Q58 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics

If $\alpha=\frac{5}{2 ! \times 3}+\frac{5 \times 7}{3 ! \times 3^2}+\frac{5 \times 7 \times 9}{4 ! \times 3^3}+\ldots .$, then $\alpha^2+4 \alpha=$

  • A
    21
  • Correct answer
    B
    23
  • C
    25
  • D
    27

Solution

Since, $$ \begin{aligned} (1+x)^n=1+n x+ & \frac{n(n-1)}{2 !} x^2+\frac{n(n-1)(n-2)}{3 !} x^3 \\ & +\frac{n(n-1)(n-2)(n-3)}{4 !} x^4+\ldots \end{aligned} $$ On comparing $\frac{n(n-1)}{2 !} x^2=\frac{5}{2 ! \times 3}$ and $\frac{n(n-1)(n-2)}{3 !} x^3=\frac{5 \times 7}{3 ! \times 3^2}$ So, $\quad x=-\frac{2}{3}$ and $n=-3 / 2$ $$ \begin{array}{ll} \therefore(1-2 / 3)^{-3 / 2}=1+1+\alpha & \Rightarrow 3^{3 / 2}=2+\alpha \\ \Rightarrow \alpha^2+4 \alpha+4=27 & \Rightarrow \alpha^2+4 \alpha=23 \end{array} $$

Hence, the correct answer is option: 'B'.

Relevant Questions