It seems to the sum of squares

Let a point $P_k$ be located somewhere on the side $\overline{AB}$ such that it's distances from vertices $B$ and $A$ are $\dfrac{ck}{n}$ and $c - \dfrac{ck}{n}$ respectively.

By Stewart's theorem , we have

$\begin{aligned} & a^2 \left( c - \dfrac{ck}{n} \right) + b^2 \dfrac{ck}{n} = c \left( \dfrac{ck}{n} \left( c - \dfrac{ck}{n} \right) + d^2 \right) \\ \implies & a^2 \left( 1 - \dfrac kn \right) + b^2 \dfrac kn = \dfrac{c^2 k}{n} - \dfrac{c^2 k^2}{n^2} + d^2 \\ \implies & d^2 = a^2 - \dfrac{\left( a^2 - b^2 + c^2 \right)}{n} k + \dfrac{c^2}{n^2} k^2 \\ \implies & d^2 = a^2 - \dfrac{2a^2}{n} k + \dfrac{c^2}{n^2} k^2 & \small {\color{#3D99F6} \left[ \because c^2 = a^2 + b^2 \right]} \end{aligned}$

Note that $d^2 = \overline{CP}_k^2$ and so we have

$\begin{aligned} \sum_{k=1}^{n-1} \overline{CP}_k^2 = \sum_{k=1}^{n-1} d^2 &= a^2 \sum_{k=1}^{n-1} (1) - \dfrac{2a^2}{n} \sum_{k=1}^{n-1} k + \dfrac{c^2}{n^2} \sum_{k=1}^{n-1} k^2 \\ &= a^2 (n-1) - \dfrac{2a^2}{n} \cdot \dfrac{(n-1)n}{2} + \dfrac{c^2}{n^2} \cdot \dfrac{(n-1)n(2n-1)}{6} \\ &= a^2 (n-1) - a^2 (n-1) + \dfrac{(n-1)(2n-1) c^2}{6n} \\ &= \boxed{\dfrac{(n-1)(2n-1) c^2}{6n}} \end{aligned}$

By cosine rule , we have:

$\begin{aligned} \overline{CP}_k^2 & = \overline{BP}_k^2 + \overline{BC}^2 - 2\overline{BP}_k \overline{BC}\cos B \\ \sum_{k=1}^{n-1} \overline{CP}_k^2 & = \sum_{k=1}^{n-1} \left(\left(\frac {kc}n \right)^2 + \overline{BC}^2 - 2\left(\frac {kc}n \right) \overline{BC}\cos B \right) \\ & = \frac {(n-1)n(2n-1)c^2}{6n^2} + (n-1){\color{#3D99F6}\overline{BC}}^2 - 2\left(\frac {n(n-1)c}{2n} \right){\color{#3D99F6}\overline{BC}}\cos B & \small \color{#3D99F6} \text{Note that }\overline{BC} = c \cos B \\ & = \frac {(n-1)(2n-1)c^2}{6n} + (n-1){\color{#3D99F6}c^2\cos^2 B} - (n-1){\color{#3D99F6}c^2\cos^2 B} \\ & = \boxed{\dfrac {(n-1)(2n-1)c^2}{6n}} \end{aligned}$

Shhh it's a trick, don't tell anybody.

If you're too lazy to calculate all the stuff, just put $n=0$ and $n=1$ .

If $n=0$ , there is no line, therefore the sum is $0$ .

If $n=1$ , there is a line whose length is half of $c$ , so the sum is $\dfrac{c^2}{4}$ .

Substitute that, and we can see that only $\dfrac{(n-1)(2n-1)c^2}{6n}$ can be the solution to this problem.

;)