The value of ${ }^{10} C_{1}+{ }^{10} C_{2}+{ }^{10} C_{3}+\ldots+{ }^{10} C_{9}$ is

If $(1 + x)^n = c_0 + c_1x + c_2x^2 + c_3x^3 + \dots + c_nx^n$,then the value of $c_0 - 3c_1 + 5c_2 - \dots + (-1)^n(2n + 1)c_n$ is

If $\sum\limits_{k=1}^{31} \binom{31}{k} \binom{31}{k-1} - \sum\limits_{k=1}^{30} \binom{30}{k} \binom{30}{k-1} = \frac{\alpha(60!)}{(30!)(31!)}$,where $\alpha \in R$,then the value of $16\alpha$ is equal to

If $\frac{{}^{11}C_1}{2} + \frac{{}^{11}C_2}{3} + \dots + \frac{{}^{11}C_9}{10} = \frac{n}{m}$ with $\gcd(n, m) = 1$,then $n + m$ is equal to

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .... + C_nx^n$,then the value of $C_0 + 2C_1 + 3C_2 + .... + (n + 1)C_n$ will be

Let $\alpha = \sum_{k=0}^n \left( \frac{({ }^n C_k)^2}{k+1} \right)$ and $\beta = \sum_{k=0}^{n-1} \left( \frac{{ }^n C_k \cdot { }^n C_{k+1}}{k+2} \right)$. If $5 \alpha = 6 \beta$,then $n$ equals: