356. Find the sum of all products of the form $1 \cdot 2, 1 \cdot 3, \ldots$ $n(n-1)$, that can be formed from the numbers $1, 2, 3, \ldots n$.
Solution. Let's take the sum of all natural numbers up to $n$ inclusive:

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

Square this equality:

$$
\begin{gathered}
S_{1}^{2}=1^{2}+2^{2}+\ldots+(n-1)^{2}+n^{2}+2[1 \cdot 2+1 \cdot 3+\ldots+ \\
+n(n-1)]
\end{gathered}
$$

Denote the sum of all double products by $\mathrm{X}$, i.e.

$$
\mathbf{X}=1 \cdot 2+1 \cdot 3+\ldots+n(n-1)
$$

Then

$$
\frac{n^{2}(n+1)^{2}}{4}=1^{2}+2^{2}+\ldots+(n-1)^{2}+n^{2}+2 \mathrm{X}
$$

or

$$
\begin{gathered}
\frac{n^{2}(n+1)^{2}}{4}=\frac{n(n+1)(2 n+1)}{6}+2 \mathrm{X} \\
2 \mathrm{X}=\frac{n^{2}(n+1)^{2}}{4}-\frac{n(n+1)(2 n+1)}{6}=\frac{n(n+1)}{12}\left(3 n^{2}-n-2\right)= \\
=\frac{n(n+1)}{12}(n-1)(3 n+2)
\end{gathered}
$$

From which

$$
\mathrm{X}=\frac{n(n+1)(n-1)(3 n+2)}{24}
$$