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:

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

If $C_j = {}^{n}C_j$,then $C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n = $

Let $X = 1({ }^{10} C _1)^2 + 2({ }^{10} C _2)^2 + 3({ }^{10} C _3)^2 + \ldots + 10({ }^{10} C _{10})^2$,where ${ }^{10} C _{ r }$ for $r \in \{1, 2, \ldots, 10\}$ denotes binomial coefficients. Then,the value of $\frac{1}{1430} X$ is:

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .... + C_nx^n$,then the value of $C_0 + 2C_1 + 3C_2 + .... + (n + 1)C_n$ will be

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 $k=$