(9) if n is an odd positive integer then the value of 1+(i)2n+(i)4n+(i)6n is :(A) - 41 (B) 0 (C) 4i (D) 4
Answers
Answered by
51
Answer:
1 + i²ⁿ + i⁴ⁿ + i⁶ⁿ = 0
Step-by-step explanation:
1 + i²ⁿ + i⁴ⁿ + i⁶ⁿ
= (1 + i²ⁿ) + i⁴ⁿ(1 + i²ⁿ)
i⁴ⁿ = 1ⁿ = 1
= (1 + i²ⁿ) + 1(1 + i²ⁿ)
= 2(1 + i²ⁿ)
i²ⁿ = (i²)ⁿ = (-1)ⁿ
n is odd
=> (-1)ⁿ = - 1
= 2( 1 - 1)
= 2 (0)
= 0
1 + i²ⁿ + i⁴ⁿ + i⁶ⁿ = 0
Answered by
0
Answer:
Answer:
1 + i ^ (2n) + i ^ (4n) + i ^ (6n) = 0
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