Evaluating $\ i+i^2+i^3+i^4+\cdots+i^{100}$

Question

$$i+i^2+i^3+i^4+\cdots+i^{100}$$

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?

Answer 1

$$ \sum_{k=1}^n i^k = \frac{i^{n+1}-1}{i-1}-1 = i\frac{i^n-1}{i-1} $$

but $i^{100} = 1$ hence $\sum_{k=1}^{100} i^k = 0$

Answer 2

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 $i^{101} = i$.

Answer 3

Just for the sake of originality: $$\sum_{k=1}^{100} i^k =\sum_{p=0}^{24} (i^{4p+1} +i^{4p+2} +i^{4p+3} +i^{4p+4}) =\sum_{p=0}^{24} i^{4p}(i +i^2 +i^3 +i^4).$$ Knowing that $i +i^2 +i^3 +i^4 =i -1 -i +1= 0$ means your sum is also zero.