Question

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

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

Answer 1

The correct option is B

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

Explanation for the correct option:

Middle term in binomial expansion:

$$\begin{gathered} \left[x^2+\left(\frac{1}{x^2}\right)+2\right]=({(x+\frac{1}{x})}^2{)}^n \\ =({(x+\frac{1}{x})}^{2n} \end{gathered}$$

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

$=(n+1)$th term.

$$\begin{gathered} T_{n+1}={}_{}{}^{2n}C_n{x}^n{\left(\frac{1}{x}\right)}^n\mathbf{[}\mathbf{\because }{\mathbf{T}}_{\mathbf{r}\mathbf{+}\mathbf{1}}{\mathbf{=}}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}}{\mathbf{x}}^{\mathbf{n}\mathbf{-}\mathbf{r}}{\mathbf{a}}^{\mathbf{r}}\mathbf{]} \\ =\frac{\left(2n\right)!}{n!n!}{x}^n\frac{1}{x^n} \\ =\frac{\left(2n\right)!}{{(n!)}^2} \end{gathered}$$

Hence, the correct option is (B)