Evaluating $\ i+i^2+i^3+i^4+\cdots+i^{100}$ $$i+i^2+i^3+i^4+\cdots+i^{100}$$
I figured out that every four terms add up to zero where $i^2=-1$, $i^3=-i$, $i^4=1$, so
$$i+i^2+i^3+i^4 = i-1-i+1 = 0$$
Thus, the whole series eventually adds up to zero. But how do I approach this problem in a more mathematical way?

A result that is useful in your case is $\sum_{k=1}^{n} z^k = \frac{z - z^{n+1}}{1 - z}$.
This follows from $(1 - z) \sum_{k=1}^{n} z^k = (1-z) (z + z^2 + \ldots + z^n) = z - z^{n+1}$, where I assumed $z \neq 1$.
In your case you get $\sum_{k=1}^{100} i^k =   \frac{i - i^{101}}{1 - i} = 0$, since $i^{101} = i$.
