Question

Find the $(r+1{)}^{th}$ term in each of the following expansion :
$\frac{1}{\sqrt[n]{n(a^n-nx)}}$

Answer 1

The general term i.e. $(r+1)th$ term in the expansion of $(1+x{)}^n$ is given by

$T_{r+1}=\frac{n(n-1)(n-2)....(n-r+1)}{r!}{x}^r$

Now,

$\frac{1}{\sqrt[n]{n(a^n-nx)}}=\frac{1}{a}{(1-\frac{nx}{a^n})}^{-\frac{1}{n}}$

$T_{r+1}=\frac{1}{a}\left[\frac{-\frac{1}{n}(-\frac{1}{n}-1)(-\frac{1}{n}-2)(-\frac{1}{n}-3)..........(-\frac{1}{n}-r+1)}{r!}{(-\frac{nx}{a^n})}^r\right]$

$=\frac{1}{a}\left[\frac{(-1)(-1-n)(-1-2n)(-1-3n)........(-1-nr+n)}{n^r.r!.a^{nr}}\right(-1{)}^r{n}^r.x^r]$

$=(-1{)}^r(-1{)}^r\frac{(n+1)(2n+1)(3n+1)..............(nr-n+1)}{r!.a^{nr+1}}{x}^r$

$=\frac{(n+1)(2n+1)(3n+1)..............(nr-n+1)}{r!}\frac{x^r}{a^{nr+1}}$