n - 3} = \frac{7}{9}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9n - 18 = 7n - 21$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2n = - 3 \Longrightarrow n = - \frac{3}{2}$
On putting the value of $n$ in Eq. (vi), we get
$\left( - \frac{3}{2} - 2 \right)x = \frac{7}{3} \Longrightarrow x = - \frac{2}{3}$
$\therefore$ From Eq. (ii),
$\left( 1 - \frac{2}{3} \right)^{- 3/2} = 1 + 1 + \frac{5}{2!3} + \frac{5 \bullet 7}{{3!3}^{2}} + \ldots$
$\Longrightarrow 3^{3/2} - 2 = \frac{5}{2!3} + \frac{5 \bullet 7}{{3!3}^{2}} + ...$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha = 3^{3/2} - 2$ [from Eq. (i)]
Now, $\alpha^{2} + 4\alpha = (3^{3/2} - {2)}^{2} + 4(3^{3/2} - 2)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 27 + 4 - 4 \bullet 3^{3/2} + 4 \bullet 3^{3/2} - 8$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 23$