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Compute \(i^600+i^599+....+i+1 = 1.\) Where \(i^2=-1\).

 Mar 31, 2019
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\(\text{I assume this is supposed to be}\\ \sum \limits_{k=0}^{600}i^k\)

 

The sum of any 4 adjacent terms of this series is 0.

 

600 is divisible by 4.

 

There are 601 terms, 600 of which will sum to zero leaving the last (or first) term i600 = i0 = 1

 Mar 31, 2019

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