Question

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

Answer 1

To find the ${(p+2)}^{th}$ from end in ${(x-\frac{1}{x})}^{2n+1}$, we find the ${(p+2)}^{th}$ term from begininng of the expansion ${(-\frac{1}{x}+x)}^{2n+1}$ i.e

$T_{p+1+1}$=${}^{2n+1}{C}_{p+1}$.${\left(\frac{-1}{x}\right)}^{2n+1-p-1}$.$x^{p+1}$

$T_{p+2}$=${}^{2n+1}{C}_{p+1}$.${\left(\frac{-1}{x}\right)}^{2n-p}$.$x^{p+1}$

$T_{p+2}$=${}^{2n+1}{C}_{p+1}$.${(-1)}^{2n-p}$.${\left(\frac{1}{x}\right)}^{2n-2p-1}$