Discussion
The value of \(i^{3} + i^{4} + i^{5} + \ldots i^{93}\), where \(i = \sqrt{-1}\) , is equal to
Answer: B
The sum is 1.
The powers of \(i\) repeat every four terms, and any four consecutive powers add to zero: \(i^{k} + i^{k+1} + i^{k+2} + i^{k+3} = i^{k}(1 + i - 1 - i) = 0\).
From \(i^{3}\) to \(i^{93}\) there are 91 terms. The first 88 (\(i^{3}\) to \(i^{90}\)) form 22 complete groups and cancel, leaving
\[ i^{91} + i^{92} + i^{93} = -i + 1 + i = 1 \]
The other options come from miscounting the leftover terms: leaving one or two terms gives \(-i\) or \(1 - i\), and a whole number of cycles gives 0.
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