Let's determine the sum of the following sequence:

$$
S_{n}=2\binom{n}{2}+6\binom{n}{3}+\cdots+(n-2)(n-1)\binom{n}{n-1}+(n-1) n\binom{n}{n}
$$
Let's replace the binomial coefficients appearing in the terms of the sequence $S_{n}$ with their values:

$$
\begin{gathered}
S_{n}=1 \cdot 2 \frac{n(n-1)}{1 \cdot 2}+2 \cdot 3 \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3}+\ldots+ \\
+k(k+1) \frac{n(n-1) \ldots(n-k)}{1 \cdot 2 \ldots(k-1) k(k+1)}+\ldots+(n-2)(n-1) n+(n-1) n \cdot 1
\end{gathered}
$$

Let's perform the simplifications that present themselves:

$$
\begin{gathered}
S_{n}=n(n-1)+n(n-1)(n-2)+\ldots+\frac{n(n-1) \ldots(n-k)}{1 \cdot 2 \ldots(k-1)}+ \\
+\ldots+(n-2)(n-1) n+(n-1) n
\end{gathered}
$$

Let's factor out $n(n-1)$, which appears in every term on the right-hand side:

$$
S_{n}=n(n-1)\left[1+(n-2)+\ldots+\frac{(n-2)(n-3) \ldots(n-k)}{1 \cdot 2 \ldots(k-1)}+\ldots+(n-2)+1\right]
$$

Notice that the terms within the square brackets can be considered as binomial coefficients, and thus

$$
S_{n}=n(n-1)\left[\binom{n-2}{0}+\binom{n-2}{1}+\ldots+\binom{n-2}{k-1}+\ldots+\binom{n-2}{n-3}+\binom{n-2}{n-2}\right]
$$

Since, by the binomial theorem,

$$
2^{n-2}=(1+1)^{n-2}=\binom{n-2}{0}+\binom{n-2}{1}+\ldots+\binom{n-2}{k-1}+\ldots+\binom{n-2}{n-3}+\binom{n-2}{n-2}
$$

it follows that

$$
s_{n}=n(n-1) 2^{n-2}
$$

Erzsébet Pásztor (Makó, József A. g. III. o. t.)