$\binom{n}{n-r} + \binom{n}{r+1}$,whenever $0 \le r \le n-1$,is equal to

The sum of the series $\frac{1}{1 \times 2} {}^{25}C_{0} + \frac{1}{2 \times 3} {}^{25}C_{1} + \frac{1}{3 \times 4} {}^{25}C_{2} + \ldots + \frac{1}{26 \times 27} {}^{25}C_{25}$ is

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .......... + C_nx^n$,then $\frac{C_1}{C_0} + \frac{2C_2}{C_1} + \frac{3C_3}{C_2} + .... + \frac{nC_n}{C_{n - 1}} = $

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 $(1 + x)^{15} = C_0 + C_1x + C_2x^2 + ...... + C_{15}x^{15},$ then $C_2 + 2C_3 + 3C_4 + .... + 14C_{15} = $

In the expansion of $(x + a)^n$,the sum of odd terms is $P$ and the sum of even terms is $Q$. Then the value of $(P^2 - Q^2)$ is: