$2 \cdot 65$ When $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ takes all permutations of $(1,2, \cdots, n)$, find the average value of the expression
$$
\left(a_{1}-a_{2}\right)^{2}+\left(a_{2}-a_{3}\right)^{2}+\cdots+\left(a_{n-1}-a_{n}\right)^{2}
$$
[Solution] The number of different permutations of $(1,2, \cdots, n)$ is $n$!. Summing the $n$! corresponding expressions $\sum_{i=1}^{n-1}\left(a_{i}-a_{i+1}\right)^{2}$ and denoting the resulting sum as $S$. For any $1 \leqslant r<s \leqslant n$, $(r-s)^{2}$ appears $2(n-1)!$ times in $S$, and there are $C_{n}^{2}$ such terms, hence we have
$$
\begin{aligned}
S & =2(n-1)!\sum_{r<s}(r-s)^{2} \\
& =2(n-1)!\sum_{s=2}^{n} \sum_{r=1}^{s-1}\left(r^{2}-2 r s+s^{2}\right) \\
& =\frac{1}{3}(n-1)!\sum_{s=1}^{n}\left(2 s^{3}-3 s^{2}+s\right) .
\end{aligned}
$$

Since
$$
\sum_{k=1}^{n} k^{2}=\frac{1}{6} n(n+1)(2 n+1), \sum_{k=1}^{n} k^{3}=\frac{1}{4} n^{2}(n+1)^{2},
$$

we get
$$
\begin{array}{l}
\underline{S}=\frac{1}{3 n}\left(\frac{1}{2} n^{2}(n+1)^{2}-\frac{1}{2} n(n+1)(2 n+1)+\frac{1}{2} n(n+1)\right) \\
=\frac{1}{6}(n+1) n(n-1)=C_{n+1}^{3} \text {, } \\
\end{array}
$$

which means the required average value is $C_{n+1}^{3}$.