## Task B-4.2.

In the expansion of $\left(\sqrt{6^{x}}+\frac{1}{\sqrt{6^{x+1}}}\right)^{n}$, the ratio of the coefficients of the third and second terms is $7: 2$. Determine $x$ such that the fourth term in the binomial expansion is equal to 2016.
## Solution.

If the ratio of the coefficients of the third and second terms is $7: 2$, then

$$
\begin{gathered}
\frac{\binom{n}{2}}{\binom{n}{1}}=\frac{7}{2} \\
\frac{n(n-1)}{2 n}=\frac{7}{2} \\
n=8 .
\end{gathered}
$$

The fourth term in the binomial expansion is

$$
\begin{array}{cc}
\binom{8}{3}\left(\sqrt{6^{x}}\right)^{5} \cdot\left(\frac{1}{\sqrt{6^{x+1}}}\right)^{3}=2016 & 1 \text{ point} \\
56 \cdot 6^{\frac{5 x}{2}} \cdot 6^{\frac{-x-3}{2}}=2016 & \\
6^{\frac{5 x}{2}} \cdot 6^{\frac{-3 x-3}{2}}=36 & 1 \text{ point} \\
\frac{5 x}{2}-\frac{3 x+3}{2}=2 & 1 \text{ point} \\
5 x-3 x-3=4 & \\
x=\frac{7}{2} . & 1 \text{ point}
\end{array}
$$