W.L.O.G we can suppose that points $ P_{1},\ P_{2}, \dots, \ P_{n}$ are arranged in this order on line $ l.$ For $ n = 3$ we get that $ \boxed{S_{3} = 1}$ (just applying Stewart's relation). Now let's try to compute $ S_{4}.$ Applying the anterior case for points $ (P_{1}, P_{2}, P_{4})$ and $ (P_{1}, P_{3}, P_{4})$ we'll obtain the following system: \[ \left \{\begin{array}{ccc} \frac {c_{1}^2\cdot P_{1}P_{3}}{d_{1}} - \frac {c_{2}^2\cdot P_{2}P_{3}}{d_{2}} + \frac {c_{4}^2\cdot P_{4}P_{3}}{d_{4}} = 1 \\ \\ \frac {c_{1}^2\cdot P_{1}P_{2}}{d_{1}} - \frac {c_{3}^2\cdot P_{3}P_{2}}{d_{3}} + \frac {c_{4}^2\cdot P_{4}P_{2}}{d_{4}} = 1 \end{array}\right. \] By substracting the first relation from the second one we'll obtain: \[ - P_{2}P_{3}\cdot\frac {c_{1}^2}{d_{1}} + P_{2}P_{3}\cdot\frac {c_{2}^2}{d_{2}} - P_{2}P_{3}\cdot\frac {c_{3}^2}{d_{3}} + P_{2}P_{3}\cdot\frac {c_{4}^2}{d_{4}} = 0 \] what rewrittes as \[ - \frac {c_{1}^2}{d_{1}} + \frac {c_{2}^2}{d_{2}} - \frac {c_{3}^2}{d_{3}} + \frac {c_{4}^2}{d_{4}} = 0. \] So $ \boxed{S_{4} = 0}$ We claim that $ S_{n} = 0, \ (\forall) n\geq 4.$ The proof is by induction on $ n.$ For $ n = 4$ it's true. We just need to prove: \[ \boxed{S_{n - 1} = 0\Longrightarrow S_{n} = 0} \] Let's say that $ n$ is odd. So $ n - 1$ is even. Applying the previous step for points $ (P_{1}, P_{2}, \dots , P_{n - 2}, P_{n})$ and $ (P_{1}, P_{2}, \dots, P_{n - 3}, P_{n - 1}, P_{n})$ we'll obtain the following system: \[ \left \{\begin{array}{lcl} - P_{1}P_{n - 1}\cdot\frac {c_{1}^2}{d_{1}} + P_{2}P_{n - 1}\cdot\frac {c_{2}^2}{d_{2}} - \dots + P_{n - 3}P_{n - 1}\cdot\frac {c_{n - 3}^2}{d_{n - 3}} - \boxed{P_{n - 2}P_{n - 1}}\cdot\frac {c_{n - 2}^2}{d_{n - 2}} + P_{n}P_{n - 1}\cdot\frac {c_{n}^2}{d_{n}} = 0 \\ \\ - P_{1}P_{n - 2}\cdot\frac {c_{1}^2}{d_{1}} + P_{2}P_{n - 2}\cdot\frac {c_{2}^2}{d_{2}} - \dots + P_{n - 3}P_{n - 2}\cdot\frac {c_{n - 3}^2}{d_{n - 3}} - \boxed{P_{n - 1}P_{n - 2}}\cdot\frac {c_{n - 1}^2}{d_{n - 1}} + P_{n}P_{n - 2}\cdot\frac {c_{n}^2}{d_{n}} = 0 \end{array}\right. \] By substracting the first relation from the second one we'll obtain: $ P_{n - 1}P_{n - 2}\cdot\frac {c_{1}^2}{d_{1}} - P_{n - 1}P_{n - 2}\cdot\frac {c_{2}^2}{d_{2}} + \dots - P_{n - 1}P_{n - 2}\cdot\frac {c_{n - 3}^2}{d_{n - 3}} + P_{n - 1}P_{n - 2}\cdot\frac {c_{n - 2}^2}{d_{n - 2}} -$ $ P_{n - 1}P_{n - 2}\cdot\frac {c_{n - 1}^2}{d_{n - 1}} + P_{n - 1}P_{n - 2}\cdot\frac {c_{n}^2}{d_{n}} = 0$ what yields: $ S_{n} = 0.$ If $ n$ is even we'll proceed as above. So $ \boxed{S_{n} = 0, \ (\forall)n\geq 4}$