$^{10}C_1 + ^{10}C_3 + ^{10}C_5 + ^{10}C_7 + ^{10}C_9 = $

Let $X = 1({ }^{10} C _1)^2 + 2({ }^{10} C _2)^2 + 3({ }^{10} C _3)^2 + \ldots + 10({ }^{10} C _{10})^2$,where ${ }^{10} C _{ r }$ for $r \in \{1, 2, \ldots, 10\}$ denotes binomial coefficients. Then,the value of $\frac{1}{1430} X$ is:

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 ${ }^{n} C_0+\frac{1}{2}{ }^{n} C_1+\frac{1}{3}{ }^{n} C_2+\ldots+\frac{1}{n+1}{ }^{n} C_{n}=\frac{1023}{10}$,then $n=$

$\sum_{k=0}^{20} \left({}^{20}C_{k}\right)^{2}$ is equal to :

$3 \cdot C_0 + 7 \cdot C_1 + 11 \cdot C_2 + \ldots + (3 + 4n) C_n =$