Question

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

• A. $21$
• B. $23$
• C. $25$
• D. $27$

Answer 1

The correct option is B

$23$

Explanation for correct option.

Step1. Finding value of ‘n’

$\mathrm{\alpha }=\frac{5}{2!\times 3}+\frac{5\times 7}{3!\times 3^2}+\frac{5\times 7\times 9}{4!\times 3^3}+.......\left(\mathrm{i}\right)$

Using expansion,

${(1+\mathrm{x})}^{\mathrm{n}}=1+\mathrm{nx}+\frac{\mathrm{n}(\mathrm{n}-1)x^2}{2!}+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)x^3}{3!}+....\left(\mathrm{ii}\right)$

From equation $\left(\mathrm{i}\right)\&\left(\mathrm{ii}\right)$

$\mathrm{n}(-1){\mathrm{x}}^2=\frac{5}{3}....\left(\mathrm{iii}\right)$

$\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2){\mathrm{x}}^3=\frac{5\times 7}{3^2}....\left(\mathrm{iv}\right)$

$\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3){\mathrm{x}}^4=\frac{5\times 7\times 9}{3^3}...\left(\mathrm{v}\right)$

On dividing equation $\left(\mathrm{iv}\right)$by $\left(\mathrm{iii}\right)$, we get

$(\mathrm{n}-2)\mathrm{x}=\frac{7}{3}...\left(\mathrm{vi}\right)$

On dividing equation $\left(v\right)$ by $\left(\mathrm{iv}\right)$, we get,

$$\begin{gathered} \Rightarrow (n-3)x=\frac{9}{3} \\ \Rightarrow x=\frac{9}{3(n-3)} \end{gathered}$$

put the value of x in equation $\left(\mathrm{vi}\right)$, we get

$\begin{array}{rcl}\Rightarrow \frac{(n-2)9}{(n-3)3}& =& \frac{7}{3}\\ & & (n-2)9=7(n-3)\\ n& =& \frac{-3}{2}\end{array}$

Now put the value of n in equation $\left(\mathrm{vi}\right)$we get,

$$\begin{gathered} \Rightarrow (-\frac{3}{2}-2)x=\frac{7}{3} \\ \Rightarrow -\frac{7}{2}x=\frac{7}{3} \\ \Rightarrow x=-\frac{2}{3} \end{gathered}$$

Step2. Finding value of ${\alpha }^2+4\alpha$

Now, equation (ii) becomes

$$\begin{gathered} \Rightarrow {(1+\mathrm{x})}^{\mathrm{n}}=1+\mathrm{nx}+\alpha \\ \Rightarrow \alpha ={(1+\mathrm{x})}^{\mathrm{n}}-1-nx \\ \Rightarrow \alpha ={(1-\frac{2}{3})}^{\frac{-3}{2}}-1-\frac{-3}{2}\times \frac{-2}{3} \\ \Rightarrow \alpha ={(1-\frac{2}{3})}^{\frac{-3}{2}}-2 \\ \Rightarrow \alpha +2={(1-\frac{2}{3})}^{\frac{-3}{2}} \\ \Rightarrow {(\alpha +2)}^2={\left(\frac{1}{3}\right)}^{-3}\Rightarrow {\alpha }^2+4\alpha +4=27 \\ \Rightarrow {\alpha }^2+4\alpha =23 \end{gathered}$$

Hence, correct option is $\mathbf{\left(}\mathit{B}\mathbf{\right)}$.