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 $$n$$.

Given that $${C_1} + 2{C_2}x + 3{C_3}{x^2} + ......... + 2n{C_{2n}}{x^{2n - 1}} = 2n{\left( {1 + x} \right)^{2n - 1}}$$
where $${C_r} = {{\left( {2n} \right)\,!} \over {r!\left( {2n - r} \right)!}}\,\,\,\,\,r = 0,1,2,\,............,2n$$
Prove that $${C_1}^2 - 2{C_2}^2 + 3{C_3}^2 - ............ - 2n{C_{2n}}^2 = {\left( { - 1} \right)^n}n{C_n}.$$