Question

Middle term in the expansion of ${(x^2+\frac{1}{x^2}+2)}^n$ is

• A. $\frac{n!}{(n!{)}^2}$
• B. $\frac{2n!}{(n!{)}^2}$
• C. $\frac{(2n-1)!}{n!}$
• D. $\frac{\left(2n\right)!}{n!}$

Answer 1

The correct option is B $\frac{2n!}{(n!{)}^2}$

The given expression is
${(x^2+\frac{1}{x^2}+2)}^n={\left[{(x+\frac{1}{x})}^2\right]}^n$
$={(x+\frac{1}{x})}^{2n}$
The binomial contains $(2n+1)$ terms in its expansion.
Therefore, the middle term is $T_{n+1}$.
$T_{n+1}{=}^{2n}{C}_n(x{)}^{2n-n}{\left(\frac{1}{x}\right)}^n$
${=}^{2n}{C}_n$
$=\frac{\left(2n\right)!}{(2n-n)!n!}=\frac{\left(2n\right)!}{(n!{)}^2}$.