Let $$p$$ be a prime and $$m$$ a positive integer. By mathematical induction on $$m$$, or otherwise, prove that whenever $$r$$ is an integer such that $$p$$ does not divide $$r$$, $$p$$ divides $${}^{np}{C_r},$$

[Hint: You may use the fact that $${\left( {1 + x} \right)^{\left( {m + 1} \right)p}} = {\left( {1 + x} \right)^p}{\left( {1 + x} \right)^{mp}}$$]

Let $$0 < {A_i} < n$$ for $$i = 1,\,2....,\,n.$$ Use mathematical induction to prove that $$$\sin {A_1} + \sin {A_2}....... + \sin {A_n} \le n\,\sin \,\,\left( {{{{A_1} + {A_2} + ...... + {A_n}} \over n}} \right)$$$

where $$ \ge 1$$ is a natural number. {You may use the fact that $$p\sin x + \left( {1 - p} \right)\sin y \le \sin \left[ {px + \left( {1 - p} \right)y} \right],$$ where $$0 \le p \le 1$$ and $$0 \le x,y \le \pi .$$}

Using mathematical induction prove that for every integer $$n \ge 1,\,\,\left( {{3^{2n}} - 1} \right)$$ is divisible by $${2^{n + 2}}$$ but not by $${2^{n + 3}}$$.

Let $$n$$ be a positive integer and $${\left( {1 + x + {x^2}} \right)^n} = {a_0} + {a_1}x + ............ + {a_{2n}}{x^{2n}}$$
Show that $$a_0^2 - a_1^2 + a_2^2...... + {a_{2n}}{}^2 = {a_n}$$