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n-1-3050-i-n-




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

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