Question

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

• A. $\frac{462}{b^5} a^6$
• B. $\frac{462 a^5}{b^6} $
• C. $\frac{ - 462 a^5}{b^6} $
• D. $\frac{ - 462 a^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}{bx^2} $
$\therefore \, T_{r + 1} = {^{11}C_{r}} (ax)^{11 - r} \left( \frac{ - 1}{bx^2} \right)^r$
$ = {^{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
${^{11}C_6 } (a)^5 \left( \frac{ -1}{b} \right)^6 = \frac{462}{b^6} a^5$