Question:medium

Evaluate \( i^2 + i^3 + \dots + i^{4000} \):

Show Hint

When working with powers of \( i \), recognize the repeating cycle \( (i, -1, -i, 1) \) and simplify by grouping terms accordingly.
Updated On: Mar 28, 2026
  • \( 1 \)
  • \( 0 \)
  • \( i \)
  • \( -i \)
Show Solution

The Correct Option is D

Solution and Explanation

The powers of \( i \) exhibit a repeating cycle of length 4: \(i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1\). This pattern repeats. The sum \( i^2 + i^3 + \dots + i^{4000} \) can be viewed as a sequence of these cycles: \[ (i^2 + i^3 + i^4 + i^1) + (i^2 + i^3 + i^4 + i^1) + \dots \text{(1000 repetitions)}. \] Each cycle sums to: \[ i^2 + i^3 + i^4 + i^1 = -1 - i + 1 + i = 0. \] Therefore, the sum of these complete cycles is \(0 \cdot 1000 = 0\). The remaining terms are \( i^2 \) and \( i^3 \): \[ i^2 + i^3 = -1 + i. \] Consequently, the total sum evaluates to: \[ 0 + (-1 + i) = -1 + i. \] The final result is: \[ \boxed{-1 + i}. \]
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