Question

The term independent of $x$ in the expansion of ${(\frac{4}{3}{x}^2-\frac{3}{2x})}^9$ is:

• A. $5^{th}$
• B. $6^{th}$
• C. $7^{th}$
• D. $8^{th}$

Answer 1

Answer: (C)

$7^{th}$

Given

${(\frac{4}{3}{x}^2-\frac{3}{2x})}^9$

General term of given expansion is

$T_{r+1}={}^9{C}_r{\left(\frac{4}{3}\right)}^{9-r}\left(x{)}^{18-2r}{\left(\frac{3}{2}\right)}^r\left(\frac{1}{x^r}\right)\right(-1{)}^r$

$T_{r+1}={}^9{C}_r{\left(\frac{4}{3}\right)}^{9-r}\left(x{)}^{18-3r}{\left(\frac{3}{2}\right)}^r\right(-1{)}^r--------\left(1\right)$

Comparing terms having power $x^0$ with $T_{r+1}={}^9{C}_r{\left(\frac{4}{3}\right)}^{9-r}\left(x{)}^{18-3r}{\left(\frac{3}{2}\right)}^r\right(-1{)}^r$ we get

$18-3r=0\Rightarrow r=6$

From eq (1)

$T_{6+1}={}^9{C}_6{\left(\frac{4}{3}\right)}^3{\left(\frac{3}{2}\right)}^6$

Hence $T_7$ term $7^{th}$ is independent of $x$