Consider the following two number sequences:

$$
\begin{array}{cccccc}
1, & 2, & 3, & \ldots, & \mathrm{n}-1, & \mathrm{n}, \\
2 \mathrm{n}-1, & 2 \mathrm{n}-3, & 2 \mathrm{n}-5, & \ldots, & 3, & 1
\end{array}
$$

Calculate the sum of the products of the numbers that are directly above or below each other.
The sum of the products is as follows:

![](https://cdn.mathpix.com/cropped/2024_05_02_a12421276cd1ef9995b6g-1.jpg?height=347&width=854&top_left_y=206&top_left_x=625)

The 1st, 2nd, 3rd, ..., $(n-1)$th, $n$th row is the sum of the first $n, n-1, n-2, \ldots, 2, 1$ odd numbers, and thus the sum of the $n$ rows is:

$$
n^{2}+(n-1)^{2}+(n-2)^{2}+\ldots+2^{2}+1^{2}=\frac{n(n+1)(2 n+1)}{6}
$$

(Lajos Kertész.)

The problem was also solved by: D. Barna, F. Beck, F. Bojedain, K. Détshy, M. Devecis, A. Erdős, J. Fekete, E. Freibauer, K. Galambos, K. Goldziher, M. Kallós, H. Kolos, A. Koós, Ö. Kornis, Gy. Krisztián, Gy. Lukhaub, I. Makk, E. Manheim, M. Roth, G. Sasvári, H. Schiffer, E. Schwartz, K. Szabó, J. Weisz.