Question

Write down and simplify :
The $(r+1{)}^{\text{th}}$ term of $(1-x{)}^{-4}$

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,

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

$=\frac{(-4)(-5)(-6)(-7)........(-r-3)}{r!}(-1{)}^r(x{)}^r$

$=(-1{)}^r(-1{)}^r\frac{\mathrm{4.5.6.7.........}(r+3)}{r!}{x}^r$

$=(-1{)}^{2r}\frac{(r+1)(r+2)(r+3)}{\mathrm{1.2.3}}{x}^r$

By cancelling like terms in numerator and denominator

$=\frac{(r+1)(r+2)(r+3)}{\mathrm{1.2.3}}{x}^r$