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.

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 ............... .

Tagent at a point $${P_1}$$ {other than $$(0, 0)$$} on the curve $$y = {x^3}$$ meets the curve again at $${P_2}$$. The tangent at $${P_2}$$ meets the curve at $${P_3}$$, and so on. Show that the abscissae of $${P_1},\,{P_2},{P_3}......{P_n},$$ form a G.P. Also find the ratio.

[area $$\left( {\Delta {P_1},{P_2},{P_3}} \right)$$]/[area $$\left( {{P_2},{P_3},{P_4}} \right)$$]