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

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 $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$ =

The constant term in the expansion of $\left(\frac{x}{2}+\frac{1}{x}+\sqrt{2}\right)^5$ is $\frac{a \sqrt{2}}{2}$,then $a=$

The number of terms which are free from radical signs in the expansion of $(y^{1/5} + x^{1/10})^{55}$ is

Let the coefficients of three consecutive terms $T_r$,$T_{r+1}$,and $T_{r+2}$ in the binomial expansion of $(a+b)^{12}$ be in a $G.P.$ and let $p$ be the number of all possible values of $r$. Let $q$ be the sum of all rational terms in the binomial expansion of $(\sqrt[4]{3}+\sqrt[3]{4})^{12}$. Then $p+q$ is equal to:

If the coefficients of $T_r, T_{r+1}, T_{r+2}$ terms of $(1+x)^{14}$ are in $A.P.$,then $r =$