Determine the sum of the first $n$ triangular numbers using higher-order arithmetic sequences.
The formula (10) of the article on higher-order arithmetic sequences (4th number, p. 190) gives the sum of an arithmetic sequence of any order. The sequence $1,3,6,10, \ldots$ is a second-order arithmetic progression, and the task in determining the sum of this sequence is merely to establish the $\alpha_{1}, \alpha_{2}, \alpha_{3}$ appearing in (10).

$$
\begin{aligned}
& \alpha_{n}^{(2)}=\alpha_{1}+\alpha_{2}\binom{n-1}{1}+\alpha_{3}\binom{n-1}{2} \text {, from which } \\
& \alpha_{1}=1 \\
& \alpha_{1}+\alpha_{2}=1+\alpha_{2}=3, \quad \alpha_{2}=2 \\
& \alpha_{1}+\binom{2}{1} \alpha_{2}+\alpha_{3}=5+\alpha_{3}=6, \quad \alpha_{3}=1
\end{aligned}
$$

Thus, if $S_{n+1}$ denotes the sum of the first $n$ triangular numbers, then

$$
S_{n+1}=1\binom{n}{1}+2\binom{n}{2}+1\binom{n}{3}=n+\frac{2 n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}=\frac{n(n+1)(n+2)}{6}
$$

If we substitute $n=1,2,3,4, \ldots$ here, we obtain the sequence $1,4,10,20, \ldots$, which indeed matches the sequence obtained by summing the triangular numbers.

Ottó Villányi (3rd class, Szentendre)