Question

Find the $(r+1{)}^{\text{th}}$ term of the following expansion :
$(1+x^2{)}^{-3}$

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,

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

$=\frac{(-3)(-4)(-5)(-6)........(-r-2)}{r!}{x}^{2r}$

$=(-1{)}^r\frac{\mathrm{3.4.5.6..........}(r+2)}{r!}{x}^{2r}$

$=(-1{)}^r\frac{(r+1)(r+2)}{1.2}{x}^{2r}$

By cancelling like terms in numerator and denominator

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