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Question Number 145954 by mathdanisur last updated on 09/Jul/21

$$1+i+i^2+i^3+...+i^{99}=?$$

Answered by mr W last updated on 09/Jul/21

$$=\frac{1-i^{100}}{1-i}=\frac{1-{(-1)}^{50}}{1-i}=\frac{1-1}{1-i}=0$$

Commented by mathdanisur last updated on 09/Jul/21

$$thankyouSer$$

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