Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

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

$$\boldsymbol{{merci}}\:\boldsymbol{{beaucoup}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com