If n is an odd positive integer, then the value of 1 + (i)^2n + (i)^4n + (i)^6n is: (A) -4i (B) 0 (C) 4i (D) 4

If n is an odd positive integer, then the value of 1 + (i)²ⁿ + (i)⁴ⁿ + (i)⁶ⁿ is:

(A) -4i

(B) 0

(C) 4i

(D) 4

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

Salim Ahmed Kawsar

(B) 0

1 + (i²)ⁿ + (i⁴)ⁿ + (i²)³ⁿ

= 1 – 1 + 1 – 1 …..(n odd positive integer)

= 0

6 views Answered 2023-02-05 15:29

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