Question

Find the $(r+1{)}^{th}$ term in the following expansion :
$(a+bx{)}^{-1}$

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,

$(a+bx{)}^{-1}=\frac{1}{a}{(1+\frac{b}{a}x)}^{-1}$

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

$=\frac{1}{a}\left[\frac{(-1)(-2)(-3)(-4)........(-r)}{r!}{\left(\frac{b}{a}\right)}^r{x}^r\right]$

$=\frac{1}{a}\left[\right(-1{)}^r\frac{\mathrm{1.2.3.4........}r}{r!}{\left(\frac{b}{a}\right)}^r{x}^r]$

$=(-1{)}^r\frac{b^r}{a^{r+1}}{x}^r$