Problem 1249

Analyzing the use of the dot product, the problem is transformed into a series summation.
Proof: Let $\overrightarrow{A B}=\boldsymbol{c}, \overrightarrow{A C}=\boldsymbol{b}, \overrightarrow{B C}=\boldsymbol{a}$. Let $\frac{1}{n} \overrightarrow{B C}=\boldsymbol{p}$, then $\overrightarrow{A P_{k}}=\overrightarrow{A B}+\overrightarrow{B P_{k}}=\boldsymbol{c}+k \boldsymbol{p}(k=0,1,2, \cdots, n)$, where $\overrightarrow{A P_{0}}=\overrightarrow{A B}, \overrightarrow{A P_{n}}=\overrightarrow{A C}$
Therefore, $\overrightarrow{A P_{k-1}} \cdot \overrightarrow{A P_{k}}=[\boldsymbol{c}+(k-1) \boldsymbol{p}] \cdot[\boldsymbol{c}+k \boldsymbol{p}]$
$$\begin{aligned} = & \boldsymbol{c}^{2}+(2 k-1) \boldsymbol{c} \cdot \boldsymbol{p}+k(k-1) \boldsymbol{p}^{2} \\ & (k=1,2, \cdots, n) \end{aligned}$$

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

Since $\quad|\boldsymbol{a}|=|\boldsymbol{b}|=|\boldsymbol{c}|=1$
and the angle between $\boldsymbol{c}$ and $\boldsymbol{a}$ is $120^{\circ}$
Therefore, $S_{n}=n-\frac{1}{2} n+\frac{n^{2}-1}{3 n}=\frac{5 n^{2}-2}{6 n}$