If $\frac{1}{n+1}{ }^n C_n+\frac{1}{n}{ }^n C_{n-1}+\ldots+\frac{1}{2}{ }^{ n } C _1+{ }^{ n } C _0=\frac{1023}{10}$ then $n$ is equal to

The coefficient of $x^{70}$ in $x^2(1+x)^{98}+x^3(1+x)^{97}+$ $x^4(1+x)^{96}+\ldots \ldots . .+x^{54}(1+x)^{46}$ is ${ }^{99} \mathrm{C}_p-{ }^{46} \mathrm{C}_{\mathrm{q}}$.

Then a possible value to $\mathrm{p}+\mathrm{q}$ 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 ${s_2} = 10 \times {2^8}$ 

$\frac{{{C_0}}}{1} + \frac{{{C_1}}}{2} + \frac{{{C_2}}}{3} + …. + \frac{{{C_n}}}{{n + 1}} = $

If the sum of the coefficients of all even powers of $x$ in the product $\left(1+x+x^{2}+\ldots+x^{2 n}\right)\left(1-x+x^{2}-x^{3}+\ldots+x^{2 n}\right)$ is $61,$ then $\mathrm{n}$ is equal to