Prove that $$\sum\limits_{r = 1}^k {{{\left( { - 3} \right)}^{r - 1}}\,\,{}^{3n}{C_{2r - 1}} = 0,} $$ where $$k = \left( {3n} \right)/2$$ and $$n$$ is an even positive integer.

Using mathematical induction, prove that
$${\tan ^{ - 1}}\left( {1/3} \right) + {\tan ^{ - 1}}\left( {1/7} \right) + ........{\tan ^{ - 1}}\left\{ {1/\left( {{n^2} + n + 1} \right)} \right\} = {\tan ^{ - 1}}\left\{ {n/\left( {n + 2} \right)} \right\}$$

For $$0 < \phi < \pi /2,$$ if
$$x = $$$$\sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\phi ,y = \sum\limits_{n = 0}^\infty {{{\sin }^{2n}}\phi ,\,\,\,\,z = \sum\limits_{n = 0}^{} {{{\cos }^{2n}}\phi {{\sin }^{2n}}\phi } } } \infty $$ then

The vertices of a triangle are $$A\left( { - 1, - 7} \right)B\left( {5,\,1} \right)$$ and $$C\left( {1,\,4} \right).$$ The equation of the bisector of the angle $$\angle ABC$$ is ............... .