2. Find all natural numbers $n$ such that some three consecutive binomial coefficients in the expansion of $(a+b)^{n}$ form an arithmetic progression.
Solution. First, note that $n \geq 2$ to have at least three terms in the expansion. Let for $n, k \in \mathbb{N}, k \geq 2$, there exists $k, k < n$ such that $2\binom{n}{k}=\binom{n}{k-1}+\binom{n}{k+1}$. The number $k$ will be chosen from the system

$$
\left\{\begin{array}{l}
n=m^{2}-2 \\
(n-2 k)^{2}=n+2
\end{array}\right.
$$

The equation $n=m^{2}-2$ is substituted into $(n-2 k)^{2}=n+2$ and from there we get $\left(m^{2}-2-2 k\right)^{2}=m^{2}$, which means $m^{2}-2-2 k=m$ or $m^{2}-2-2 k=-m$. We have two possibilities $k_{1}=\frac{m(m-1)}{2}-1$ and $k_{2}=\frac{m(m+1)}{2}-1$.

It is sufficient to prove the statement for one of them, for example for $k_{1}$. We have $k_{1}=\frac{m^{2}-m-2}{2}<\frac{m^{2}-2}{2}<m^{2}-2<n$. Thus, for $k_{1}$ it holds that $k_{1}<n$ and $(n-2 k)^{2}=n+2$.

The latter equality is equivalent to the equality

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

Finally, all the required numbers are of the form $m^{2}-2, m=3,4, \ldots$