The value of 1i - 1i2 - 1i3 - - - 1i102 is

The correct option is E 1 i 1i + 1i^2 + 1i^3 + .... + 1i^102 (GP)Therefore, S_n = 1i{1 (1i )^102}1 (1i ) = (1 + 1)i 1 (∵ i^2 = 1 and ...

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Summation by Sigma Method

The value of ...

Question

The value of $\frac{1}{i} + \frac{1}{i^{2}} + \frac{1}{i^{3}} + . . . . + \frac{1}{i^{102}}$ is

- 1 - i

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- 1 + i

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1 - i

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1 + i

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1 - 2 i

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Solution

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1 + (1 + i) + (1 + i)^{2} + (1 + i)^{3} =

i = \sqrt{- 1}

i^{2} = - 1

i^{3} = - i

i^{4} = 1

i^{127}

{(\frac{1 + i}{1 - i})}^{2} + {(\frac{1 - i}{1 + i})}^{2}

(1 + i + i^{2} + i^{3} + i^{4} + i^{5}) (1 + i) =

i^{3} = \frac{1}{i} = \frac{1}{i} \times \frac{i}{i} = \frac{- i}{1} = - i

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