If $C_{x} \equiv^{25} C_{x}$ and $\mathrm{C}_{0}+5 \cdot \mathrm{C}_{1}+9 \cdot \mathrm{C}_{2}+\ldots .+(101) \cdot \mathrm{C}_{25}=2^{25} \cdot \mathrm{k}$ then $\mathrm{k}$ is equal to

Let n and k be positive integers such that $n \ge \frac{{k(k + 1)}}{2}$. The number of solutions $({x_1},{x_2},....{x_k})$, ${x_1} \ge 1,{x_2} \ge 2,....{x_k} \ge k,$ all integers, satisfying ${x_1} + {x_2} + .... + {x_k} = n$, is

In the expansion of ${(x + a)^n}$, the sum of odd terms is $P$ and sum of even terms is $Q$, then the value of $({P^2} - {Q^2})$ will be

The value of $^{4n}{C_0}{ + ^{4n}}{C_4}{ + ^{4n}}{C_8} + ....{ + ^{4n}}{C_{4n}}$ is

If ${a_k} = \frac{1}{{k(k + 1)}},$ for $k = 1,\,2,\,3,\,4,.....,\,n$, then ${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = $

$\sum_{\mathrm{k}=0}^{20}\left({ }^{20} \mathrm{C}_{\mathrm{k}}\right)^{2}$ is equal to :