Question Number 70132 by RAKESH MANDA last updated on 01/Oct/19
$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{3050}} {\sum}}\:{i}^{{n}} \\ $$
Commented by mathmax by abdo last updated on 01/Oct/19
$${S}\:=\sum_{{n}=\mathrm{0}} ^{\mathrm{3050}} \:{i}^{{n}} −\mathrm{1}\:=\frac{\mathrm{1}−{i}^{\mathrm{3051}} }{\mathrm{1}−{i}}−\mathrm{1}\:=\frac{\mathrm{1}−{i}\left({i}^{\mathrm{2}} \right)^{\mathrm{1525}} }{\mathrm{1}−{i}}−\mathrm{1} \\ $$$$=\frac{\mathrm{1}−{i}\left(−\mathrm{1}\right)^{\mathrm{1525}} }{\mathrm{1}−{i}}−\mathrm{1}\:=\frac{\mathrm{1}+{i}}{\mathrm{1}−{i}}−\mathrm{1}\:=\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1} \\ $$$$=\frac{\mathrm{1}+\mathrm{2}{i}−\mathrm{1}}{\mathrm{2}}−\mathrm{1}\:={i}−\mathrm{1} \\ $$
Answered by naka3546 last updated on 01/Oct/19
$${i}−\mathrm{1} \\ $$