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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 answerfrown...... Can anybody crack the code? laugh

Thanks to everyone who tries! I'll give out likes to the person who can finish it first!!! angel

 Dec 15, 2018

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 




i^97 = i

i^98 = -1

i^99 = - i


So...the sum  is just  -1



cool cool cool

 Dec 15, 2018

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

 Dec 15, 2018
edited by PartialMathematician  Dec 15, 2018

Nope, PM....your method is just as good   (and maybe better) than mine...!!!



cool cool cool

CPhill  Dec 15, 2018

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

PartialMathematician  Dec 15, 2018

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! laugh

 Dec 15, 2018

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