Find i³⁵ and i⁻¹⁴. Then find the sum i + i² + i³ + … + i¹⁰.

Step 1. $i^{35}$. $35 = 4 \times 8 + 3$, so $i^{35} = i^3 = -i$.

Step 2. $i^{-14}$. $i^{-14} = \dfrac{1}{i^{14}}$, and $14 = 4 \times 3 + 2$, so $i^{14} = -1$ and $i^{-14} = -1$.

Step 3. The sum. The first eight powers make two groups of four, which add up to $0$. What is left is $i^9 + i^{10} = i - 1$.