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Question

Question: If \(i = \sqrt{- 1}\) and n is a positive integer, than \[i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3}...

If i=−1i = \sqrt{- 1} and n is a positive integer, than

in+in+1+in+2+in+3=i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3} =

A

1

B

I

C

ini^{n}

D

0

Answer

0

Explanation

Solution

Sol. in+in+1+in+2+in+3=in(1+i+i2+i3)=in(1+i−1−i)=o.i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3} = i^{n}(1 + i + i^{2} + i^{3}) = i^{n}(1 + i - 1 - i) = o. Trick: Since the sum of four consecutive powers of i is always zero.

⇒ in+in+1+in+2+in+3=0,n∈I.i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3} = 0,n \in I.