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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

$$anoldquestion$$

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

$$\underset{k=0}{\sum ^n}{(-1)}^n\frac{{\stackrel{n}{C}}_0}{3k+1}={\int }_0^1\underset{k=0}{\sum ^n}{(-1)}^k{C}_0^k{x}^{3k}={\int }_0^1{(1-x^3)}^{n}dx$$$$=\frac{1}{3}{\int }_0^1{u}^{-\frac{2}{3}}{(1-u)}^{n}dx=\frac{\mathrm{\Gamma }\left(\frac{1}{3}\right)\mathrm{\Gamma }(n+1)}{3\mathrm{\Gamma }(n+\frac{4}{3})}=\frac{\mathrm{\Gamma }\left(\frac{4}{3}\right)\mathrm{\Gamma }(n+1)}{\mathrm{\Gamma }(n+\frac{4}{3})}=\mathrm{\Phi }\left(n\right)$$$$n=0\mathrm{\Phi }\left(0\right)=1$$$$n=1\mathrm{\Phi }\left(1\right)=\frac{3}{4}$$$$...$$

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

$${(1-x^3)}^n=\underset{k=0}{\sum ^n}{(-1)}^k{C}_k^n{x}^{3k}$$$${\int }_0^1{(1-x^3)}^{n}dx=\underset{k=0}{\sum ^n}{(-1)}^k\frac{C_k^n}{3k+1}$$$$\Rightarrow \underset{k=0}{\sum ^n}{(-1)}^k\frac{C_k^n}{3k+1}=\frac{B(\frac{1}{3},n+1)}{3}$$

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

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

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

$$yes,{that}^{\prime }scorrect.$$

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