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Question Number 135532 by mr W last updated on 13/Mar/21

Commented by mr W last updated on 13/Mar/21

an old question

$${an}\:{old}\:{question} \\ $$

Answered by Dwaipayan Shikari last updated on 13/Mar/21

Σ_(k=0) ^n (−1)^n (C_0 ^n /(3k+1))=∫_0 ^1 Σ_(k=0) ^n (−1)^k C_0 ^k x^(3k) =∫_0 ^1 (1−x^3 )^n dx =(1/3)∫_0 ^1 u^(−(2/3)) (1−u)^n dx=((Γ((1/3))Γ(n+1))/(3Γ(n+(4/3))))=((Γ((4/3))Γ(n+1))/(Γ(n+(4/3))))=Φ(n) n=0 Φ(0)=1 n=1 Φ(1)=(3/4) ...

$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\overset{{n}} {{C}}_{\mathrm{0}} }{\mathrm{3}{k}+\mathrm{1}}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} {C}_{\mathrm{0}} ^{{k}} {x}^{\mathrm{3}{k}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{{n}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{−\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{1}−{u}\right)^{{n}} {dx}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left({n}+\mathrm{1}\right)}{\mathrm{3}\Gamma\left({n}+\frac{\mathrm{4}}{\mathrm{3}}\right)}=\frac{\Gamma\left(\frac{\mathrm{4}}{\mathrm{3}}\right)\Gamma\left({n}+\mathrm{1}\right)}{\Gamma\left({n}+\frac{\mathrm{4}}{\mathrm{3}}\right)}=\Phi\left({n}\right) \\ $$$${n}=\mathrm{0}\:\:\Phi\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${n}=\mathrm{1}\:\:\:\Phi\left(\mathrm{1}\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$... \\ $$

Answered by mr W last updated on 13/Mar/21

(1−x^3 )^n =Σ_(k=0) ^n (−1)^k C_k ^n x^(3k) ∫_0 ^1 (1−x^3 )^n dx=Σ_(k=0) ^n (−1)^k (C_k ^n /(3k+1)) ⇒Σ_(k=0) ^n (−1)^k (C_k ^n /(3k+1))=((B((1/3),n+1))/3)

$$\left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} {C}_{{k}} ^{{n}} {x}^{\mathrm{3}{k}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{{n}} {dx}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{{C}_{{k}} ^{{n}} }{\mathrm{3}{k}+\mathrm{1}} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{{C}_{{k}} ^{{n}} }{\mathrm{3}{k}+\mathrm{1}}=\frac{{B}\left(\frac{\mathrm{1}}{\mathrm{3}},{n}+\mathrm{1}\right)}{\mathrm{3}} \\ $$

Commented by Dwaipayan Shikari last updated on 13/Mar/21

(1/3)B((1/3),n+1)=((Γ((1/3))Γ(n+1))/(3Γ(n+(4/3)))) =((Γ((4/3))Γ(n+1))/(Γ(n+(4/3)))) :)

$$\left.\frac{\mathrm{1}}{\mathrm{3}}{B}\left(\frac{\mathrm{1}}{\mathrm{3}},{n}+\mathrm{1}\right)=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left({n}+\mathrm{1}\right)}{\mathrm{3}\Gamma\left({n}+\frac{\mathrm{4}}{\mathrm{3}}\right)}\:=\frac{\Gamma\left(\frac{\mathrm{4}}{\mathrm{3}}\right)\Gamma\left({n}+\mathrm{1}\right)}{\Gamma\left({n}+\frac{\mathrm{4}}{\mathrm{3}}\right)}\:\:\:\::\right) \\ $$

Commented by mr W last updated on 13/Mar/21

yes, that′s correct.

$${yes},\:{that}'{s}\:{correct}. \\ $$

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