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TS EAMCET · Maths · Complex Number

\(i^2+i^3+\ldots+i^{4000}=\)

  1. A 1
  2. B 0
  3. C \(i\)
  4. D \(-i\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-i\)

Step-by-step Solution

Detailed explanation

\(\because 1^n+i^{n+1}+i^{n+2}+i^{n+3}=0, \forall n \in\) Integer So, \(i^2+i^3+i^4+\ldots+i^{4000}\) \[ =-i+\left[i+i^2+i^3+\ldots+i^{4000}\right]=-i \]