Question

Find the general term in the expansion of $(1-x{)}^{-3}.$

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,

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

$=\frac{-3(-4)(-5)(-6).......(-2-r)}{r!}(-1{)}^r{x}^r$

$=(-1{)}^r(-1{)}^r\frac{\mathrm{3.4.5.6.....}(r+2)}{\mathrm{1.2.3.4.5........}(r-1).r}{x}^r$

$=\frac{\mathrm{3.4.5.6.....}(r+2)}{\mathrm{1.2.3.4.5........}(r-1).r}{x}^r$ $[\because (-1{)}^r(-1{)}^r=(-1{)}^{2r}=1]$

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

By removing like factors from numerator and denominator

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