Let $S = \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$ up to $13$ terms. If $13S = \frac{2^{k}}{n!}$ where $k \in N$,then $n + k$ is equal to

If $\binom{40}{0} + \binom{41}{1} + \binom{42}{2} + \dots + \binom{60}{20} = \frac{m}{n} \binom{60}{20}$,where $m$ and $n$ are coprime,then $m+n$ is equal to

If $(1 + x + x^2)^{25} = a_0 + a_1x + a_2x^2 + ..... + a_{50}x^{50}$,then $a_0 + a_2 + a_4 + ..... + a_{50}$ is :

The value of $\frac{1}{1! 50!} + \frac{1}{3! 48!} + \frac{1}{5! 46!} + \dots + \frac{1}{49! 2!} + \frac{1}{51! 1!}$ is $.............$.

Let $\alpha = \sum_{k=0}^n \left( \frac{({ }^n C_k)^2}{k+1} \right)$ and $\beta = \sum_{k=0}^{n-1} \left( \frac{{ }^n C_k \cdot { }^n C_{k+1}}{k+2} \right)$. If $5 \alpha = 6 \beta$,then $n$ equals:

Suppose $\sum_{r=0}^{2023} r \cdot ^{2023}C_r = 2023 \times \alpha \times 2^{2022}$. Then the value of $\alpha$ is $............$