Question

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

• A. ${}^{2n+1}{C}_n.x$
• B. ${}^{2n+1}{C}_n$
• C. $(-1{)}^n{.}^{2n+1}{C}_n$
• D. $(-1{)}^n{.}^{2n+1}{C}_n.x$

Answer 1

The correct option is D $(-1{)}^n{.}^{2n+1}{C}_n.x$

Let the middle term of ${(x-\frac{1}{x})}^{2n+1}$ be $M$.

The middle term is $\frac{1}{2}(2n+1+1)$ i.e $(n+1{)}^{th}$ term.

$M=T_{n+1}$

We k.n.t $T_r$ of expansion $(a+b{)}^n$ is $T_r{=}^n{C}_{r-1}{a}^{n-r+1}{b}^{r-1}$

$\therefore M=T_{n+1}{=}^{2n+1}{C}_n{x}^{2n+1-n}{(-\frac{1}{x})}^n=(-1{)}^n{.}^{2n+1}{C}_n.x^{n+1-n}=(-1{)}^n{.}^{2n+1}{C}_n.x$

Option (D) is correct.