Question

The coefficient of $x^{-7}$ in the expansion of ${\left[ax-\frac{1}{b{x}^2}\right]}^{11}$ will be

• A. $\frac{462}{b^5}{a}^6$
• B. $\frac{462a^5}{b^6}$
• C. $\frac{-462a^5}{b^6}$
• D. $\frac{-462a^6}{b^5}$

Answer 1

Answer: (B)

Suppose $x^{-7}$ occurs in $(r+1{)}^{th}$ term.
we have $T_{r+1}={}^n{C}_r{x}^{n-r}{a}^r$ in $(x+a{)}^n$.
In the given question, n = 1, x = ax, a = $\frac{-1}{b{x}^2}$
$\therefore T_{r+1}={}^{11}{C}_r(ax{)}^{11-r}{\left(\frac{-1}{b{x}^2}\right)}^r$
$={}^{11}{C}_r{a}^{11-r}{b}^{-r}{x}^{11-3r}(-1{)}^r$
This term contains $x^{-7}$ if $11-3r=-7$
$\Rightarrow r=6$
Therefore, coefficient of $x^{-7}$ is
${}^{11}{C}_6(a{)}^5{\left(\frac{-1}{b}\right)}^6=\frac{462}{b^6}{a}^5$