Question

The coefficient of $a^{-6}{b}^4$ in the expansion of ${(\frac{1}{a}-\frac{2b}{3})}^{10}$ is

• A. $\frac{120}{27}$
• B. $\frac{1120}{27}$
• C. $\frac{2010}{27}$
• D. $\frac{1128}{27}$

Answer 1

The correct option is B $\frac{1120}{27}$

Given : ${(\frac{1}{a}-\frac{2b}{3})}^{10}$
General term is
$T_{r+1}={}^{10}{C}_r{\left(\frac{1}{a}\right)}^{10-r}{\left(\frac{-2b}{3}\right)}^r$
Now, we need coefficient of $a^{-6}{b}^4$, which occur when $r=4$
$\therefore T_5={}^{10}{C}_4{(-\frac{2}{3})}^4{a}^{-6}{b}^4\Rightarrow T_5=\frac{10\cdot 9\cdot 8\cdot 7\cdot 16}{4!\cdot 3^4}{a}^{-6}{b}^4\Rightarrow T_5=\frac{10\cdot 7\cdot 16}{3\cdot 3^2}{a}^{-6}{b}^4\Rightarrow T_5=\frac{1120}{27}{a}^{-6}{b}^4$
Hence, the required coefficient $=\frac{1120}{27}$