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

If $\sum\limits_{ k =1}^{31}\left({ }^{31} C _{ k }\right)\left({ }^{31} C _{ k -1}\right)-\sum\limits_{ k =1}^{30}\left({ }^{30} C _{ k }\right)\left({ }^{30} C _{ k -1}\right)=\frac{\alpha(60 !)}{(30 !)(31 !)}$

Where $\alpha \in R$, then the value of $16 \alpha$ is equal to

The coefficient of $x ^{301}$ in $(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+\ldots . .+x^{500}$ is:

Let $(1+2 x)^{20}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$.Then $3 a_0+2 a_1+3 a_2+2 a_3+3 a_4+2 a_5+\ldots+2 a_{19}+3 a_{20}$ equals

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

The coefficient of $t^{50}$ in $(1 + t^2)^{25} (1 + t^{25}) (1 + t^{40}) (1 + t^{45}) (1 + t^{47})$ is