Given $${s_n} = 1 + q + {q^2} + ...... + {q^2};$$
$${S_n} = 1 + {{q + 1} \over 2} + {\left( {{{q + 1} \over 2}} \right)^2} + ........ + {\left( {{{q + 1} \over 2}} \right)^n}\,\,\,,q \ne 1$$
Prove that $${}^{n + 1}{C_1} + {}^{n + 1}{C_2}{s_1} + {}^{n + 1}{C_3}{s_2} + ..... + {}^{n + 1}{C_n}{s_n} = {2^n}{S_n}$$

If $${\left( {1 + x} \right)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ..... + {C_n}{x^n}$$ then show that the sum of the products of the $${C_i}s$$ taken two at a time, represented $$\sum\limits_{0 \le i < j \le n} {\sum {{C_i}{C_j}} } $$ is equal to $${2^{2n - 1}} - {{\left( {2n} \right)!} \over {2{{\left( {n!} \right)}^2}}}$$

Use mathematical Induction to prove : If $$n$$ is any odd positive integer, then $$n\left( {{n^2} - 1} \right)$$ is divisible by 24.

Prove that $${7^{2n}} + \left( {{2^{3n - 3}}} \right)\left( {3n - 1} \right)$$ is divisible by 25 for any natural number $$n$$.