Compute $i^{600} + i^{599} + \cdots + i + 1$, where $i^2=-1$.
So the sum is $S_{601} = 1 \frac{1-i^{601}}{1-i}$. Now I need to simplify this expression. I notice that $i^{601} = i^{600} \cdot i = (i^4)^{150} \cdot i = 1^{150} \cdot i = i$, since $i^4=1$. So the numerator becomes $1-i$. The denominator is $1-i$ as well, so the fraction is just 1. Therefore, the sum is $S_{601} = 1 \cdot 1 = 1$. 