Discussion
Given that \(i^{2} = -1\) . Then \(i^{13} + i^{14} + i^{15} + \ldots + i^{2026}\) is equal to
Answer: D
The sum is \(i - 1\).
Powers of \(i\) repeat in blocks of four, \(i, -1, -i, 1\), and each block of four consecutive powers adds to 0. From \(i^{13}\) to \(i^{2026}\) there are \(2026 - 13 + 1 = 2014\) terms, and \(2014 = 4 \times 503 + 2\).
So 503 complete blocks contribute 0, leaving the last two terms:
\[ \begin{aligned} i^{2025} + i^{2026} &= i^{4 \cdot 506 + 1} + i^{4 \cdot 506 + 2} \\ &= i + (-1) = i - 1 \end{aligned} \]
Equivalently, the first two terms \(i^{13} = i\) and \(i^{14} = -1\) are the ones left over once the complete blocks cancel. The other options use the wrong remainder or sign.
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