Example 7. Find the sum of the first $n$ terms of the sequence $1 \cdot 2 \cdots \mathrm{k}, 2 \cdot 3 \cdots(\mathrm{k}+1), 3 \cdot 4 \cdots$ $(k+2), \cdots$.

 untranslated part: $(k+2), \cdots$ 

Note: The untranslated part is due to the direct translation of the mathematical expression which is already in a form that is commonly used in both languages. However, if you prefer a more explicit translation, it would be: $(k+2), \ldots$
Analysis: When $k=1$, the sequence is a beautiful sequence; but when $k$ is any natural number, the sequence remains a beautiful sequence. Therefore, setting $\mathrm{k}=1$ does not provide a solution, so. Below, we will explore the case where $\mathrm{k}=2$.
When $k=2$, the sequence is
$1 \cdot 2, 2 \cdot 3, 3 \cdot 4, \cdots, n \cdot (n+1), \cdots$.

Let the sum of the first $\mathrm{n}$ terms be $\mathrm{S}_{\mathbf{A}}$.
(1) Since $a_{0}=n(n+1)$
$$
=\frac{1}{3}[n(n+1)(n+2)-(n-1) n(n+1)] \text {, so }
$$

$\mathrm{S} .=1 \cdot 2+2 \cdot 3+\cdots+n \cdot(n+1)$
$$
\begin{aligned}
= & \frac{1}{3}[(1 \cdot 2 \cdot 3-0 \cdot 1 \cdot 2)+(2 \cdot 3 \cdot 4-1 \cdot 2 \cdot 3) \\
& +\cdots+n(n+1)(n+2)-(n-1) n(n \\
& +1)] \\
= & \frac{1}{3} n(n+1)(n+2) .
\end{aligned}
$$
(2) Since $a_{4}=n(n+1)$
$=2!\frac{(n+1) n}{2!}=2 \mathbf{C}_{\mathbf{2}+1}^{2}$, so
$$
\begin{aligned}
\mathrm{S} & =1 \cdot 2+2 \cdot 3+\cdots+\mathrm{n}(\mathrm{n}+1) \\
& =2\left(\mathbf{C}_{2}^{2}+\mathbf{C}_{3}^{2}+\cdots+\mathbf{C}_{n+1}^{2}\right) \\
& =2 \mathbf{C}_{n+2}^{3}=\frac{1}{3} n(n+1)(n+2) .
\end{aligned}
$$
(3) Since $[1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5+\cdots$
$$
\begin{aligned}
& +n(n+1)(n+2)]-[1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5 \\
& +\cdots+n(n+1)(n+2)] \\
= & 1 \cdot 2 \cdot 3+(2 \cdot 3 \cdot 4-1 \cdot 2 \cdot 3)+(3 \cdot 4 \cdot 5-2 \cdot 3 \cdot 4) \\
& +\cdots+[n(n+1)(n+2)-(n-1) n(n+1)] \\
& -n(n+1)(n+2) \\
= & \underset{-n \cdot 2 \cdot 2 \cdot 2 \cdot 3+3 \cdot 3 \cdot 4+\cdots+3 n(n+1)}{n(n+1)(n+2)} \\
& =3[1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+n(n+1)] \\
& -n(n+1)(n+2)=0,
\end{aligned}
$$

i.e., $3[1 \cdot 2+2 \cdot 3+\cdots+n(n+1)]$
$=\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)$.
So $S_{2}=1 \cdot 2+2 \cdot 3+\cdots n(n+1)$
$$
=\frac{1}{3} n(n+1)(n+2) \text {. }
$$

From this, we obtain the following three solutions:
Solution one Because $a_{n}=n(n+1) \cdots(n+k-1)$
$$
\begin{aligned}
= & \frac{1}{k+1}[n(n+1) \cdots(n+k)-(n-1) n \cdots \\
& (n+k-1)],
\end{aligned}
$$

So $S_{0}=\frac{1}{k+1}\{[i \cdot 2 \cdots k(k+1)-0 \cdot 1 \cdot 2 \cdots k]$
$$
\begin{array}{l}
\because[2 \cdot 3 \cdots(k+1)(k+2)-1 \cdot 2 \cdots k(k+1)] \\
+[3 \cdot 4 \cdots(k+2)(k+3)-2 \cdot 3 \cdots(k+1)(k+2)] \\
+\cdots+[n(n+1) \cdots(n+k)-(n-1) n \cdots \\
(n+k-1)]=\frac{1}{k+1} n(n+1) \cdots(n+k) .
\end{array}
$$

Solution two Because
$$
\begin{array}{l}
a_{2}=k!\frac{(n+k-1)(n+k-2) \cdots(n+1) n}{k!} \\
=k!C_{n+k-1}^{k},
\end{array}
$$

So $S_{\mathbf{n}}=\mathrm{k}!\left(\mathbf{C}_{k}^{k}+\mathbf{C}_{k+1}^{k}+\mathbf{C}_{k+2}+\cdots\right.$ $\left.+C_{i+k-1}^{*}\right)$
$$
=k!\quad C_{:+k}^{k+1}=\frac{1}{k+1} n(n+1) \cdots(n+k) .
$$

Solution three Because $[1 \cdot 2 \cdots k(k+1)+2 \cdot 3 \cdots(k+2)$
$$
\begin{array}{l}
+\cdots+n(n+1) \cdots(n+k)]-[1 \cdot 2 \cdots k(k+1)+2 \cdot 3 \\
\cdots(k+1)(k+2]+\cdots+n(n+1) \cdots(n+k-1) \\
(n+k)]=(k+1) \cdot 1 \cdot 2 \cdots k+(k+1) \cdot 2 \cdot 3 \cdots(k+1) \\
+\cdots+(k+1) n(n+1) \cdots(n+k-1)-n(n+1) \\
\cdots(n+k-1)(n+k)=0
\end{array}
$$
$$
\begin{array}{l}
\text { i.e., }(k+1)[1 \cdot 2 \cdots k+2 \cdot 3 \cdots(k+1)+\cdots+n(n+1) \\
\cdots(n+k-1)]=n(n+1) \cdots(n+k),
\end{array}
$$

So $S_{0}=\frac{1}{k+1} n(n+1) \cdots(n+k)$.