If the coefficient of $x^{10}$ in the binomial expansion of $\left(\frac{\sqrt{x}}{5^{1/4}}+\frac{\sqrt{5}}{x^{1/3}}\right)^{60}$ is $5^k l$,where $l, k \in N$ and $l$ is coprime to $5$,then $k$ is equal to

The coefficient of $x^2$ in the expansion of $(2x^2 + \frac{1}{x})^{10}, x \neq 0$, is:

In the expansion of ${\left( 2{x^2} - \frac{1}{x} \right)^{12}}$,the term independent of $x$ is: (in $^{th}$)

Arrange the expansion of $\left(x^{1/2} + \frac{1}{2x^{1/4}}\right)^n$ in decreasing powers of $x$. Suppose the coefficients of the first three terms form an arithmetic progression. Then,the number of terms in the expansion having integer power of $x$ is

If $C_0, C_1, C_2, \ldots, C_8$ are the binomial coefficients in the expansion of $(1+x)^8$,then $\sum_{r=1}^8 r^3 \frac{C_r}{C_{r-1}} =$

If the coefficient of $x$ in the expansion of $\left(x^2+\frac{k}{x}\right)^5$ is $270$,then $k$ is equal to