Let's calculate the sum of the first $n$ terms of the following series:

$$
0 \cdot 1^{2}+1 \cdot 2^{2}+2 \cdot 3^{2}+3 \cdot 4^{2}+4 \cdot 5^{2}+\ldots=0+4+18+48+100+\ldots
$$
The given series can also be written as:

$$
\begin{aligned}
& (1-1) 1^{2}+(2-1) 2^{2}+(3-1) 3^{2}+\ldots+(n-1) n^{2}= \\
= & \left(1^{3}+2^{3}+3^{3}+\ldots+n^{3}\right)-\left(1^{2}+2^{2}+3^{2}+\ldots+n^{2}\right)= \\
= & \frac{n^{2}(n+1)^{2}}{4}-\frac{n(n+1)(2 n+1)}{6}=\binom{n+1}{3} \cdot \frac{3 n+2}{2}
\end{aligned}
$$

(Mihály Fekete, Zenta.)

The problem was also solved by: L. Bánó, I. Csada, K. Epstein, H. Fodor, A. Haar, P. Heimlich, E. Kirchknopf, J. Kiss, P. Merse, A. Paunz, S. Pichler, P. Sárközy, Gy. Schuster, O. Schwarz, B. Tóth, J. Tóth, V. Wáhl.