## Task B-4.3.

Determine all numbers $n$ for which

$$
\binom{2022}{n+1}-\binom{2021}{n} \leqslant \frac{1}{2023}\binom{2023}{n+2}
$$
## Solution.

The number $n$ is an integer for which $0 \leqslant n \leqslant 2021$.

Let's write the binomial coefficients using factorials and apply their properties. Then we have:

$$
\begin{aligned}
& \frac{2022!}{(n+1)!(2021-n)!}-\frac{2021!}{n!(2021-n)!} \leqslant \frac{1}{2023} \cdot \frac{2023!}{(n+2)!(2021-n)!} \\
& \frac{2022 \cdot 2021!}{(n+1) \cdot n!}-\frac{2021!}{n!} \leqslant \frac{1}{2023} \cdot \frac{2023 \cdot 2022 \cdot 2021!}{(n+2)(n+1) \cdot n!} \\
& \frac{2022}{n+1}-1 \leqslant \frac{2022}{(n+2)(n+1)} \\
& 2022(n+2)-(n+1)(n+2) \leqslant 2022 \\
& 2022 n+4044-n^{2}-3 n-2-2022 \leqslant 0 \\
& -n^{2}+2019 n+2020 \leqslant 0 \\
& -n^{2}+2020 n-n+2020 \leqslant 0 \\
& (2020-n)(n+1) \leqslant 0
\end{aligned}
$$

The solution to this inequality is all natural numbers $n \geqslant 2020$, and 1 point

since $0 \leqslant n \leqslant 2021$, the final solution is $n \in\{2020,2021\}$.