Let's calculate the value of $S_{n}=(1 \cdot 2)+(1 \cdot 3)+\ldots+(1 \cdot n)+(2 \cdot 3)+(2 \cdot 4+\ldots+(2 \cdot n)+\ldots+(n-1) \cdot n$.
I. solution: Write the sum in the following form:

$$
\begin{aligned}
S_{n}=1 \cdot 2+1 \cdot 3 & +1 \cdot 4+\ldots+1(n-1)+1 n+ \\
+2 \cdot 3 & +2 \cdot 4+\ldots+2(n-1)+2 n+ \\
+ & 3 \cdot 4+\ldots+3(n-1)+3 n+ \\
+ & (n-1) n
\end{aligned}
$$

We sum the terms column by column

$$
\begin{aligned}
& s_{1}=0 \\
& s_{2}=1 \cdot 2 \quad=2 \\
& s_{3}=1 \cdot 3+2 \cdot 3 \quad=(1+2) 3=9 \\
& s_{4}=1 \cdot 4+2 \cdot 4+3 \cdot 4=(1+2+3) 4=24 \\
& \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
& s_{n}=(1+2+3+\ldots+n-1) n
\end{aligned}
$$

That is,

$$
S_{n}=0+1 \cdot 2+(1+2) 3+\ldots+\frac{(n-1) n^{2}}{2}
$$

Since the $n$-th term is a cubic polynomial expression in $n$, this is a third-order arithmetic sequence. (This can also be easily proven by forming the third difference sequence of the sequence.)

Our task is thus solvable according to formula (10) if we determine the constants $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ in the expression $s_{n}=\alpha_{1}+\alpha_{2}\binom{n-1}{1}+\alpha_{3}\binom{n-1}{2} \alpha_{4}\binom{n-1}{3}$. By setting $n=1,2,3,4$ in turn, we get $\alpha_{1}=0, \alpha_{2}=2, a_{3}=5, a_{4}=3$, so

$$
S_{n}=2\binom{n}{2}+5\binom{n}{3}+3\binom{n}{4}
$$

II. solution: Complete the sequence rearranged in the previous manner as follows:

$$
\begin{gathered}
1 \cdot 1+1 \cdot 2+1 \cdot 3+1 \cdot 4+\ldots+1(n-1)+1 n+ \\
2 \cdot 1+2 \cdot 2+2 \cdot 3+2 \cdot 4+\ldots+2(n-1)+2 n+ \\
\ldots \\
n \cdot 1+n \cdot 2+n \cdot 3+n \cdot 4+\ldots+n(n-1)+n \cdot n
\end{gathered}
$$

We see that the first $n$ square numbers are placed along one diagonal, to the right of which are the elements of the original sequence, and to the left are the same elements. Since the sum of the entire expression is $\binom{n+1}{2}^{2}$, if we subtract the sum of the square numbers from this, we get twice the desired sum. The sum of the square numbers is $\frac{n(n+1)(2 n+1)}{6}$, so

$$
2 S_{n}=\left(\frac{n(n+1)}{2}\right)^{2}-\frac{n(n+1)(2 n+1)}{6}
$$

Factoring out $n(n+1)$,

$$
\begin{aligned}
& \quad S_{n}=n(n+1)\left[\frac{n(n+1)}{8}-\frac{2 n+1}{12}\right]=n(n+1)\left[\frac{3 n(n+1)-4 n-2}{24}\right]= \\
& =n(n+1)\left[\frac{3 n^{2}+3 n-4 n-2}{24}\right]=\frac{n(n+1)(n-1)(3 n+2)}{24}
\end{aligned}
$$

III. solution: Let the sum of the squares of the first $n$ numbers be $S_{n}^{2}$, and the sum of the cubes be $S_{n}^{3}$.

Consider again the rearranged form of the sequence used in the first solution. In this form, the sum of the terms in the $(k-1)$-th column is $k\binom{k}{2}=\frac{1}{2}\left(k^{3}-k^{2}\right)$, which needs to be summed for $k=2,3, \ldots, n$, i.e., since it gives 0 for $k=1$, we can sum from 1 to $n$. Thus we get $S_{n}=\frac{1}{2}\left(S_{n}^{3}-S_{n}^{2}\right)$.

Now complete the sequence in the same way as in the previous solution.

$$
\begin{aligned}
& 1 \cdot 1+1 \cdot 2+1 \cdot 3+\ldots+1 \cdot n \\
& 2 \cdot 1+2 \cdot 2+2 \cdot 3+\ldots+2 \cdot n \\
& \vdots \\
& n \cdot 1+n \cdot 2+n \cdot 3+\ldots+n \cdot n
\end{aligned}
$$

Now let's see what the excess is that was added to the $k$-th row. Originally, it was:

$$
k(k+1)+k(k+2)+\ldots+k n
$$

the new row is

$$
k \cdot 1+k \cdot 2+\ldots k \cdot k+k \cdot(k+1)+\ldots+k \cdot n
$$

the excess is

$$
k(1+2+\ldots+k)=k\binom{k+1}{2}=\frac{k^{3}+k^{2}}{2}
$$

Summing this for $k=1, \ldots n$, we get $\frac{1}{2}\left(S_{n}^{3}+S_{n}^{2}\right)$. This must be subtracted from the total sum to get $S_{n}$.

$$
S_{n}=\binom{n+1}{2}^{2}-\frac{1}{2}\left(S_{n}^{3}+S_{n}^{2}\right)
$$

but as we already found,

$$
S_{n}=\frac{1}{2}\left(S_{n}^{3}-S_{n}^{2}\right)
$$

Adding the two equations and dividing by 2, we get:

$$
S_{n}=\frac{1}{2}\left[\binom{n+1}{2}^{2}-S_{n}^{2}\right]=\frac{1}{2}\left[\binom{n+1}{2}^{2}-\frac{n(n+1)(2 n+1)}{6}\right]
$$

which can be factored as before.

Although this solution is more cumbersome than the previous one, its interesting feature is that it gives the sum of the cubes of the first $n$ numbers. If we do not add the two equations but subtract one from the other, we get:

$$
S_{n}^{3}=\binom{n+1}{2}^{2}
$$

We could get the same result by comparing the first equation with the starting equation of the first solution.