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