10:22 ** Let the side length of the equilateral $\triangle A B C$ be 1, and there are $n$ equal division points on side $B C$. Along the direction from point $B$ to point $C$, they are sequentially $P_{1}, P_{2}, \cdots, P_{n-1}$. Find $S_{n}=\overrightarrow{A B} \cdot \overrightarrow{A P_{1}}+\overrightarrow{A P_{1}} \cdot \overrightarrow{A P_{2}}+\cdots+\overrightarrow{A P_{n-1}} \cdot \overrightarrow{A C}$ (expressed in terms of $n$).
Let $\overrightarrow{A B}=c, \overrightarrow{A C}=b, \overrightarrow{B C}=a$, and let $\frac{1}{n} \overrightarrow{B C}=p$. Then, $\overrightarrow{A P_{k}}=\overrightarrow{A B}+\overrightarrow{B P_{k}}= \boldsymbol{c}+k \boldsymbol{p} (k=0,1,2, \cdots, n)$, where $\overrightarrow{A P_{0}}=\overrightarrow{A B}$ and $\overrightarrow{A P_{n}}=\overrightarrow{A C}$.

Therefore, $\overrightarrow{A P_{k-1}} \cdot \overrightarrow{A P_{k}}=[\boldsymbol{c}+(k-1) \boldsymbol{p}] \cdot (\boldsymbol{c}+k \boldsymbol{p})=\boldsymbol{c}^{2}+(2 k-1) \boldsymbol{c} \cdot \boldsymbol{p}+k(k-1) \boldsymbol{p}^{2} (k=0,1,2, \cdots, n)$. Since $S_{n}=\overrightarrow{A B} \cdot \overrightarrow{A P_{1}}+\overrightarrow{A P_{1}} \cdot \overrightarrow{A P_{2}}+\cdots+\overrightarrow{A P_{n-1}} \cdot \overrightarrow{A C}$, we have
$$
\begin{array}{r}
S_{n}=n \boldsymbol{c}^{2}+\sum_{k=1}^{n}(2 k-1) \boldsymbol{c} \cdot \boldsymbol{p}+\left[\sum_{k=1}^{n} k(k-1)\right] \boldsymbol{p}^{2}=n \boldsymbol{c}^{2}+n^{2} \boldsymbol{c} \cdot \boldsymbol{p}+ \\
\frac{n(n+1)(n-1)}{3} \boldsymbol{p}^{2}=n \boldsymbol{c}^{2}+n \boldsymbol{c} \cdot(n \boldsymbol{p})+\frac{n^{2}-1}{3 n}(n \boldsymbol{p})^{2}=n \boldsymbol{c}^{2}+n \boldsymbol{c} \cdot \boldsymbol{a}+\frac{n^{2}-1}{3 n} a^{2}.
\end{array}
$$

Given that $|\boldsymbol{a}|=|\boldsymbol{b}|=|\boldsymbol{c}|=1$ and the angle between $\boldsymbol{c}$ and $\boldsymbol{a}$ is $60^{\circ}$, we have $S_{n}=n-\frac{1}{2} n+\frac{n^{2}-1}{3 n}=\frac{5 n^{2}-2}{6 n}$.