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

A possible value of $^{\prime}x^{\prime}$, for which the ninth term in the expansion of $\left\{3^{\log _{3} \sqrt{25^{x-1}+7}}+3^{\left(-\frac{1}{8}\right) \log _{3}\left(5^{x-1}+1\right)}\right\}^{10}$ in the increasing powers of $3^{\left(-\frac{1}{8}\right) \log _{3}\left(5^{x-1}+1\right)}$ is equal to $180$ , is:

The coefficient of $t^{50}$ in $(1 + t^2)^{25} (1 + t^{25}) (1 + t^{40}) (1 + t^{45}) (1 + t^{47})$ is

The sum, of the coefficients of the first $50$ terms in the binomial expansion of $(1-x)^{100}$, is equal to

Let $m, n \in N$ and $\operatorname{gcd}(2, n)=1$. If $30\left(\begin{array}{l}30 \\ 0\end{array}\right)+29\left(\begin{array}{l}30 \\ 1\end{array}\right)+\ldots+2\left(\begin{array}{l}30 \\ 28\end{array}\right)+1\left(\begin{array}{l}30 \\ 29\end{array}\right)= n .2^{ m }$ then $n + m$ is equal to (Here $\left.\left(\begin{array}{l} n \\ k \end{array}\right)={ }^{ n } C _{ k }\right)$