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

In the expansion of

$(2x + 1).(2x + 5) . (2x + 9) . (2x + 13)...(2x + 49),$ find the coefficient of $x^{12}$ is :-

If $n$ be a positive integer such that $n \ge 3$, then the value of the sum to $n$ terms of the series $1 . n - \frac{{\left( {n\, - \,1} \right)}}{{1\,\,!}} (n - 1) + \frac{{\left( {n\, - \,1} \right)\,\,\left( {n\, - \,2} \right)}}{{2\,\,!}} (n - 2) $$-  \frac{{\left( {n\, - \,1} \right)\,\,\left( {n\, - \,2} \right)\,\,\left( {n\, - \,3} \right)}}{{3\,\,!}} (n - 3) + ......$ is 

Let $a =$ Minimum $\{x^2 + 2x + 3, x \in R\}$ and $b = \mathop {\lim }\limits_{\theta  \to 0} \frac{{1 - \cos \theta }}{{{\theta ^2}}}$ The value of $\sum\limits_{r = 0}^n {{a^r}.{b^{n - r}}} $ is

If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1+x)^{10}$, then $\sum \limits_{r=1}^{10} r^3\left(\frac{a_r}{a_{r-1}}\right)^2$ is equal to

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