I understand that $\lambda$ is given as $np$ when $E[X]=np$. Though, I'm uncertain as to how the expression simplify's $i$ in the second expression for both the numerator and the denominator. I believe it has something to do with the $\sum$ function, though a clarification on how it's done would be greatly appreciated!
$\sum^{\infty}_{i=0}\frac{i^2e^{-\lambda}\lambda^i}{i!}$ simplify to: $\lambda\sum^{\infty}_{i=1}\frac{ie^{-\lambda}\lambda^{i-1}}{(i-1)!}$
One may see that $$\begin{align} \sum^{\infty}_{i=0}\frac{i^2e^{-\lambda}\lambda^i}{i!}&=\frac{0^2e^{-\lambda}\lambda^0}{0!}+\sum^{\infty}_{i=1}\frac{i\times i \times e^{-\lambda}\lambda^i}{i \times(i-1)!}\\[1ex]&=\sum^{\infty}_{i=1}\frac{ie^{-\lambda}\lambda^i}{(i-1)!}\\[1ex]&=\lambda\sum^{\infty}_{i=1}\frac{ie^{-\lambda}\lambda^{i-1}}{(i-1)!}. \end{align}$$ the term from $i=0$ vanishing in the first series.