Question

Write down and simplify :
The $5^{\text{th}}$ term of $(3a-2b{)}^{-1}$

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,

$(3a-2b{)}^{-1}=(3a{)}^{-1}{(1-\frac{2b}{3a})}^{-1}=\frac{1}{3a}{(1-\frac{2b}{3a})}^{-1}$

For $T_5,$ $r=4$

$T_5=\frac{1}{3a}\left[\frac{(-1)(-1-1)(-1-2)(-1-3)}{4!}{(-\frac{2b}{3a})}^4\right]$

$=\frac{1}{3a}\left[\frac{(-1)(-2)(-3)(-4)}{4!}\right(-1{)}^4\left(\frac{16b^4}{81a^4}\right)]$

$=\frac{16b^4}{243a^5}$