If $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \ldots + (101) \cdot C_{25} = 2^{25} \cdot k$,then $k$ is equal to:

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:

If $a$ and $d$ are two complex numbers,then the sum to $(n + 1)$ terms of the following series $a{C_0} - (a + d){C_1} + (a + 2d){C_2} - \dots$ is

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

$2 \cdot {}^{20}C_0 + 5 \cdot {}^{20}C_1 + 8 \cdot {}^{20}C_2 + 11 \cdot {}^{20}C_3 + \dots + 62 \cdot {}^{20}C_{20}$ is equal to

The sum of the series $1 + \frac{1}{2} {}^{n}C_{1} + \frac{1}{3} {}^{n}C_{2} + \dots + \frac{1}{n+1} {}^{n}C_{n}$ is equal to