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docker41 [41]
3 years ago
12

Compute

i+i^2+i^3+\cdots+i^{258}+i^{259}" alt="i+i^2+i^3+\cdots+i^{258}+i^{259}" align="absmiddle" class="latex-formula">.
Mathematics
2 answers:
artcher [175]3 years ago
7 0

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.

xz_007 [3.2K]3 years ago
3 0

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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