If $\left({ }^{30} C _1\right)^2+2\left({ }^{30} C _2\right)^2+3\left({ }^{30} C _3\right)^2+\ldots \ldots+30\left({ }^{30} C _{30}\right)^2=$ $\frac{\alpha 60 !}{(30 !)^2}$, then $\alpha$ is equal to

The coefficient of $x^{49}$ in the expansion of $(x - 1)$$\left( {x\, - \,\frac{1}{2}\,} \right)$$\left( {x\, - \,\frac{1}{{{2^2}}}\,} \right)$ .....$\left( {x\, - \,\frac{1}{{{2^{49}}}}\,} \right)$ is equal to

If ${C_r}$ stands for $^n{C_r}$, the sum of the given series $\frac{{2(n/2)!(n/2)!}}{{n!}}[C_0^2 - 2C_1^2 + 3C_2^2 - ..... + {( - 1)^n}(n + 1)C_n^2]$, Where $n$ is an even positive integer, is

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 number of terms in the expansion of $(1 +x)^{101}  (1 +x^2 - x)^{100}$ in powers of $x$ is

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