Question

Find the $(r+1{)}^{\text{th}}$ term of the following expansion :
$(2-x{)}^{-2}$

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,

$(2-x{)}^{-2}=\frac{1}{4}{(1-\frac{1}{2}x)}^{-2}$

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

$=\frac{1}{4}\left[\frac{(-2)(-3)(-4)(-5)........(-r-1)}{r!}\right(-1{)}^r{\left(\frac{1}{2}\right)}^r{x}^r]$

$=\frac{1}{4}\left[\right(-1{)}^r(-1{)}^r\frac{\mathrm{2.3.4.5........}(r+1)}{r!}{\left(\frac{1}{2}\right)}^r{x}^r]$

By cancelling like terms in the numerator and denominator, we get

$=(-1{)}^{2r}\frac{(r+1)}{2^{r+2}}{x}^r$

$=\frac{(r+1)}{2^{r+2}}{x}^r$