In the expansion of

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

The coefficient of $x^{91}$ in the series $^{100}{C_1}\,{2^8}.\,{\left( {1\, - \,x} \right)^{99}}\, + {\,^{100}}{C_2}\,{2^7}.\,{\left( {1\, - \,x} \right)^{98}}\, + {\,^{100}}{C_3}\,{2^6}.\,{\left( {1\, - \,x} \right)^{97}}\, + \,....\, + {\,^{100}}{C_9}\,{\left( {1\, - \,x} \right)^{91}}$ is equal to -

What is the sum of the coefficients of ${({x^2} - x - 1)^{99}}$

Let ${ }^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1+ x )^{ n }.$

If $\sum_{ k =0}^{10}\left(2^{2}+3 k \right){ }^{ n } C _{ k }=\alpha .3^{10}+\beta \cdot 2^{10}, \alpha, \beta \in R$ then $\alpha+\beta$ is equal to ....... .

The sum of coefficients in ${(1 + x - 3{x^2})^{2134}}$ is

Let ${\left( {1 + x + {x^2}} \right)^{20}}\left( {2x + 1} \right) = {a_0} + {a_1}{x^1} + {a_2}{x^2} + ... + {a_{41}}{x^{41}}$ , then $\frac{{{a_0}}}{1} + \frac{{{a_1}}}{2} + .... + \frac{{{a_{41}}}}{{42}}$ is equal to