If $(1 -x + x^2)^n = a_0 + a_1x + a_2x^2 + ....... + a_{2n}x^{2n}$, then $a_0 + a_2 + a_4 +........+ a_{2n}$ is equal to

If $n$ be a positive integer such that $n \ge 3$, then the value of the sum to $n$ terms of the series $1 . n - \frac{{\left( {n\, - \,1} \right)}}{{1\,\,!}} (n - 1) + \frac{{\left( {n\, - \,1} \right)\,\,\left( {n\, - \,2} \right)}}{{2\,\,!}} (n - 2) $$-  \frac{{\left( {n\, - \,1} \right)\,\,\left( {n\, - \,2} \right)\,\,\left( {n\, - \,3} \right)}}{{3\,\,!}} (n - 3) + ......$ is 

$\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) = $

$\sum\limits_{n = 0}^4 {{{\left( {1009 - 2n} \right)}^4}\left( \begin{gathered}
  4 \hfill \\
  n \hfill \\ 
\end{gathered}  \right)} {\left( { - 1} \right)^n}$   is

If $\sum\limits_{ k =1}^{31}\left({ }^{31} C _{ k }\right)\left({ }^{31} C _{ k -1}\right)-\sum\limits_{ k =1}^{30}\left({ }^{30} C _{ k }\right)\left({ }^{30} C _{ k -1}\right)=\frac{\alpha(60 !)}{(30 !)(31 !)}$

Where $\alpha \in R$, then the value of $16 \alpha$ is equal to

For integers $n$ and $r$, let $\left(\begin{array}{l} n \\ r \end{array}\right)=\left\{\begin{array}{ll}{ }^{n} C _{ r }, & \text { if } n \geq r \geq 0 \\ 0, & \text { otherwise }\end{array}\right.$

The maximum value of $k$ for which the sum $\sum_{i=0}^{k}\left(\begin{array}{c}10 \\ i\end{array}\right)\left(\begin{array}{c}15 \\ k-i\end{array}\right)+\sum_{i=0}^{k+1}\left(\begin{array}{c}12 \\ i\end{array}\right)\left(\begin{array}{c}13 \\ k+1-i\end{array}\right)$ exists, is equal to ...... .