The ${(p+2)}^{th}$ term from the end in ${\left(x-\frac{{\displaystyle 1}}{{\displaystyle x}}\right)}^{2n+1}$, is 

Given ${\left(x-\frac{{\displaystyle 1}}{{\displaystyle x}}\right)}^{2n+1}$

To find ${(p+2)}^{th}$ term from the end of the expansion of  ${\left(x-\frac{{\displaystyle 1}}{{\displaystyle x}}\right)}^{2n+1}$

Formula:- In the Binomial expansion of ${(x+a)}^n$, the r^th  term from the end is ${(n-r+2)}^{th}$term from the beginning.

$\therefore {(p+2)}^{th}$ term from the end $={(2n-p+1)}^{th}$ term from the beginning

$\begin{array}{l}\therefore T_{2n-p+1}{=}^{(2n+1)}{C}_{2n-p}.{\left(x\right)}^{(2n+1)(2n-p)}.{(-\frac{1}{x})}^{2n-p}\\ ={(-1)}^{2n-p}{.}^{(2n+1)}{C}_{2n-p}.x^{1+p}.{\left(\frac{1}{x}\right)}^{2n-p}\\ ={(-1)}^p{.}^{(2n+1)}{C}_{2n-p}.x^{2p-2n+1}\\ \left[\because {(-1)}^{2n-p}={(-1)}^{2n}.{(-1)}^{-p}=1.\frac{1}{{(-1)}^p}={(-1)}^p\right]\end{array}$