In the expansion of $(1 + x)^5$,the sum of the coefficients of the terms is

Using the Binomial Theorem,evaluate $(96)^{3}$.

If $\frac{(1-px)^{-1}}{(1-qx)}=a_0+a_1x+a_2x^2+a_3x^3+\ldots$,then $a_n=$

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

If the sum of the coefficients in the expansion of $({\alpha ^2}{x^2} - 2\alpha x + 1)^{51}$ vanishes,then the value of $\alpha$ is

If the sum of the coefficients in the expansion of $(\alpha x^2 - 2x + 1)^{35}$ is equal to the sum of the coefficients in the expansion of $(x - \alpha y)^{35}$,then $\alpha = $