Question

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

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,

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

$=\frac{(-3)(-5)(-7)(-9)........(-1-2r)}{2^r.r!}(-1{)}^r{.2}^r.(x{)}^r$

$=(-1{)}^r(-1{)}^r\frac{\mathrm{3.5.7.9......}(2r+1)}{r!}{x}^r$

$=(-1{)}^{2r}\frac{\mathrm{3.5.7.9......}(2r+1)}{r!}{x}^r$ $[\because (-1{)}^{2r}=1]$

$=\frac{\mathrm{3.5.7.9......}(2r+1)}{r!}{x}^r$