Question

Find the $(r+1{)}^{th}$ term in each of the following expansion :
$\frac{1}{\sqrt[3]{3(1-3x{)}^2}}$

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,

$\frac{1}{\sqrt[3]{3(1-3x{)}^2}}=(1-3x{)}^{-\frac{2}{3}}$

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

$=\frac{(-2)(-5)(-8)(-11)........(1-3r)}{3^r.r!}(-1{)}^r{.3}^r.x^r$

$=(-1{)}^r.(-1{)}^r\frac{\mathrm{2.5.8.11.......}(3r-1)}{r!}{x}^r$

$=(-1{)}^{2r}\frac{\mathrm{2.5.8.11.......}(3r-1)}{r!}{x}^r$ $[\because (-1{)}^{2r}=1]$

$=\frac{\mathrm{2.5.8.11.......}(3r-1)}{r!}{x}^r$