## Task B-4.2.

Determine the natural number $n$, so that the ratio of the seventh term counting from the beginning to the seventh term counting from the end, in the expansion of the binomial $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n}$, is equal to $\frac{1}{6}$.
## Solution.

In the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n}=\left(2^{\frac{1}{3}}+3^{-\frac{1}{3}}\right)^{n}$, the seventh term from the beginning is

$$
\binom{n}{6}\left(2^{\frac{1}{3}}\right)^{n-6}\left(3^{-\frac{1}{3}}\right)^{6}
$$

and the seventh term from the end is

$$
\binom{n}{6}\left(2^{\frac{1}{3}}\right)^{6}\left(3^{-\frac{1}{3}}\right)^{n-6}
$$

Then,

$$
\begin{aligned}
& \frac{\binom{n}{6}\left(2^{\frac{1}{3}}\right)^{n-6}\left(3^{-\frac{1}{3}}\right)^{6}}{\binom{n}{6}\left(2^{\frac{1}{3}}\right)^{6}\left(3^{-\frac{1}{3}}\right)^{n-6}}=\frac{1}{6} \\
& 2^{\frac{n-12}{3}} \cdot 3^{\frac{n-12}{3}}=\frac{1}{6} \\
& 6^{\frac{n-12}{3}}=6^{-1} \\
& n=9
\end{aligned}
$$

$$
1 \text { point }
$$