Can anyone explain the method used for finding the answer to this question? I was practising questions on principles on mathematics. I stumbled onto this question and I don't know where to start. Can anyone please help??
If $P_1\,P_2\,\ldots\,P_n$  is a regular polygon in the $(x,y)$-plane, each side of length $a>0$ (so the $P_i$   are the corners of an $n$-sided figure with sides of equal length $a>0$ ). Find the sum
$$
 S=\sum_{j=2}^{n}\;(\overline{P_1P_j})^2=(\overline{P_1P_2})^2 +(\overline{P_1P_3} )^2 +\ldots+(\overline{P_1P_n})^2 ;
$$ 
 here $\overline{P1 Pj}$   stands for the length of the line form the point $P_1$   to the point $P_j$   (your expression for $S$  will be a function of $a$ , $n$  and a well-known trigonometric function).
Exemple for $n=4$:

 A: Hardmath's hint is excellent. You may also want to use (prove) that :
1) We can express 
$$P_{k+1}=ae^{\frac{2k\pi i}{n}}\;\;,\;\;k=0,1,2,...,n-1$$
2) If $\,z_1=r_1e^{it_1}\,\,,\,z_2=r_2e^{it_2}\,$ , then
$$dist(z_2,z_2)=\overline{z_1z_2}=\sqrt{r_1^2+r_2^2-2r_1r_2\cos(t_2-t_1)}$$
A: Here you have a geometric approach...
As showed in the picture, $O$ represents the (circum)center of the polygon, $\alpha=\frac{2\pi}{n}$ denotes the angle $\widehat{P_1OP_2}$ and $\beta=(k-1)\alpha$ is the angle $\widehat{P_1OP_k}$ where $2\leq k\leq n$ is fixed.

The Law of cosines on the triangle $P_1OP_2$ implies that $\overline{OP_1}^2=\frac{a^2}{2(1-\cos\alpha)}$, and the same reasoning on the triangle $P_1OP_k$ shows that
$$
  \overline{P_1P_k}^2=2\overline{OP_1}^2(1-\cos\beta)
=
  \frac{a^2}{\left(1-\cos{\frac{2\pi}n}\right)}\left(1-\cos{\frac{(k-1)2\pi}n}\right)
$$
Therefore the answer is
$$
  \sum_{k=2}^n{\overline{P_1P_k}^2}
=
  \frac{a^2}{\left(1-\cos{\frac{2\pi}n}\right)}
  \sum_{k=1}^{n-1}{\left(1-\cos{\frac{2k\pi}n}\right)}
$$
Using the fact that $\cos x$ is the real part of $e^{ix}$ and that $\sum_{k=0}^{n-1}e^{i\frac{2k\pi}{n}}=0$ you have that the above sum is equal to
$$
\frac{na^2}{1-\cos{\frac{2\pi}{n}}}
$$
