Express in closed form the sum

$$
1 \cdot 2^{2} \cdot 3+2 \cdot 3^{2} \cdot 4+\ldots+(n-2)(n-1)^{2} n
$$

where $n$ is a natural number greater than three.
I. solution. Let's denote the sought sum by $S_{n}$ and transform it as follows:

$$
\begin{gathered}
S_{n}=(2-1) 2^{2}(2+1)+(3-1) 3^{2}(3+1)+\ldots+(n-1-1)(n-1)^{2}(n-1+1)= \\
=2^{2}\left(2^{2}-1\right)+3^{2}\left(3^{2}-1\right)+\ldots+(n-1)^{2}\left((n-1)^{2}-1\right)= \\
=2^{4}-2^{2}+3^{4}-3^{2}+\ldots+(n-1)^{4}-(n-1)^{2}
\end{gathered}
$$

Thus, $S_{n}$ can be written as:

$$
S_{n}=1^{4}+2^{4}+\ldots+(n-1)^{4}-\left(1^{2}+2^{2}+\ldots+(n-1)^{2}\right)
$$

The sum in parentheses is known, $\frac{(n-1) n(2 n-1)}{6}$, so to find the closed form of $S_{n}$, we need the closed form of the sum

$$
1^{4}+2^{4}+\ldots+k^{4}
$$

For this, consider the following equalities:

$$
\begin{aligned}
2^{5}= & (1+1)^{5}=1^{5}+5 \cdot 1^{4}+10 \cdot 1^{3}+10 \cdot 1^{2}+5 \cdot 1+1 \\
3^{5}=(2+1)^{5} & =2^{5}+5 \cdot 2^{4}+10 \cdot 2^{3}+10 \cdot 2^{2}+5 \cdot 2+1 \\
& \vdots \\
& (k+1)^{5}=k^{5}+5 \cdot k^{4}+10 \cdot k^{3}+10 \cdot k^{2}+5 \cdot k+1
\end{aligned}
$$

Adding these equalities and rearranging, we get

$$
\begin{gathered}
(k+1)^{5}=1^{5}+5\left(1^{4}+2^{4}+\ldots+k^{4}\right)+10\left(1^{3}+2^{3}+\ldots+k^{3}\right)+10\left(1^{2}+2^{2}+\ldots+k^{2}\right)+ \\
\\
+5(1+2+\ldots+k)+k
\end{gathered}
$$

From this, using the known formulas for the sum of the first $k$ positive integers, the sum of their squares, and the sum of their cubes, after some calculations, we get

$$
1^{4}+2^{4}+\ldots+k^{4}=\frac{k(k+1)(2 k+1)\left(3 k^{2}+3 k-1\right)}{30}
$$

Thus,

$$
S_{n}=\frac{(n-1) n(2 n-1)\left(3 n^{2}-3 n-1\right)}{30}-\frac{(n-1) n(2 n-1)}{6}
$$

and therefore

$$
S_{n}=\frac{(n-2)(n-1) n(n+1)(2 n-1)}{10}
$$

II. solution. Let's seek $S_{n}$ as a fifth-degree polynomial in $n$, i.e.,

$$
p(n)=a n^{5}+b n^{4}+c n^{3}+d n^{2}+e n+f
$$

Our task is to determine the coefficients of the polynomial $p$. Consider $p$ at $n-1$:

$$
p(n-1)=a(n-1)^{5}+b(n-1)^{4}+c(n-1)^{3}+d(n-1)^{2}+e(n-1)+f
$$

Performing the operations and arranging in descending powers of $n$:

$$
\begin{aligned}
p(n-1)= & a \cdot n^{5}+(-5 a+b) n^{4}+(10 a-4 b+c) n^{3}+(-10 a+6 b-3 c+d) n^{2}+ \\
& +(5 a-4 b+3 c-2 d+e) n+(-a+b-c+d-e+f)
\end{aligned}
$$

From the written forms of $p(n)$ and $p(n-1)$:

$$
\begin{gathered}
p(n)-p(n-1)=5 a n^{4}+(-10 a+4 b) n^{3}+(10 a-6 b+3 c) n^{2}+ \\
+(-5 a+4 b-3 c+2 d) n+(a-b+c-d+e)
\end{gathered}
$$

On the other hand, from the original sum:

$$
S_{n}-S_{n-1}=(n-2)(n-1)^{2} n=n^{4}-4 n^{3}+5 n^{2}-2 n
$$

(1) and (2) are equal for all $n>3$ natural numbers if and only if the coefficients of the same powers of $n$ in the two polynomials are equal, i.e.,

$$
\begin{aligned}
5 a & =1 \\
-10 a+4 b & =-4 \\
10 a-6 b+3 c & =5 \\
-5 a+4 b-3 c+2 d & =-2 \\
a-b+c-d+e & =0
\end{aligned}
$$

From this system of equations, $a=\frac{1}{5}, b=-\frac{1}{2}, c=0, d=-\frac{1}{2}, e=\frac{1}{5}$, and thus the polynomial

$$
p(n)=\frac{1}{5} n^{5}-\frac{1}{2} n^{4}+\frac{1}{2} n^{2}-\frac{1}{5} n
$$

satisfies $p(n)-p(n-1)=S_{n}-S_{n-1}$ for all $n>3$ integers. (In fact, $p(n)-p(n-1)=(n-2)(n-1)^{2} \cdot n$ holds for all integers, and even for all real $n$.)

Since, in addition, $p(4)=108=S_{4}$, it is indeed true that $p(n)=S_{n}$ for all admissible values of $n$.

Remark. In the second solution, it was necessary to verify the initial values, since from the mere existence of the polynomial $p$ - which is guaranteed by the solvability of the found system of equations - only the following can be inferred: if $S_{n}$ can be represented as a fifth-degree polynomial in $n$, then this polynomial can only be $p(n)$.