Question

If the term independent of $x$ in the expansion of ${\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9$ is $k$, then $18k$ is equal to

• A. $5$
• B. $9$
• C. $7$
• D. $11$

Answer 1

The correct option is C

$7$

Explanation for the correct option:

Finding the value of $18k$:

By Binomial Expansion:

$$\begin{gathered} {T_{r+1=}}^9{C}_r{\left(\frac{3}{2}{x}^2\right)}^{9-r}{\left(\frac{-1}{3x}\right)}^r \\ {=}^9{C}_r\frac{3^{9-r}}{2^{9-r}}{\left(\frac{-1}{3}\right)}^r{x}^{18-3r} \end{gathered}$$

Now, The term is independent of $x$ when the exponent of $x$ is $0$.

$$\begin{gathered} 18-3r=0 \\ \Rightarrow 3r=18 \\ \Rightarrow r=6 \end{gathered}$$

Substitute value of $r$ in $T_{r+1}$.Then,

$$\begin{gathered} T_7{=}^9{C}_6\frac{3^3}{2^3}{\left(\frac{-1}{3}\right)}^6{x}^{18-6r} \\ =84\left(\frac{1}{216}\right) \\ =\frac{7}{18}...\left(1\right) \end{gathered}$$

Also, The term independent of $x$ is $k$.

$$\begin{gathered} T_7=k \\ \Rightarrow \frac{7}{18}=k\left[\mathrm{from}1\right] \\ \Rightarrow 18k=7 \end{gathered}$$

Hence, Option (C) is the correct answer.