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Question Number 142305 by anonymo last updated on 29/May/21
Answered by Olaf_Thorendsen last updated on 01/Jun/21
![(1+i)^(80) = Σ_(p=0) ^(80) C_p ^(80) i^p (1+i)^(80) = Σ_(k=0) ^(80) C_(2k) ^(80) i^(2k) +Σ_(k=0) ^(39) C_(2k+1) ^(80) i^(2k+1) (1+i)^(80) = Σ_(k=0) ^(80) (−1)^k C_(2k) ^(80) +Σ_(k=0) ^(39) (−1)^k C_(2k+1) ^(80) i Re[(1+i)^(80) ] = Σ_(k=0) ^(80) (−1)^k C_(2k) ^(80) Re[((√2)e^(i(π/4)) )^(80) ] = Σ_(k=0) ^(80) (−1)^k C_(2k) ^(80) 2^(40) = Σ_(k=0) ^(80) (−1)^k C_(2k) ^(80)](Q142460.png)
$$\left(\mathrm{1}+{i}\right)^{\mathrm{80}} \:=\:\underset{{p}=\mathrm{0}} {\overset{\mathrm{80}} {\sum}}\mathrm{C}_{{p}} ^{\mathrm{80}} {i}^{{p}} \\ $$$$\left(\mathrm{1}+{i}\right)^{\mathrm{80}} \:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{80}} {\sum}}\mathrm{C}_{\mathrm{2}{k}} ^{\mathrm{80}} {i}^{\mathrm{2}{k}} +\underset{{k}=\mathrm{0}} {\overset{\mathrm{39}} {\sum}}\mathrm{C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{80}} {i}^{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\left(\mathrm{1}+{i}\right)^{\mathrm{80}} \:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{80}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \mathrm{C}_{\mathrm{2}{k}} ^{\mathrm{80}} +\underset{{k}=\mathrm{0}} {\overset{\mathrm{39}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \mathrm{C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{80}} {i} \\ $$$$\mathrm{Re}\left[\left(\mathrm{1}+{i}\right)^{\mathrm{80}} \right]\:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{80}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \mathrm{C}_{\mathrm{2}{k}} ^{\mathrm{80}} \\ $$$$\mathrm{Re}\left[\left(\sqrt{\mathrm{2}}\mathrm{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{80}} \right]\:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{80}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \mathrm{C}_{\mathrm{2}{k}} ^{\mathrm{80}} \\ $$$$\mathrm{2}^{\mathrm{40}} \:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{80}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \mathrm{C}_{\mathrm{2}{k}} ^{\mathrm{80}} \\ $$