Let $a_0, a_1, a_2, \ldots, a_n \in \mathbb{R}$ be in an arithmetic progression and let $C_0, C_1, C_2, \ldots, C_n$ be the binomial coefficients. Then $\sum_{k=0}^n a_k \cdot C_k =$

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 $\binom{n}{k}$ 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:

Let $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. Then the sum $\frac{1}{2^{10}} \sum_{k=0}^{10} \binom{10}{k} k^2$ lies in the interval

If $1^2 \cdot \binom{15}{1} + 2^2 \cdot \binom{15}{2} + 3^2 \cdot \binom{15}{3} + \ldots + 15^2 \cdot \binom{15}{15} = 2^m \cdot 3^n \cdot 5^k$, where $m, n, k \in N$, then $m + n + k$ is equal to:

$\sum\limits_{k = 0}^{10} {^{20}{C_k} = }$