Let's find the sum of the following series:

$$
2(1 \cdot 2)+2(1 \cdot 3)+\ldots+2 \cdot(1 \cdot n)+2 \cdot(2 \cdot 3)+\ldots+2 \cdot(2 \cdot n)+\ldots+2(n-1) n
$$
Minthogy

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

therefore, the sought sum

$$
\begin{gathered}
S=(1+2+2+\ldots+n)^{2}-\left(1^{2}+2^{2}+3^{2}+\ldots+n^{2}\right)= \\
=\left(\frac{n(1+n)}{2}\right)^{3}-\frac{n(n+1)(2 n+1)}{6}= \\
=\frac{n(n+1)[3 n(n+1)-(4 n+2)]}{12}=\frac{n(n+1)\left(3 n^{2}-n-2\right)}{12}
\end{gathered}
$$

(Julius Schwarz, Budapest.)

The problem was also solved by. Ádámffy E., Csada I., Dömény I., Erdős V., Fodor H., Földes R., Fuchs I., Füstös P., Haar A., Harsányi Z., Heimlich P., Jánosy Gy., Kiss J., Krampera Gy., Kräuter F., Kürti I., Messer P., Miklóssy K., Rosenberg J., Ruvald S., Sárközy P., Schwarz O., Singer D., Sonnenfeld J., Steiner D., Tandlich E.