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

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

$2{C_0} + \frac{{{2^2}}}{2}{C_1} + \frac{{{2^3}}}{3}{C_2} + .... + \frac{{{2^{11}}}}{{11}}{C_{10}}$ = . . . 

If ${(1 - x + {x^2})^n} = {a_0} + {a_1}x + {a_2}{x^2} + .... + {a_{2n}}{x^{2n}}$, then ${a_0} + {a_2} + {a_4} + .... + {a_{2n}} = $

If $\sum\limits_{r = 0}^{25} {\left\{ {^{50}{C_r}.{\,^{50 - r}}{C_{25 - r}}} \right\} = K\left( {^{50}{C_{25}}} \right)} $, then $K$ is equal to

Coefficient of $x^{19}$ in the polynomial $(x-1) (x-2^1) (x-2^2) .... (x-2^{19})$ is