## Task B-4.1.

If in each term of the binomial expansion $(x-y)^{n}, n \geqslant 2, x, y \neq 0$, we substitute $x=672 y$, the sum of the third and fourth terms will be equal to 0. Determine the 2018th term in the expansion of this binomial.
## Solution.

The sum of the third and fourth term in the binomial expansion is

$$
\begin{array}{cc}
\binom{n}{2} x^{n-2}(-y)^{2}+\binom{n}{3} x^{n-3}(-y)^{3}=0 . & 1 \text { point } \\
\frac{1}{2} n(n-1) \cdot(672 y)^{n-2} y^{2}-\frac{1}{6} n(n-1)(n-2) \cdot(672 y)^{n-3} y^{3}=0 . & 1 \text { point }
\end{array}
$$

Since $n \geqslant 2$, we can divide the last equation by $\frac{1}{2} n(n-1)$. We get

$$
\begin{gathered}
(672 y)^{n-2} y^{2}-\frac{1}{3}(n-2) \cdot(672 y)^{n-3} y^{3}=0 \\
672^{n-2} y^{n}-\frac{1}{3}(n-2) \cdot 672^{n-3} y^{n}=0 \\
672^{n-3} y^{n}\left(672-\frac{n-2}{3}\right)=0
\end{gathered}
$$

Since $y \neq 0$, it must be $672-\frac{n-2}{3}=0$ or $n=672 \cdot 3+2=2018$. This means 1 point that the 2018th term in the expansion of the given binomial is

$$
\binom{2018}{2017} x^{1}(-y)^{2017}=2018 \cdot 672 y \cdot(-y)^{2017}=-1356096 y^{2018}
$$

Note: The result is accepted in any of these forms.