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

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 $\frac{P}{2Q}$ is equal to

If $\sum_{r=1}^{30} \frac{r^2({}^{30}C_r)^2}{{}^{30}C_{r-1}} = \alpha \times 2^{29}$,then $\alpha$ 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 $n=$

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