Find the coefficient of $x^7$ in the expansion of $(a{x}^2+\frac{1}{bx}{)}^{11}$.

Solution

1. The general term of the expansion $(a{x}^2+\frac{1}{bx}{)}^{11}$ is given by $T_{r+1}=\left(\genfrac{}{}{0ex}{}{11}{r}\right)(a{x}^2{)}^{11-r}(b^{-1}{x}^{-1}{)}^r$.
2. Simplifying the powers of $x$, we have $x^{2(11-r)}\cdot x^{-r}=x^{22-3r}$.
3. Setting the exponent equal to 7, we solve $22 - 3r = 7$, which gives $3r = 15$, so $r=5$.
4. Substituting $r=5$ into the general term, we get $\left(\genfrac{}{}{0ex}{}{11}{5}\right)a^{11-5}{b}^{-5}=330a^6{b}^{-5}$.
5. Hence the answer is (A).