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)

$\frac{462 a^5}{b^6}$

Answer 2

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 ${^{11}C_6 } (a)^5 \left( \frac{ -1}{b} \right)^6 = \frac{462}{b^6} a^5$