Task A-4.2. (8 points)

The third term in the expansion of $\left(2 \cdot \sqrt[n]{2^{-1}}+\frac{4}{\sqrt[4-n]{4}}\right)^{6}$ is 240. Determine $n$.
## Solution.

The third term in the expansion of the given binomial is $\binom{6}{2} \cdot\left(2^{\frac{n-1}{n}}\right)^{4} \cdot\left(4^{\frac{3-n}{4-n}}\right)^{2}$.

By equating it to 240, we get $15 \cdot 16^{\frac{n-1}{n}} \cdot 16^{\frac{3-n}{4-n}}=240$, which simplifies to $16^{\frac{n-1}{n}+\frac{3-n}{4-n}}=16$, equivalent to the equation $\frac{n-1}{n}+\frac{3-n}{4-n}=1$.

From here, we obtain the quadratic equation $n^{2}-4 n+4=0$ with the single solution $n=2$.

For 4 points: correct method with a calculation error, must be accurate (*).

For 8 points: correct result and method.