Limit + Trignometry

Let

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

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

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

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

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

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

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

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

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

$\displaystyle \lim _{ n\to \infty } \frac { 3f(n) }{ (n+1)(n+2) } =\frac { 3\left( \frac { (n-1)(n-2) }{ 3 } \right) }{ (n+1)(n+2) } =1\\ \therefore \quad p=1,q=1\quad \\ i.e\quad \boxed { p+q=2 }$