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
$${\tan ^{ - 1}}\left( {1/3} \right) + {\tan ^{ - 1}}\left( {1/7} \right) + ........{\tan ^{ - 1}}\left\{ {1/\left( {{n^2} + n + 1} \right)} \right\} = {\tan ^{ - 1}}\left\{ {n/\left( {n + 2} \right)} \right\}$$

If $$\sum\limits_{r = 0}^{2n} {{a_r}{{\left( {x - 2} \right)}^r}\,\, = \sum\limits_{r = 0}^{2n} {{b_r}{{\left( {x - 3} \right)}^r}} } $$ and $${a_k} = 1$$ for all $$k \ge n,$$ then show that $${b_n} = {}^{2n + 1}{C_{n + 1}}$$

Let $$p \ge 3$$ be an integer and $$\alpha $$, $$\beta $$ be the roots of $${x^2} - \left( {p + 1} \right)x + 1 = 0$$ using mathematical induction show that $${\alpha ^n} + {\beta ^n}.$$
(i) is an integer and (ii) is not divisible by $$p$$