Question

Compute i^1+i^2+i^3+...+ i^97 + i^98+i^99.

Answer 1

If $i=\sqrt{-1}$, then $i^2=-1$, $i^3=-i$, and $i^4=1$.

We can break up the given sum into 25 groups (after adding and subtracting $i^{100}$):

$(i+i^2+i^3+i^4)+(i^5+i^6+i^7+i^8)+\cdots+(i^{97}+i^{98}+i^{99}+i^{100})-i^{100}$

We have

$i+i^2+i^3+i^4=i-1-i+1=0$

and in each group, we can pull this out as a factor. For example,

$i^{97}+i^{98}+i^{99}+i^{100}=i^{96}(i+i^2+i^3+i^4)=0$

So the entire sum reduces to the remaining term,

$-i^{100}=-i^{25\cdot4}=-(i^4)^{25}=-1^{25}=\boxed{-1}$