Let's calculate the sum $S_{n}=1 \cdot 2^{2}+2 \cdot 3^{2}+3 \cdot 4^{2}+\ldots+n(n+1)^{2}$.
We can determine that the terms of the expression to be summed form a third-order arithmetic progression. Indeed,

| 4 |  | 18 |  | 48 |  |  | 100 |  | 180 | $\ldots$ |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | 14 |  | 30 |  | 52 |  | 80 | $\ldots$ |  |  |
|  | 16 |  | 22 |  |  | 28 | $\ldots$ |  |  |  |
|  |  | 6 |  | 6 | $\ldots$ |  |  |  |  |  |

Therefore, we are essentially dealing with a case similar to problem 231, but now we need to determine $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ from (10).

$$
\begin{aligned}
4 & =\alpha_{1} \\
18 & =4+\alpha_{2}, \quad \alpha_{2}=14 \\
48 & =4+14\binom{2}{1}+\alpha_{3}, \quad \alpha_{3}=16 \\
100 & =4+14\binom{3}{1}+16\binom{3}{2}+\alpha_{4}, \quad \alpha_{4}=6
\end{aligned}
$$

Thus, based on (10) (where $S_{n+1}$ in that context corresponds to $S_{n}$ here),

$$
\begin{gathered}
S_{n}=4 n+14\binom{n}{2}+16\binom{n}{3}+6\binom{n}{4}= \\
=\frac{n}{24}\left\{96+168 n-168+64\left(n^{2}-3 n+2\right)+6\left(n^{3}-6 n^{2}+11 n-6\right)\right\}= \\
=\frac{n}{12}\left\{3 n^{3}+14 n^{2}+21 n+10\right\}=\frac{n(n+1)(n+2)(3 n+5)}{12} .
\end{gathered}
$$

If $n=1,2,3,4$, then it gives the sequence $4,22,70,170$, which is the sequence that results from summing the sequence $4,18,48,100$.