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!!!

Web2.0CalcUser123 Dec 15, 2018

#1**+3 **

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

CPhill Dec 15, 2018

#2**+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

PartialMathematician Dec 15, 2018

#4**0 **

Thanks, CPhill, both of our answers used the same steps. They were both really good!

PartialMathematician
Dec 15, 2018

#5**+2 **

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!

Web2.0CalcUser123 Dec 15, 2018