EAMCET Engineering 2008 Solved Paper

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