Compute i+i^2+i^3+\cdots+i^{258}+i^{259}.

Question

Compute i+i^2+i^3+\cdots+i^{258}+i^{259}.

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Neala 4 years 2021-07-20T13:36:05+00:00 2 Answers 15 views 0

Answers ( )

  1. Helga
    0
    2021-07-20T13:37:11+00:00
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    9514 1404 393

    Answer:

      -1

    Step-by-step explanation:

    Alternate terms have a sum of zero:

      i^n +i^(n+2) = (i^n)(1 +i^2) = (i^n)(1 -1) = 0

    So, the sum to i^256 is zero, and i^257 +i^259 = 0. The value of the sum is then i^258 = i^(258 mod 4) = i^2 = -1

    The given expression evaluates to -1.

  2. thnahdat
    0
    2021-07-20T13:37:49+00:00
    Reply

    Answer:

    -1

    Step-by-step explanation:

    Note that i+i^2+i^3+i^4 = i-1-i+1 = 0, and this means that every 4 terms, the terms cancel to 0. Therefore, by taking modulo 4, we only need to find i^{257}+i^{258}+i^{259} = i-1-i = -1 since $i$ has a cycle of 4.

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