If $C_0, C_1, C_2, \ldots, C_n$ are the binomial coefficients in the expansion of $(1+x)^n$,then the value of $\sum_{r=0}^{n} r^3 \cdot C_r$ when $n=5$ is

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

Let $a_0, a_1, a_2, \ldots, a_n \in \mathbb{R}$ be in an arithmetic progression and let $C_0, C_1, C_2, \ldots, C_n$ be the binomial coefficients. Then $\sum_{k=0}^n a_k \cdot C_k =$

If $(1+x-2x^2)^6 = 1+a_1x+a_2x^2+\ldots+a_{12}x^{12}$, then the value of $a_2+a_4+a_6+\ldots+a_{12}$ is

$\sum_{r=0}^{10} {}^{40-r} C_5$ is equal to

$C_0 - C_1 + C_2 - C_3 + \dots + (-1)^n C_n$ is equal to