4A. Determine all natural numbers $n$ for which there exist three consecutive coefficients in the expansion of $(a+b)^{n}$ that form an arithmetic progression.
Solution. Let $\binom{n}{k-1},\binom{n}{k}$, and $\binom{n}{k+1}$ be three consecutive coefficients in the expansion of $(a+b)^{n}$, where $0<k<n$. A necessary and sufficient condition for these three coefficients to form an arithmetic progression is:

$$
2\binom{n}{k}=\binom{n}{k-1}+\binom{n}{k+1}
$$

from which, after simplification, we get:

$$
4 k^{2}-4 n k+n^{2}-n-2=0
$$

Thus, the required numbers are all those $n \in \mathbb{N}$ for which equation (1) has at least one integer solution $k$ such that $0<k<n$.

The solutions to the equation are $k_{1 / 2}=\frac{n \pm \sqrt{n+2}}{2}$.

Since $0<k_{1}<n$ or $0<k_{2}<n$ must hold, we get $-n<\sqrt{n+2}<n$. The condition $-n<\sqrt{n+2}$ is always satisfied for $n \in \mathbb{N}$, and the condition $\sqrt{n+2}<n$ is satisfied for $n \geq 3$.

Conversely, if $n \geq 3$, then $0<k_{1}<k_{2}<n$ holds.

For at least one solution to equation (1) to be an integer, it is necessary that $n+2$ is a perfect square, i.e., $n+2=s^{2}$ for some $s \in \mathbb{N}$.

Conversely, if $n+2=s^{2}$ for some $s \in \mathbb{N}$, the solutions to equation (1) are: $k_{1 / 2}=\frac{s(s \pm 1)}{2}-1$, and these two numbers are integers because $s(s-1)$ and $s(s+1)$ are products of consecutive numbers, and thus divisible by 2.

Finally, we conclude that the required numbers are: $n=s^{2}-2$, for $s=3,4,5, \ldots$.