Question

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

Answer 1

If
`i=√(-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}`