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.

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 $$p$$

Using induction or otherwise, prove that for any non-negative integers $$m$$, $$n$$, $$r$$ and $$k$$ ,
$$\sum\limits_{m = 0}^k {\left( {n - m} \right)} {{\left( {r + m} \right)!} \over {m!}} = {{\left( {r + k + 1} \right)!} \over {k!}}\left[ {{n \over {r + 1}} - {k \over {r + 2}}} \right]$$