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Compute i+i^2+i^3.............i^258+i^259

 Jan 15, 2023
 #2
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Because \((i, i^2, i^3, i^4) = (i, -1, -i, 1)\), we have  \(i + i^2 + i^3 \cdots + i^{258} + i^{259} = 64(i - 1 - i + 1) + i^{257} + i^{258} + i^{259}\). So, the desired sum is \(i^{257} + i^{258} + i^{259} = i - 1 - i = \boxed{-1}\).

 Jan 16, 2023

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