The Stars of 2017

Let's call {2017, 2} the star obtained by skipping just a single vertex (that is going to the second vertex from the current one). The next one, {2017, 3}, will be obtained by going to every third vertex, etc. The last possible start will have lines going to the last vertex before reaching half way around and will be called {2017, 1008}. The sum of the internal angles of a star designated {p, q} is

$p\times 180-q\times 360$

This can be shown by using triangles made by two vertices following each other within the star and the center of the star. The star can be decomposed into $p$ such triangles, each contributing $180^\circ$ . The sum of the internal angles of the star is then obtained by subtracting the angle made by the triangles at the center. This angle is $360^\circ$ times the number of times the line making up the star, and consequently the line of triangles, wraps itself around the center. This number is $q$ .

We can list the angles totals for the stars in order, starting with {2017, 2}:

$(2017-4)\times 180, (2017-6)\times 180, (2017-8)\times 180, (2017-10)\times 180,... 3\times 180, 1\times 180$

Basically what we need to do is add all the odd numbers starting with 1 and ending with 2013, then multiply the answer by 180. The sum of all odd numbers from 1 to $n$ is

$(\frac{n+1}{2})^2$

So our total of the angles will be $(\frac{2013+1}{2})^2\times 180=1007^2\times 180$ .

This total will need to be divided by $p=2017$ , since there are 2017 angles per star, and by the total number of stars, which is $1007$ .