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

The sum to $(n + 1)$ terms of the following series $\frac{{{C_0}}}{2} - \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} - \frac{{{C_3}}}{5} + $..... is

If the sum of the coefficients of all even powers of $x$ in the product $\left(1+x+x^{2}+\ldots+x^{2 n}\right)\left(1-x+x^{2}-x^{3}+\ldots+x^{2 n}\right)$ is $61,$ then $\mathrm{n}$ is equal to

The sum of the coefficients of three consecutive terms in the binomial expansion of $(1+ x )^{ n +2}$, which are in the ratio $1: 3: 5$, is equal to

Coefficients of ${x^r}[0 \le r \le (n - 1)]$ in the expansion of ${(x + 3)^{n - 1}} + {(x + 3)^{n - 2}}(x + 2)$$ + {(x + 3)^{n - 3}}{(x + 2)^2} + ... + {(x + 2)^{n - 1}}$

If ${a_1},{a_2},{a_3},{a_4}$ are the coefficients of any four consecutive terms in the expansion of ${(1 + x)^n}$, then $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$ =