the imaginary number i is defined such that i^2 = -1. what does i + i^2 + i^3 + ....i ^49 equal?
A-Best: i^3 = -i and i^4 = 1. Then the results are cycled through again:
i^5 = i = i^1
i^6 = -1 = i^2 etc.
So the first 48 terms are 12*(i - 1 - i + 1)
and the 49th term is i
The sum of the first 48 is 12*0 = 0, so the final answer is
i
A: S49 = i * (1 - i^49) / (1 - i)
S49 = i(1 - i) / 1 - i)
S49 = i
A: Every four iterations of this summation is equal to zero:
i + (-1) + (-i) + 1 = 0
49/4 = 12 r1, which only leaves i after all of the x4 iterations have been canceled.
Therefore, i^49 is equivalent to i.
