Question

Write down and simplify :
The $(r+1{)}^{\text{th}}$ term of $(1-x{)}^{-2}$

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,

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

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

$=(-1{)}^r(-1{)}^r\frac{\mathrm{2.3.4.5.......}(r+1)}{r!}{x}^r$

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

By cancelling like terms in numerator and denominator

$=(r+1)x^r$