Question

The term independent of ${}^{\prime }{x}^{\prime }$ in expansion of ${(4x^4-\frac{5}{2x^2})}^9$ is

• A. $T_5$
• B. $T_6$
• C. $T_7$
• D. $T_8$

Answer 1

The correct option is C $T_7$

Now,

$\begin{array}{c}{T}_{r+1}={}^9{C}_r{\left(\frac{-5}{2x^2}\right)}^r{\left(4x^4\right)}^{9-r}\\ ={}^9{C}_r{\left(\frac{-5}{2}\right)}^r{x}^{-2r}{\left(4\right)}^{9-r}{x}^{36-4r}\\ ={}^9{C}_r{\left(\frac{-5}{2}\right)}^r{\left(4\right)}^{9-r}{x}^{36-6r}\end{array}$

For term independent of $x$ $36-6r=0$

$\Rightarrow r=6$

hence we get the term $T_7$

So, option $C$ is correct answer.