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Question Number 136098 by SOMEDAVONG last updated on 18/Mar/21

$$\mathrm{I}).\mathrm{compute}\underset{\mathrm{k}=0}{\sum ^{1000}}{\mathrm{C}}_{2000}^{2\mathrm{k}}.$$

Answered by Olaf last updated on 18/Mar/21

$${(1-1)}^{2000}=0=\underset{p=0}{\sum ^{2000}}{(-1)}^p{\mathrm{C}}_p^{2000}$$$$0=\underset{k=0}{\sum ^{1000}}{(-1)}^{2k}{\mathrm{C}}_{2k}^{2000}+\underset{k=0}{\sum ^{999}}{(-1)}^{2k+1}{\mathrm{C}}_{2k+1}^{2000}$$$$0=\underset{k=0}{\sum ^{1000}}{\mathrm{C}}_{2k}^{2000}-\underset{k=0}{\sum ^{999}}{\mathrm{C}}_{2k+1}^{2000}$$$$\underset{k=0}{\sum ^{1000}}{\mathrm{C}}_{2k}^{2000}=\underset{k=0}{\sum ^{999}}{\mathrm{C}}_{2k+1}^{2000}$$$$$$$${(1+1)}^{2000}=2^{2000}=\underset{p=0}{\sum ^{2000}}{\mathrm{C}}_p^{2000}$$$$2^{2000}=\underset{k=0}{\sum ^{1000}}{\mathrm{C}}_{2k}^{2000}+\underset{k=0}{\sum ^{999}}{\mathrm{C}}_{2k+1}^{2000}$$$$2^{2000}=\underset{k=0}{\sum ^{1000}}{\mathrm{C}}_{2k}^{2000}+\underset{k=0}{\sum ^{1000}}{\mathrm{C}}_{2k}^{2000}$$$$2^{2000}=2\underset{k=0}{\sum ^{1000}}{\mathrm{C}}_{2k}^{2000}$$$$\underset{k=0}{\sum ^{1000}}{\mathrm{C}}_{2k}^{2000}=\frac{1}{2}\left(2^{2000}\right)=2^{1999}$$

Commented by SOMEDAVONG last updated on 18/Mar/21

$$$$$$\mathbb{THANKS}\mathbb{TEACHER}$$

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