Question

Term independent of $x$ in the expansion of ${(x-\frac{1}{x^2})}^{3n}$ (where $n$ is even natural number)

• A. $\frac{3n!}{n!\times 2n!}$
• B. $\frac{3n!}{n!}$
• C. $\frac{3n!}{2n!}$
• D. $\frac{3n!}{(n!{)}^2}$

Answer 1

The correct option is A $\frac{3n!}{n!\times 2n!}$

$T_{r+1}=(-1{)}^r\cdot {}^{3n}{C}_r{x}^{3n-r}\times {\left(\frac{1}{x^2}\right)}^r$
$\Rightarrow T_{r+1}=(-1{)}^r\cdot {}^{3n}{C}_r{x}^{3n-3r}$
For independent term
Put $3n-3r=0\Rightarrow r=n.$

So, $T_{n+1}$ is free from $x.$
And, $T_{n+1}=(-1{)}^n\cdot {}^{3n}{C}_n={}^{3n}{C}_n$