4.221 For what natural numbers $n$ and $k$, do the binomial coefficients $C_{n}^{k-1}, C_{n}^{k}, C_{n}^{k+1}$ form an arithmetic sequence?
[Solution] The necessary and sufficient condition for three binomial coefficients to form an arithmetic sequence is that the following equation holds: $\square$
$$C_{n}^{k-1}-2 C_{n}^{k}+C_{n}^{k+1}=0$$

Let $k-1 \geqslant 0$ and $k+1 \leqslant n$, i.e., $1 \leqslant k \leqslant n-1$.
Multiplying both sides of (1) by the positive number $\frac{(k+1)!(n-k+1)!}{n!}$, we get

i.e., $\square$
$$k(k+1)-2(k+1)(n-k+1)+(n-k)(n-k+1)=0,$$

or
$$\begin{array}{l}
n^{2}-4 n k+4 k^{2}-n-2=0 \\
n=(n-2 k)^{2}-2 \\
=u^{2}-2 .
\end{array}$$

where $u$ is a natural number satisfying the relation $u=n-2 k$ (or $u=2 k-n$), from which we can obtain
$$k=k_{1}=\frac{n-u}{2}=\frac{u^{2}-u}{2}-1=C_{u}^{2}-1,$$

or $k=k_{2}=\frac{n+u}{2}=C_{u+1}^{2}-1$.
To make $n$ positive, we should have $u \geqslant 2$. However, when $u=2$, the values of $k_{1}$ and $k_{2}$ do not satisfy inequality (2). When $u \geqslant 3$, we get
$$k_{1}=C_{u}^{2}-1 \geqslant 1 \text {, and } k_{1}=\frac{n-u}{2}2$ when $u \geqslant 3$, the pairs $(n, k)$ determined by $n=u^{2}-2$ and $k=C_{u}^{2}-1$ or $k=C_{u+1}^{2}-1$ satisfy (1).