$\frac{{^nC_0}}{1} + \frac{{^nC_2}}{3} + \frac{{^nC_4}}{5} + \frac{{^nC_6}}{7} + \dots = $

If $(1 + x)^n = \sum\limits_{r = 0}^n {{C_r}{x^r}} $,then $\left( {1 + \frac{{{C_1}}}{{{C_0}}}} \right)\left( {1 + \frac{{{C_2}}}{{{C_1}}}} \right)....\left( {1 + \frac{{{C_n}}}{{{C_{n - 1}}}}} \right) = $

If $n$ is a positive integer,then $\sum_{r=1}^n r^2 \cdot C_r = (\ldots \ldots \ldots) 2^{n-2}$

If $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$ for $n \in N$,then $C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n+1} =$

Let $(1+x)^{10} = \sum_{r=0}^{10} c_{r} x^{r}$ and $(1+x)^{7} = \sum_{r=0}^{7} d_{r} x^{r}$. If $P = \sum_{r=0}^{5} c_{2r}$ and $Q = \sum_{r=0}^{3} d_{2r+1}$, then $\frac{P}{Q}$ is equal to:

If $(\frac{1}{^{15}C_{0}}+\frac{1}{^{15}C_{1}})(\frac{1}{^{15}C_{1}}+\frac{1}{^{15}C_{2}})...(\frac{1}{^{15}C_{12}}+\frac{1}{^{15}C_{13}}) = \frac{a^{13}}{^{14}C_{0} \cdot ^{14}C_{1} \cdot ... \cdot ^{14}C_{12}}$, then $30a$ is equal to: