For $2 \le r \le n$,$\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$ is equal to

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

Let $m, n \in \mathbb{N}$ and $\operatorname{gcd}(2, n)=1$. If $30\binom{30}{0} + 29\binom{30}{1} + \ldots + 2\binom{30}{28} + 1\binom{30}{29} = n \cdot 2^m$,then $n + m$ is equal to (Here $\binom{n}{k} = {^nC_k}$)

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

Let $(1 + x)^{10} = \sum_{r=0}^{10} C_r x^r$ and $(1 + x)^7 = \sum_{r=0}^7 d_r x^r$. If $P = \sum_{r=0}^5 C_{2r}$ and $Q = \sum_{r=0}^3 d_{2r+1}$,then $\frac{P}{2Q}$ is equal to

If $3 \times { }^5 C_0 + 8 \times { }^5 C_1 + 13 \times { }^5 C_2 + 18 \times { }^5 C_3 + 23 \times { }^5 C_4 + 28 \times { }^5 C_5 = k \times 2^n$,where $n$ is a power of $2$,find $k$. Specifically,if $3 \times { }^5 C_0 + 8 \times { }^5 C_1 + 13 \times { }^5 C_2 + 18 \times { }^5 C_3 + 23 \times { }^5 C_4 + 28 \times { }^5 C_5 = k \times 2^4$,then $k=$