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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

$${(1+i)}^{80}=\underset{p=0}{\sum ^{80}}{\mathrm{C}}_p^{80}{i}^p$$$${(1+i)}^{80}=\underset{k=0}{\sum ^{80}}{\mathrm{C}}_{2k}^{80}{i}^{2k}+\underset{k=0}{\sum ^{39}}{\mathrm{C}}_{2k+1}^{80}{i}^{2k+1}$$$${(1+i)}^{80}=\underset{k=0}{\sum ^{80}}{(-1)}^k{\mathrm{C}}_{2k}^{80}+\underset{k=0}{\sum ^{39}}{(-1)}^k{\mathrm{C}}_{2k+1}^{80}i$$$$\mathrm{Re}\left[{(1+i)}^{80}\right]=\underset{k=0}{\sum ^{80}}{(-1)}^k{\mathrm{C}}_{2k}^{80}$$$$\mathrm{Re}\left[{\left(\sqrt{2}{\mathrm{e}}^{i\frac{\pi }{4}}\right)}^{80}\right]=\underset{k=0}{\sum ^{80}}{(-1)}^k{\mathrm{C}}_{2k}^{80}$$$$2^{40}=\underset{k=0}{\sum ^{80}}{(-1)}^k{\mathrm{C}}_{2k}^{80}$$

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