Find the sum of all distinct two-factor products that can be made from the numbers $1,2,3, \ldots, n$, i.e., $[1 \cdot 2,1 \cdot 3, \ldots, 1 \cdot n, 2 \cdot 3,2 \cdot 4, \ldots,(n-1) n]$.
Let us form the sum $S^{\prime}$ from the $n \cdot n$ two-factor products, whose first $n$ terms have the first factor 1, the second factor being the numbers $1,2, \ldots, n$; in the $n+1, n+2, \ldots, 2 n$-th terms, the first factor is 2, the second factor is the numbers $1,2, \ldots, n$; in the $2 n+1,2 n+2, \ldots, 3 n$-th terms, the first factor is 3, the second factor is the numbers $1,2, \ldots, n$, and so on, finally in the $(n-1) n+1,(n-1) n+2, \ldots,(n-1) n+n=n \cdot n$-th term, the first factor is $n$, the second factor is the numbers $1,2, \ldots, n$. This sum can be written, by factoring out the equal first factors, then the equal sums of the second factors, and finally applying the formula for the sum of an arithmetic series, as follows:

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

In this sum, every product required by our problem appears, but there are also unnecessary terms. On the one hand, these are the products of two equal numbers, i.e., the squares from $1^{2}$ to $n^{2}$, whose sum, according to the known formula, is

\[
S^{\prime \prime}=1^{2}+2^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}
\]

On the other hand, those in which the first factor is greater than the second, because these products also appear with the factors in the (increasing) reversed order. Thus, every required product appears twice in $S^{\prime}$, and there are no other unnecessary terms. Therefore, the desired sum is half the difference between $S^{\prime}$ and $S^{\prime \prime}$:

\[
\begin{gathered}
S=\frac{1}{2}\left(S^{\prime}-S^{\prime \prime}\right)=\frac{n(n+1)}{4}\left(\frac{n(n+1)}{2}-\frac{2 n+1}{3}\right)= \\
=\frac{n(n+1)\left(3 n^{2}-n-2\right)}{24}=\frac{n(n+1) 3(n-1)\left(n+\frac{2}{3}\right)}{24}= \\
=\frac{(n-1) n(n+1)(3 n+2)}{24}
\end{gathered}
\]

(After simplification, we calculated the roots of the quadratic polynomial $3 n^{2}-n-2$, and then we factored the polynomial using its roots and the leading coefficient 3.)

Sász Domokos (Bp. V., Eötvös J. g. III. o. t.)

Remark: The relationship $S=\left(S^{\prime}-S^{\prime \prime}\right) / 2$ can also be derived from the identity for the square of a polynomial:

\[
\begin{aligned}
(1+2+\ldots+n)^{2}=\left(1^{2}+\right. & \left.2^{2}+\ldots+n^{2}\right)+2[1 \cdot 2+1 \cdot 3+\ldots+ \\
& +(n-1) \cdot n]
\end{aligned}
\]

i.e., $S^{\prime}=S^{\prime \prime}+2 S$