The sum of the coefficients in the expansion of ${(1 + x - 3{x^2})^{3148}}$ is

Let $(1 + x)(1 + x + x^2)(1 + x + x^2 + x^3)\,\, ......\,\,$$(1 + x + x^2 + ..... + x^{30}) = $$a_0 + a_1x + a_2x^2$ .....$+$ $a_{465}x^{465}$, then sum of $a_0 + a_2 + a_4 + ......... +$ is

The sum of coefficients in the expansion of ${(1 + x + {x^2})^n}$ is

A possible value of $^{\prime}x^{\prime}$, for which the ninth term in the expansion of $\left\{3^{\log _{3} \sqrt{25^{x-1}+7}}+3^{\left(-\frac{1}{8}\right) \log _{3}\left(5^{x-1}+1\right)}\right\}^{10}$ in the increasing powers of $3^{\left(-\frac{1}{8}\right) \log _{3}\left(5^{x-1}+1\right)}$ is equal to $180$ , 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 -

The value of $\frac{1}{1 ! 50 !}+\frac{1}{3 ! 48 !}+\frac{1}{5 ! 46 !}+\ldots .+\frac{1}{49 ! 2 !}+\frac{1}{51 ! 1 !}$ is $.............$.