The number $111......1 $ ( $ 91$ times) is

Let ${s_1} = \mathop \sum \limits_{j = 1}^{10} j\left( {j - 1} \right)\left( {\begin{array}{*{20}{c}}{10}\\j\end{array}} \right)\;,$$\;{s_2} = \mathop \sum \limits_{j = 1}^{10} j\;\left( {\begin{array}{*{20}{c}}{10}\\j\end{array}} \right)\;and,$${s_3} = \mathop \sum \limits_{j = 1}^{10} {j^2}\left( {\begin{array}{*{20}{c}}{10}\\j\end{array}} \right)\;,\;$

Statement $-1$:${s_3} = 55 \times {2^9}$

Statement $-2$: ${s_1} = 90 \times {2^8}\;$ and ${s_2} = 10 \times {2^8}$ 

Let ${\left( {1 + x + {x^2}} \right)^{20}}\left( {2x + 1} \right) = {a_0} + {a_1}{x^1} + {a_2}{x^2} + ... + {a_{41}}{x^{41}}$ , then $\frac{{{a_0}}}{1} + \frac{{{a_1}}}{2} + .... + \frac{{{a_{41}}}}{{42}}$ is equal to 

The sum of the co-efficients of all odd degree terms in the expansion of  ${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5},\left( {x > 1} \right)$ 

Let $(1+2 x)^{20}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$.Then $3 a_0+2 a_1+3 a_2+2 a_3+3 a_4+2 a_5+\ldots+2 a_{19}+3 a_{20}$ equals

If number of terms in the expansion of ${(x - 2y + 3z)^n}$ are $45$, then $n=$