Powers of i follow a cycle of length 4: i^0=1, i^1=i, i^2=-1, i^3=-i, i^4=1, ...

Quick computation: i^n = i^(n mod 4). So i^{2025} = i^(2025 mod 4) = i^1 = i.

Sum formula: i + i^2 + i^3 + ... + i^{4n} = 0 (every complete cycle sums to zero).

For partial sums: i + i^2 + ... + i^k, compute k mod 4 and add only the remaining terms to the complete cycles (which contribute 0).

Negative powers: i^(-1) = 1/i = i^3 = -i (multiply numerator and denominator by i). i^(-2) = -1. i^(-3) = i.