Question

Find the$(p+2)$th term from the end in ${(x-\frac{1}{x})}^{2n+1}$.

Answer 1

We know that the $r^{th}$ term from beginning in expansion of $(x+a{)}^n$ is given by ${}^n{C}_{r-1}{x}^{n+1-r}{a}^{r-1}$

and $r^{th}$ term from beginning is $(n-r+2{)}^{th}$ term from end

$\therefore$ coefficient of $(n-r{)}^{th}$ term can be found by putting $n-r+2$ instead of $r$ in above equation, from which we get ${}^n{C}_{r-1}{a}^{n+1-r}{x}^{r-1}$ $\because {}^n{C}_{r-1}{=}^n{C}_{n-r+1}$

$\therefore$ The $r^{th}$ term from end in expansion of $(x+a{)}^n$ is ${}^n{C}_{r-1}{a}^{n+1-r}{x}^{r-1}$

In the given question the term and power are $p+2\&2n+1$ respectively,

$\therefore$ Replace $r\Rightarrow p+2$ and $n\Rightarrow 2n+1$

On putting it in the equation we get $(p+2{)}^{th}$ term of ${(x-\frac{1}{x})}^{2n+1}$ equal to ${}^{2n+1}{C}_{p+1}{(-\frac{1}{x})}^{2n-p}(x{)}^{p+1}$

$\because (-1{)}^{2n+p}=(-1{)}^p$
$\Rightarrow$ the required coefficient is $(-1{)}^p{}^{2n+1}{C}_{p+1}{x}^{2p-2n+1}$