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.