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