Question

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

• A. $\frac{n}{\left[\right(n/2)!{]}^2}$
• B. $\frac{\left(2n\right)!}{\left[\right(n/2)!{]}^2}$
• C. $\frac{\mathrm{1.3.5}...(2n+1)}{n!}{2}^n$
• D. $\frac{\left(2n\right)!}{(n!{)}^2}$

Answer 1

Answer: (D)

$\frac{\left(2n\right)!}{(n!{)}^2}$

Since, ${(x^2+\frac{1}{x^2}+2)}^n={\left[{(x+\frac{1}{x})}^2\right]}^n={(x+\frac{1}{x})}^{2n}$
Thus the number of terms in the expansion of ${(x+\frac{1}{x})}^{2n}$ is $2n+1$ which is odd. Therefore, $(n+1)th$ term will be middle term.
Therefore, $T_{n+1}{=}^{2n}{C}_n(x{)}^{2n-n}{\left(\frac{1}{x}\right)}^n$
${=}^{2n}{C}_n{x}^n\cdot \frac{1}{x^n}$
${=}^{2n}{C}_n=\frac{\left(2n\right)!}{n!n!}=\frac{\left(2n\right)!}{(n!{)}^2}$.