$2.{}^{20}{C_0} + 5.{}^{20}{C_1} + 8.{}^{20}{C_2} + 11.{}^{20}{C_3} + ......62.{}^{20}{C_{20}}$ is equal to

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

If the sum of the coefficients in the expansion of ${(1 - 3x + 10{x^2})^n}$ is $a$ and if the sum of the coefficients in the expansion of ${(1 + {x^2})^n}$ is $b$, then

The sum of the series $aC_0 + (a + b)C_1 + (a + 2b)C_2 + ..... + (a + nb)C_n$ is where $Cr's$ denotes combinatorial coefficient in the expansion of $(1 + x)^n, n \in N$

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

Let $\alpha=\sum_{k=0}^n\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right)$ and $\beta=\sum_{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right)$. If $5 \alpha=6 \beta$, then $n$ equals