SOLUTION: The imaginary number i is defined such that i^2 = –1. What does i + i^2 + i^3 + .... + i^23 equal? The answer is supposed to be -1, but I thought it should be -i according to anoth

Algebra.Com
Question 176697: The imaginary number i is defined such that i^2 = –1. What does i + i^2 + i^3 + .... + i^23 equal? The answer is supposed to be -1, but I thought it should be -i according to another source. Is the answer key wrong?
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!








, and so on repeating the sequence every 4 increments of the exponent.

The sum of the first 4 is: , so the sum of the each subsequent 4 must also be zero.

Use integer division is 5 with a remainder of 3. So in your string of 23 terms there are 5 sets of 4 terms each of which sum to zero, so it is only the last three terms that make any difference:



And that is your sum. It is your 'other source' that is incorrect.

RELATED QUESTIONS

The imaginary number i is defined i^2 = -1. What is the value of... (answered by josgarithmetic)
How do I complete this square? 2x^2+7x+3=0 The answer is supposed to be -3, 1/2... (answered by ReadingBoosters)
(2+i)(3-i)(2-3i) i=imaginary unit I know we're supposed to do FOIL, but in what... (answered by solver91311,robertb)
hope someone could help (-i)^4 "i" is an imaginary number What is... (answered by scott8148)
What is the conjugate of {{{-2/i^3}}} ? I understand what a conjugate is, and I... (answered by macston)
What is the simplify form of (6^(1/2)+i)(6^(1/2)-i). i is the imaginary number -1^(1/2) (answered by Alan3354)
i know i^2 =-1 so i^26 must be -1. but i can write i^26 as (i^4)^(26/4) and we know i^4... (answered by Edwin McCravy)
We use the Greek letter iota, i, to represent the square root of -1 ( ). We call i an... (answered by MathLover1)
I know that i^2 is -1 but what is i^3 and i^4... (answered by longjonsilver)