$2{C_0} + \frac{2^2}{2}{C_1} + \frac{2^3}{3}{C_2} + \dots + \frac{2^{11}}{11}{C_{10}} = \dots$

The first $3$ terms in the expansion of $(1 + ax)^n$ $(n \ne 0)$ are $1, 6x$ and $16x^2$. Then the value of $a$ and $n$ are respectively

For $x \in R, x \neq -1,$ if $(1 + x)^{2016} + x(1 + x)^{2015} + x^2(1 + x)^{2014} + \dots + x^{2016} = \sum_{i = 0}^{2016} a_i x^i,$ then $a_{17}$ is equal to

Let $n$ be a positive integer. Let $A = \sum_{k=0}^{n} (-1)^{k} {}^{n}C_{k} \left[ \left(\frac{1}{2}\right)^{k} + \left(\frac{3}{4}\right)^{k} + \left(\frac{7}{8}\right)^{k} + \left(\frac{15}{16}\right)^{k} + \left(\frac{31}{32}\right)^{k} \right]$. If $63A = 1 - \frac{1}{2^{30}}$,then $n$ is equal to ......

The coefficient of $x^{101}$ in the expression $(5+x)^{500} + x(5+x)^{499} + x^{2}(5+x)^{498} + \ldots + x^{500}$ for $x > 0$ is:

If $x$ is so small that $x^3$ and higher powers of $x$ may be neglected,then $\frac{(1 + x)^{3/2} - (1 + \frac{1}{2}x)^3}{(1 - x)^{1/2}}$ may be approximated as