Imaginary periodicity
i 0 + i 1 + i 2 + . . . + i 2 0 1 1 + i 2 0 1 2
The sum is equal to:
From Campinense's Olympiad Mathematics 2012 , Brazil
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2 solutions
i 0 = 1
i 1 = i i 2 = − 1 i 3 = − i i 4 = 1 i 5 = i 1 = i i 6 = i 2 = − 1 i 0 + i 1 + i 2 + i 3 = i 4 + i 5 + i 6 + i 7 = 0 i 0 + i 1 + i 2 + . . . + i 2 0 1 2 = 0 + 0 + . . . + 0 + i 2 0 1 2 = i 0 = 1
Since its an geometric progression with first term =1 and comnon ratio=i sum=1[(i^2013) - 1]/(i-1) Which comes out to be 1 *using the property: i^4=1