Question

The $5^{th}$ term of $(3a-2b{)}^{-1}$ is

• A. $\frac{16b^4}{243a^5}$
• B. $\frac{18b^4}{243a^5}$
• C. $\frac{16b^4}{245a^5}$
• D. $\frac{18b^4}{245a^5}$

Answer 1

The correct option is A $\frac{16b^4}{243a^5}$

We know for the expansion of ${(1-x)}^{-n}$ the genral term is

$T_{r+1}$=${}^{n+r-1}{C}_r{x}^r$

We make our eqn in the form of ${(1-x)}^{-n}$

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

Therefore, $T_{4+1}$=$\frac{1}{3a}$${}^{1+4-1}{C}_4$${\left(\frac{2b}{3a}\right)}^4$

$\Rightarrow T_5$=$\frac{b^4}{a^5}$.1.$\frac{2^4}{3^5}$

$\Rightarrow T_5$=$\frac{16b^4}{243b^5}$