Simplify\(i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}\)
Need help ASAP on this! Due soon, and I couldn't get the answer...... Can anybody crack the code?
Thanks to everyone who tries! I'll give out likes to the person who can finish it first!!!
Note that
i^1 + i^2 + i^3 + i^4 =
i - 1 - i + 1 = 0
And each of the 4 successive terms have the same pattern and will also sum to 0
So...... i^1 to i^96 inclusive will have a sum of 0
So
i^97 = i
i^98 = -1
i^99 = - i
So...the sum is just -1
Note: i^1 = i, i^2 = -1, i^3 = -i, and i^4 = 1.
Different powers of i still have the same properties. You can divide the power by 4 and determine the remainder to see if it equals i^1. i^2, i^3, or i^4.
As you can see, i - 1 -i + 1 cancel out each other, equalling 0. The last three terms of your sequence are i^97, i^98, and i^99, which are the same as i^1, i^2, and i^3. i - 1 -i = -1, so the answer is \(\boxed{-1}\).
Hope this helps, (CPhill will give a more vivid answer)
- PM
Thanks, CPhill, both of our answers used the same steps. They were both really good!
Thank you both! Your answers were both REALLY good! I hit the like button for both of you, and I wouldn't have been able to solve it! Thanks!