Calculation of $EX$ using the binomial expansion formula is easy:
\begin{align} EX &= \sum_{x=0}^{n}x\frac{n!}{(n-x)!x!}p^{x}(1-p)^{n-x}\\& = np \sum_{x=1}^{n}\frac{(n-1)!}{(n-x)!(x-1)!}p^{x-1}(1-p)^{n-x}\\& = np(p+(1-p))^{n-1}\\& = np \end{align}
How do I calculate $EX^2$ using the binomial expansion formula?
This is a common exercise many of the textbooks would provide. There is not much to tell here but the approach to ease the derivation is to use the fact that ${n\choose x}=\frac nx \cdot\frac{n-1}{x-1}{{n-2}\choose{x-2}}$ and writing $x^2=x(x-1) +x$ to utilize the former. You only need to incorporate them in $\mathbb EX^2.$
