If $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$ for $n \in N$,then $C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n+1} =$

The mean of the values $0, 1, 2, \dots, n$ having corresponding weights $^nC_0, ^nC_1, ^nC_2, \dots, ^nC_n$ respectively is

Let $\binom{n}{k}$ denote ${}^{n}C_{k}$ and $\left[\begin{array}{c} n \\ k \end{array}\right]=\begin{cases} \binom{n}{k}, & \text{if } 0 \leq k \leq n \\ 0, & \text{otherwise} \end{cases}$. If $A_{k}=\sum_{i=0}^{9}\binom{9}{i}\left[\begin{array}{c} 12 \\ 12-k+i \end{array}\right]+\sum_{i=0}^{8}\binom{8}{i}\left[\begin{array}{c} 13 \\ 13-k+i \end{array}\right]$ and $A_{4}-A_{3}=190p$,then $p$ is equal to:

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 $(1+x)^n = \sum_{r=0}^n C_r x^r$,then the value of $C_0 + (C_0 + C_1) + (C_0 + C_1 + C_2) + \ldots + (C_0 + C_1 + C_2 + \ldots + C_n)$ is

The value of $^{15}C_0^2 - ^{15}C_1^2 + ^{15}C_2^2 - ... - ^{15}C_{15}^2$ is