Let $${s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}$$,

$${{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j}$$ and

$${{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{C_j}.} }$$

Statement-1 : $${{S_3} = 55 \times {2^9}}$$.
Statement-2 : $${{S_1} = 90 \times {2^8}}$$ and $${{S_2} = 10 \times {2^8}}$$.

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