$\sum\limits_{k = 0}^{10} {^{20}{C_k} = } $

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

$\left( {\left( {\begin{array}{*{20}{c}}
{21}\\
1
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
1
\end{array}} \right)} \right) + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
2
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
2
\end{array}} \right)} \right)$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
3
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
3
\end{array}} \right)} \right) + \;.\;.\;.$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
{10}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
{10}
\end{array}} \right)} \right) = $

If ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $, then $\frac{{{t_n}}}{{{S_n}}}$ is equal to

The coefficient of $x^{256}$ in the expansion of $(1-x)^{101}\left(x^{2}+x+1\right)^{100}$ is:

For natural numbers $m,n$ ,if ${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n} = 1 + {a_1}y + {a_2}{y^2} + \ldots \;$ and $a_1= a_2=10,$ then $(m,n)$ =______.