$\frac{{^nC_0}}{1} + \frac{{^nC_2}}{3} + \frac{{^nC_4}}{5} + \frac{{^nC_6}}{7} + \dots = $

If $a$ and $d$ are two complex numbers,then the sum to $(n + 1)$ terms of the following series $a{C_0} - (a + d){C_1} + (a + 2d){C_2} - \dots$ is

If $^{2017}C_0 + ^{2017}C_1 + ^{2017}C_2 + ...... + ^{2017}C_{1008} = \lambda^2$ where $\lambda > 0$,then the remainder when $\lambda$ is divided by $33$ is:

The sum of the coefficients of $x^r$ (where $r=0, 1, 2, \ldots, 15$) in the expansion of $(3x-1)^{15}$ is equal to the sum of the binomial coefficients of which of the following expansions?
$(a)\ (1+x)^{15}$
$(b)\ (1+x)^{16}+(1-x)^{16}$
$(c)\ (1+x)^{16}-(1-x)^{16}$

For $r=0, 1, \ldots, 10$,let $A_{r}, B_{r}$ and $C_{r}$ denote,respectively,the coefficient of $x^{r}$ in the expansions of $(1+x)^{10}$,$(1+x)^{20}$ and $(1+x)^{30}$. Then $\sum_{r=1}^{10} A_r(B_{10} B_r - C_{10} A_r)$ is equal to

If $\binom{10}{2} + \binom{10}{3} + \binom{11}{4} + \binom{12}{5} + \binom{13}{6} = \binom{14}{r}$,then $r = \dots$