Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively in the expansion of $\left(\frac{1}{3}x^{1/3} + \frac{1}{2x^{2/3}}\right)^{18}$. Then $\left(\frac{n}{m}\right)^{1/3}$ is:
Step-by-Step Solution
Key Concept: Use $T_{r+1} = \binom{18}{r}\left(\frac{x^{1/3}}{3}\right)^{18-r}\left(\frac{1}{2x^{2/3}}\right)^r$ and pick $r=6$ (7th term) and $r=12$ (13th term). The $m$ and $n$ values are the constant numerical coefficients (x-part cancels when computing the ratio).
$T_7 = \binom{18}{6}\left(\frac{x^{1/3}}{3}\right)^{12}\left(\frac{1}{2x^{2/3}}\right)^6 = \binom{18}{6}\cdot\frac{1}{3^{12}}\cdot\frac{1}{2^6}$, so $m=\binom{18}{6}\cdot 3^{-12}\cdot 2^{-6}$.
$T_{13}=\binom{18}{12}\cdot\frac{1}{3^6}\cdot\frac{1}{2^{12}}\cdot x^{-6}$, so $n=\binom{18}{12}\cdot 3^{-6}\cdot 2^{-12}$.
Since $\binom{18}{6}=\binom{18}{12}$: $\frac{n}{m}=\frac{2^{-12}\cdot3^{-6}}{3^{-12}\cdot2^{-6}}=\frac{3^6}{2^6}=\left(\frac{3}{2}\right)^6$.
$\left(\frac{n}{m}\right)^{1/3}=\left(\frac{3}{2}\right)^2=\frac{9}{4}$.
Correct Answer: 4