The remainder left out when $${8^{2n}} - {\left( {62} \right)^{2n + 1}}$$ is divided by 9 is :

Statement - 1 : $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r} = \left( {n + 2} \right){2^{n - 1}}.} $$
Statement - 2 : $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r} = {{\left( {1 + x} \right)}^n} + nx{{\left( {1 + x} \right)}^{n - 1}}.} $$

In the binomial expansion of $${\left( {a - b} \right)^n},\,\,\,n \ge 5,$$ the sum of $${5^{th}}$$ and $${6^{th}}$$ terms is zero, then $$a/b$$ equals

The sum of the series $${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .....\, - \,.....\, + {}^{20}{C_{10}}$$ is