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1-i-i-2-i-3-i-99-




Question Number 145954 by mathdanisur last updated on 09/Jul/21
1+i+i^2 +i^3 +...+i^(99) =?
$$\mathrm{1}+{i}+{i}^{\mathrm{2}} +{i}^{\mathrm{3}} +…+{i}^{\mathrm{99}} =? \\ $$
Answered by mr W last updated on 09/Jul/21
=((1−i^(100) )/(1−i))=((1−(−1)^(50) )/(1−i))=((1−1)/(1−i))=0
$$=\frac{\mathrm{1}−{i}^{\mathrm{100}} }{\mathrm{1}−{i}}=\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{50}} }{\mathrm{1}−{i}}=\frac{\mathrm{1}−\mathrm{1}}{\mathrm{1}−{i}}=\mathrm{0} \\ $$
Commented by mathdanisur last updated on 09/Jul/21
thank you Ser
$${thank}\:{you}\:{Ser} \\ $$

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