Question

If the expansion of ${\left[\frac{3\sqrt{x}}{7}-\frac{5}{2x\sqrt{x}}\right]}^{13n}$contains a term independent of $x$ in ${14}^{th}$ term, then $n$ should be

• A. $10$
• B. $5$
• C. $6$
• D. $4$
• E. $11$

Answer 1

The correct option is D

$4$

Explanation for the correct option:

Given expansion is ${\left[\frac{3\sqrt{x}}{7}-\frac{5}{2x\sqrt{x}}\right]}^{13n}$ and ${14}^{th}$ term is independent of $x$

The general term for the given expansion will be $T_{r+1}={}^{13n}C_r{\left(\frac{3\sqrt{x}}{7}\right)}^{13n-r}{\left(-\frac{5}{2x\sqrt{x}}\right)}^r$

$$\begin{gathered} \begin{array}{crcl}& & =& {}^{13n}C_r{\left(-\frac{5}{2}\right)}^r{\left(\frac{3}{7}\right)}^{13n-r}{\left(x\right)}^{-r}{\left(x\right)}^{\frac{1}{2}\left(13n-r-r\right)} \\ \\ & T_{r+1}& =& {}^{13n}C_r{\left(-\frac{5}{2}\right)}^r{\left(\frac{3}{7}\right)}^{13n-r}{\left(x\right)}^{\frac{13n}{2}-2r}\end{array} \end{gathered}$$

Substitute $r=13$

$\begin{array}{crcl}\Rightarrow & T_{14}& =& {}^{13n}C_{13}{\left(-\frac{5}{2}\right)}^{13}{\left(\frac{3}{7}\right)}^{13n-13}{\left(x\right)}^{\frac{13n}{2}-26}\end{array}$

$T_{14}$ term is independent of $x\Rightarrow x^0$

$\begin{array}{crcl}\Rightarrow & \frac{13n}{2}-26& =& 0\\ \Rightarrow & \frac{13n}{2}& =& 26\\ \Rightarrow & 13n& =& 52\\ \Rightarrow & n& =& \frac{52}{13}\\ \Rightarrow & n& =& 4\end{array}$

Hence the correct answer is option (D) i.e. $4$