Question

Compute
`i+i^2+i^3+\cdots+i^(258)+i^(259)`.

Answer 1

-1

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.

Answer 2

9514 1404 393

Answer:

-1

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.