Question

Find the $(r+1{)}^{\text{th}}$ term of the following expansion :
$(1-x{)}^{-5}$

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,

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

$=\frac{(-5)(-6)(-7)(-8)........(-r-4)}{r!}(-1{)}^r(x{)}^r$

$=(-1{)}^r(-1{)}^r\frac{\mathrm{5.6.7.8.........}(r+4)}{r!}{x}^r$

$=(-1{)}^{2r}\frac{(r+1)(r+2)(r+3)(r+4)}{\mathrm{1.2.3.4}}{x}^r$

By cancelling like terms in numerator and denominator

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