The sum of the coefficients of even powers of $x$ in the expansion of ${(1 + x + x^2 + x^3)^5}$ is

The coefficient of $x^{50}$ in the expansion of $(1+x)^{100}+2x(1+x)^{99}+3x^2(1+x)^{98}+\dots+101x^{100}$ is:

If $(1 - x + 2x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$,where $n \in N$,$x \in R$,and $a_0, a_1, a_2$ are in Arithmetic Progression $(A.P.)$,then there exists:

If $(1 + x - 3x^2)^{2145} = a_0 + a_1x + a_2x^2 + \dots$,then $a_0 - a_1 + a_2 - a_3 + \dots$ ends with:

Find the coefficient of $a^{4}$ in the product $(1+2a)^{4}(2-a)^{5}$ using the binomial theorem.

If $C_0, C_1, C_2, \ldots, C_{n}$ are the binomial coefficients in the expansion of $(1+x)^{n}$,then $(C_0+C_1)-(C_2+C_3)+(C_4+C_5)-(C_6+C_7)+\ldots=$