If the Coefficient of $x^{30}$ in the expansion of $\left(1+\frac{1}{x}\right)^6\left(1+x^2\right)^7\left(1-x^3\right)^8 ; x \neq 0$ is $\alpha$, then $|\alpha|$ equals

If ${a_1},{a_2},{a_3},{a_4}$ are the coefficients of any four consecutive terms in the expansion of ${(1 + x)^n}$, then $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$ =

$\mathrm{b}=1+\frac{{ }^1 \mathrm{C}_0+{ }^1 \mathrm{C}_1}{1 !}+\frac{{ }^2 \mathrm{C}_0+{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2}{2 !}+\frac{{ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3}{3 !}+\ldots$

Let $\mathrm{a}=1+\frac{{ }^2 \mathrm{C}_2}{3 !}+\frac{{ }^3 \mathrm{C}_2}{4 !}+\frac{{ }^4 \mathrm{C}_2}{5 !}+\ldots$, Then $\frac{2 b}{a^2}$ is equal to.........................

Let $m, n \in N$ and $\operatorname{gcd}(2, n)=1$. If $30\left(\begin{array}{l}30 \\ 0\end{array}\right)+29\left(\begin{array}{l}30 \\ 1\end{array}\right)+\ldots+2\left(\begin{array}{l}30 \\ 28\end{array}\right)+1\left(\begin{array}{l}30 \\ 29\end{array}\right)= n .2^{ m }$ then $n + m$ is equal to (Here $\left.\left(\begin{array}{l} n \\ k \end{array}\right)={ }^{ n } C _{ k }\right)$

Let $C _{ r }$ denote the binomial coefficient of $x ^{ r }$ in the expansion of $(1+x)^{10}$. If $\alpha, \beta \in R$. $C _{1}+3 \cdot 2 C _{2}+5 \cdot 3 C _{3}+\ldots$ upto $10$ terms $=\frac{\alpha \times 2^{11}}{2^{\beta}-1}\left( C _{0}+\frac{ C _{1}}{2}+\frac{ C _{2}}{3}+\ldots . .\right.$ upto 10 terms $)$ then the value of $\alpha+\beta$ is equal to

${C_1} + 2{C_2} + 3{C_3} + 4{C_4} + .... + n{C_n} = $