Find the Function !

Let

$x=-i.cot\frac { k\pi }{ n }$

$\large\large\large\large\therefore \quad \frac { x+1 }{ x-1 } ={ e }^{ i\frac { 2\pi }{ n } }$

$\large\Rightarrow \quad { \left( \frac { x+1 }{ x-1 } \right) }^{ n }=1$

$\therefore \quad { (x+1) }^{ n }-{ (x-1) }^{ n }=0$

$\therefore \quad { C }_{ 1 }{ x }^{ n-1 }+{ C }_{ 3 }{ x }^{ n-3 }+....+{ C }_{ n }{ x }^{ 0 }=0$

Note that root of this equation is $-i.cot\left( \frac { k\pi }{ n } \right)$

$\therefore \quad \displaystyle \sum _{ k=1 }^{ n-1 }{ { \left( i.cot\frac { k\pi }{ n } \right) }^{ 2 } } =(\sum _{ k=1 }^{ n-1 }{ { \left( -i.cot\frac { k\pi }{ n } \right) }^{ 2 } } )-2\sum _{ 1\le p \le }{ \sum _{ q\le (n-1) }{ \left( \left( i.cot\frac { p\pi }{ n } \right) .\left( i.cot\frac { q\pi }{ n } \right) \right) } }$

$\large\qquad \qquad \qquad \qquad \qquad =\quad 0-\frac { 2.{ C }_{ 3 } }{ { C }_{ 1 } }$

$\large\large\therefore \quad \boxed { f\left( n \right) =\frac { (n-1)(n-2) }{ 3 } }$

In this problem I showed that $g(n) \; = \; \sum_{m=1}^{n-1} \cot^2 \tfrac{m\pi}{2n} \; = \; \tfrac13(n-1)(2n-1)$ By symmetry it is clear that $f(2n) \; = \; 2g(n)$ and hence $f(50) \; = \; \tfrac23 \times 24 \times 49 = \boxed{784}$