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Question Number 136098 by SOMEDAVONG last updated on 18/Mar/21
I).compute Σ_(k=0) ^(1000) C_(2000) ^(2k) .
$$\left.\mathrm{I}\right).\mathrm{compute}\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{1000}} {\sum}}\mathrm{C}_{\mathrm{2000}} ^{\mathrm{2k}} . \\ $$
Answered by Olaf last updated on 18/Mar/21
(1−1)^(2000) = 0 = Σ_(p=0) ^(2000) (−1)^p C_p ^(2000) 0 = Σ_(k=0) ^(1000) (−1)^(2k) C_(2k) ^(2000) +Σ_(k=0) ^(999) (−1)^(2k+1) C_(2k+1) ^(2000) 0 = Σ_(k=0) ^(1000) C_(2k) ^(2000) −Σ_(k=0) ^(999) C_(2k+1) ^(2000) Σ_(k=0) ^(1000) C_(2k) ^(2000) = Σ_(k=0) ^(999) C_(2k+1) ^(2000) (1+1)^(2000) = 2^(2000) = Σ_(p=0) ^(2000) C_p ^(2000) 2^(2000) = Σ_(k=0) ^(1000) C_(2k) ^(2000) +Σ_(k=0) ^(999) C_(2k+1) ^(2000) 2^(2000) = Σ_(k=0) ^(1000) C_(2k) ^(2000) +Σ_(k=0) ^(1000) C_(2k) ^(2000) 2^(2000) = 2Σ_(k=0) ^(1000) C_(2k) ^(2000) Σ_(k=0) ^(1000) C_(2k) ^(2000) = (1/2)(2^(2000) ) = 2^(1999)
$$\left(\mathrm{1}−\mathrm{1}\right)^{\mathrm{2000}} \:=\:\mathrm{0}\:=\:\underset{{p}=\mathrm{0}} {\overset{\mathrm{2000}} {\sum}}\left(−\mathrm{1}\right)^{{p}} \mathrm{C}_{{p}} ^{\mathrm{2000}} \\ $$$$\mathrm{0}\:=\:\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{1000}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{2}{k}} \mathrm{C}_{\mathrm{2}{k}} ^{\mathrm{2000}} +\underset{{k}=\mathrm{0}} {\overset{\mathrm{999}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{2}{k}+\mathrm{1}} \mathrm{C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{2000}} \\ $$$$\mathrm{0}\:=\:\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{1000}} {\sum}}\mathrm{C}_{\mathrm{2}{k}} ^{\mathrm{2000}} −\underset{{k}=\mathrm{0}} {\overset{\mathrm{999}} {\sum}}\mathrm{C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{2000}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\mathrm{1000}} {\sum}}\mathrm{C}_{\mathrm{2}{k}} ^{\mathrm{2000}} \:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{999}} {\sum}}\mathrm{C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{2000}} \\ $$$$ \\ $$$$\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2000}} \:=\:\mathrm{2}^{\mathrm{2000}} \:=\:\:\underset{{p}=\mathrm{0}} {\overset{\mathrm{2000}} {\sum}}\mathrm{C}_{{p}} ^{\mathrm{2000}} \\ $$$$\mathrm{2}^{\mathrm{2000}} \:=\:\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{1000}} {\sum}}\mathrm{C}_{\mathrm{2}{k}} ^{\mathrm{2000}} +\underset{{k}=\mathrm{0}} {\overset{\mathrm{999}} {\sum}}\mathrm{C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{2000}} \\ $$$$\mathrm{2}^{\mathrm{2000}} \:=\:\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{1000}} {\sum}}\mathrm{C}_{\mathrm{2}{k}} ^{\mathrm{2000}} +\underset{{k}=\mathrm{0}} {\overset{\mathrm{1000}} {\sum}}\mathrm{C}_{\mathrm{2}{k}} ^{\mathrm{2000}} \\ $$$$\mathrm{2}^{\mathrm{2000}} \:=\:\:\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\mathrm{1000}} {\sum}}\mathrm{C}_{\mathrm{2}{k}} ^{\mathrm{2000}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\mathrm{1000}} {\sum}}\mathrm{C}_{\mathrm{2}{k}} ^{\mathrm{2000}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}^{\mathrm{2000}} \right)\:=\:\mathrm{2}^{\mathrm{1999}} \\ $$
Commented by SOMEDAVONG last updated on 18/Mar/21
THANKS TEACHER
$$ \\ $$$$\mathbb{THANKS}\:\:\:\mathbb{TEACHER} \\ $$

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