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Question-187947




Question Number 187947 by thean last updated on 24/Feb/23
Answered by witcher3 last updated on 24/Feb/23
(1+i)^(2000) =Σ_(k=0) ^(2000) i^k C_(2000) ^k =(Σ_(k=0) ^(1000) (−1)^k C_(2000) ^(2k) +iΣ_(k=0) ^(999) (−1)^k C_(2000) ^(2k+1) )  S=Re(1+i)^(2000) =2^(1000) e^(i500π) =(2)^(1000)
$$\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{2000}} =\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2000}} {\sum}}\mathrm{i}^{\mathrm{k}} \mathrm{C}_{\mathrm{2000}} ^{\mathrm{k}} =\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{1000}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{C}_{\mathrm{2000}} ^{\mathrm{2k}} +\mathrm{i}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{999}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{C}_{\mathrm{2000}} ^{\mathrm{2k}+\mathrm{1}} \right) \\ $$$$\mathrm{S}=\mathrm{Re}\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{2000}} =\mathrm{2}^{\mathrm{1000}} \mathrm{e}^{\mathrm{i500}\pi} =\left(\mathrm{2}\right)^{\mathrm{1000}} \\ $$

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