Calculate i^21 + i^22 + i^23 + i^24?

Based on the fact that i^2 = -1 and i^4 = 1, we'll
get:

i^21 = i^(20+1) = i^20*i = [(i^4)^5]*i = 1*i =
i

i^22 = i^(20+2) = i^20*i^2 = [(i^4)^5]*i^2 = 1*(-1)
=-1

i^23 = i^(20+3) = i^20*i^3 = [(i^4)^5]*i^3 = 1*(-i) =
-i

i^24 = i^(20+1) = i^20*i^4 = [(i^4)^5]*i^4 = 1*1
=1

i^21 + i^22 + i^23 + i^24 = i - 1 - i +
1=0

Therefore, the result of addition of
powers of i is i^21 + i^22 + i^23 + i^24 = 0.