Question

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">.

Answer 1

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.

Answer 2

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.