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Question Number 159259 by pticantor last updated on 14/Nov/21
montre que   Σ_(k=0) ^n C_(2n) ^(2k) =2^(2n−1)
$$\boldsymbol{{montre}}\:\boldsymbol{{que}}\: \\ $$$$\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\sum}}\boldsymbol{{C}}_{\mathrm{2}\boldsymbol{{n}}} ^{\mathrm{2}\boldsymbol{{k}}} =\mathrm{2}^{\mathrm{2}\boldsymbol{{n}}−\mathrm{1}} \\ $$
Answered by mr W last updated on 14/Nov/21
(1+x)^(2n) =Σ_(k=0) ^n C_(2k) ^(2n) x^(2k) +Σ_(k=0) ^(n−1) C_(2k+1) ^(2n) x^(2k+1)    ..(i)  (1−x)^(2n) =Σ_(k=0) ^n C_(2k) ^(2n) x^(2k) −Σ_(k=0) ^(n−1) C_(2k+1) ^(2n) x^(2k+1)    ..(ii)    Σ_(k=0) ^n C_(2k) ^(2n) x^(2k) =(((1+x)^(2n) +(1−x)^(2n) )/2)  let x=1  Σ_(k=0) ^n C_(2k) ^(2n) =(((1+1)^(2n) +(1−1)^(2n) )/2)=(2^(2n) /2)=2^(2n−1)
$$\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{\mathrm{2}{k}} ^{\mathrm{2}{n}} {x}^{\mathrm{2}{k}} +\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{2}{n}} {x}^{\mathrm{2}{k}+\mathrm{1}} \:\:\:..\left({i}\right) \\ $$$$\left(\mathrm{1}−{x}\right)^{\mathrm{2}{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{\mathrm{2}{k}} ^{\mathrm{2}{n}} {x}^{\mathrm{2}{k}} −\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{2}{n}} {x}^{\mathrm{2}{k}+\mathrm{1}} \:\:\:..\left({ii}\right) \\ $$$$ \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{\mathrm{2}{k}} ^{\mathrm{2}{n}} {x}^{\mathrm{2}{k}} =\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} +\left(\mathrm{1}−{x}\right)^{\mathrm{2}{n}} }{\mathrm{2}} \\ $$$${let}\:{x}=\mathrm{1} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{\mathrm{2}{k}} ^{\mathrm{2}{n}} =\frac{\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}{n}} +\left(\mathrm{1}−\mathrm{1}\right)^{\mathrm{2}{n}} }{\mathrm{2}}=\frac{\mathrm{2}^{\mathrm{2}{n}} }{\mathrm{2}}=\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \\ $$
Commented by pticantor last updated on 15/Nov/21
merci beaucoup
$$\boldsymbol{{merci}}\:\boldsymbol{{beaucoup}} \\ $$

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