Question:
The value of \(i^{3} + i^{4} + i^{5} + \ldots i^{93}\) , where \(i = \sqrt{-1}\) , is equal to
The value of \(i^{3} + i^{4} + i^{5} + \ldots i^{93}\) , where \(i = \sqrt{-1}\) , is equal to
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Group powers in sets of 4; sum of each set is 0. Handle remaining terms individually.
Updated On: Apr 25, 2026
- i
- 1
- -i
- -1
- 0
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The Correct Option is B
Solution and Explanation
Step 1: Concept:
• Powers of \(i\) repeat in a cycle of 4: \[ i^1 = i,\quad i^2 = -1,\quad i^3 = -i,\quad i^4 = 1 \]
• After every 4 terms, the pattern repeats.
Step 2: Calculation:
• We need to find: \[ i^3 + i^4 + i^5 + \dots + i^{93} \]
• Total number of terms: \[ 93 - 3 + 1 = 91 \]
• Since powers repeat every 4 terms: \[ 91 = 22 \times 4 + 3 \]
• Sum of any 4 consecutive powers of \(i\) is: \[ i^3 + i^4 + i^5 + i^6 = (-i + 1 + i - 1) = 0 \]
• So, 22 full cycles contribute: \[ 22 \times 0 = 0 \]
• Remaining 3 terms: \[ i^3 + i^4 + i^5 = (-i + 1 + i) = 1 \]
Step 3: Final Answer:
• Total sum = \(1\)
• Correct Option: (B)
• Powers of \(i\) repeat in a cycle of 4: \[ i^1 = i,\quad i^2 = -1,\quad i^3 = -i,\quad i^4 = 1 \]
• After every 4 terms, the pattern repeats.
Step 2: Calculation:
• We need to find: \[ i^3 + i^4 + i^5 + \dots + i^{93} \]
• Total number of terms: \[ 93 - 3 + 1 = 91 \]
• Since powers repeat every 4 terms: \[ 91 = 22 \times 4 + 3 \]
• Sum of any 4 consecutive powers of \(i\) is: \[ i^3 + i^4 + i^5 + i^6 = (-i + 1 + i - 1) = 0 \]
• So, 22 full cycles contribute: \[ 22 \times 0 = 0 \]
• Remaining 3 terms: \[ i^3 + i^4 + i^5 = (-i + 1 + i) = 1 \]
Step 3: Final Answer:
• Total sum = \(1\)
• Correct Option: (B)
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