The value of $\frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \dots$ is equal to

Let $S_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $S_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
Assertion $(A) : S_3 = 55 \times 2^9$
Reason $(R) : S_1 = 90 \times 2^8$ and $S_2 = 10 \times 2^8$

If for $3 \leq r \leq 30$, $\binom{30}{30-r} + 3\binom{30}{31-r} + 3\binom{30}{32-r} + \binom{30}{33-r} = \binom{m}{r}$, then $m$ equals:

If $\sum\limits_{K = 1}^{12} {12K \cdot {^{12}C_K} \cdot {^{11}C_{K - 1}}} $ is equal to $\frac{{12 \times 21 \times 19 \times 17 \times \dots \times 3}}{{11!}} \times {2^{12}} \times p$,then $p$ is:

$\frac{1}{1!(n - 1)!} + \frac{1}{3!(n - 3)!} + \frac{1}{5!(n - 5)!} + \dots = $

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