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Find-i-4-i-8-i-12-i-16-i-20-i-24-i-100-




Question Number 207885 by hardmath last updated on 29/May/24
Find:  i^4  + i^8  + i^(12)  + i^(16)  + i^(20)  + i^(24)  +...+ i^(100)  = ?
$$\mathrm{Find}: \\ $$$$\boldsymbol{\mathrm{i}}^{\mathrm{4}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{8}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{12}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{16}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{20}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{24}} \:+…+\:\boldsymbol{\mathrm{i}}^{\mathrm{100}} \:=\:? \\ $$
Commented by mr W last updated on 29/May/24
do you mean i=(√(−1)) ?
$${do}\:{you}\:{mean}\:{i}=\sqrt{−\mathrm{1}}\:? \\ $$
Commented by hardmath last updated on 29/May/24
yes professor...  answer: 25
$$\mathrm{yes}\:\mathrm{professor}… \\ $$$$\mathrm{answer}:\:\mathrm{25} \\ $$
Commented by mr W last updated on 29/May/24
i^4 =i^8 =i^(16) =...=i^(100) =1  Σ=1+1+1+...+1_(25 times) =25
$${i}^{\mathrm{4}} ={i}^{\mathrm{8}} ={i}^{\mathrm{16}} =…={i}^{\mathrm{100}} =\mathrm{1} \\ $$$$\Sigma=\underset{\mathrm{25}\:{times}} {\mathrm{1}+\mathrm{1}+\mathrm{1}+…+\mathrm{1}}=\mathrm{25} \\ $$
Commented by hardmath last updated on 29/May/24
  Professor, is there a golden rule for these types of examples?
$$ \\ $$Professor, is there a golden rule for these types of examples?
Commented by mr W last updated on 29/May/24
i=(√(−1))  i^2 =−1  i^4 =(−1)^2 =1  i^8 =i^4 ×i^4 =1×1=1  ...
$${i}=\sqrt{−\mathrm{1}} \\ $$$${i}^{\mathrm{2}} =−\mathrm{1} \\ $$$${i}^{\mathrm{4}} =\left(−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${i}^{\mathrm{8}} ={i}^{\mathrm{4}} ×{i}^{\mathrm{4}} =\mathrm{1}×\mathrm{1}=\mathrm{1} \\ $$$$… \\ $$
Commented by hardmath last updated on 29/May/24
thank you dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Commented by Frix last updated on 29/May/24
i^(4n) =1  i^(4n+1) =i  i^(4n+2) =−1  i^(4n+3) =−i
$$\mathrm{i}^{\mathrm{4}{n}} =\mathrm{1} \\ $$$$\mathrm{i}^{\mathrm{4}{n}+\mathrm{1}} =\mathrm{i} \\ $$$$\mathrm{i}^{\mathrm{4}{n}+\mathrm{2}} =−\mathrm{1} \\ $$$$\mathrm{i}^{\mathrm{4}{n}+\mathrm{3}} =−\mathrm{i} \\ $$

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