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Question Number 27380 by abdo imad last updated on 05/Jan/18

$$findthevalueof{S}_n=\sum _{k=0}^{k=n}\frac{{(-1)}^k{C}_n^k}{2k+1}.$$

Commented by abdo imad last updated on 07/Jan/18

$$letintroducethepolynomialp\left(x\right)=\sum _{k=0}^{k=n}\frac{{(-1)}^k{C}_n^k}{2k+1}{x}^{2k+1}$$$$S_n=p\left(1\right)wehave{p}^{,}\left(x\right)=\sum _{k=0}^n{C}_n^k{(-1)}^k{x}^{2k}$$$$=\sum _{k=0}^{k=n}{C}_n^k{(-x^2)}^k={(1-x^2)}^n$$$$p\left(x\right)={\int }_0^x{(1-t^2)}^{n}dt+\lambda and\lambda =p\left(0\right)=0$$$$S_n={\int }_0^1{(1-t^2)}^{n}dtandthech.t=sinxgive$$$$S_n={\int }_0^{\frac{\pi }{2}}{(1-{sin}^{2}x)}^{n}cosxdx={\int }_0^{\frac{\pi }{2}}{cosx}^{2n+1}dxletput$$$$I_n={\int }_0^{\frac{\pi }{2}}{\left(cosx\right)}^{n}dx\left(wallisintegral\right)integrationbyparts$$$$give{I}_n=\frac{n-1}{n}{I}_{n-2}\Rightarrow I_{2n+1}=\frac{2n}{2n+1}{I}_{2n-1}$$$$\prod _{k=1}^n{I}_{2k+1}=\prod _{k=1}^n\frac{2k}{2k+1}\prod _{k=1}^n{I}_{2k-1}$$$$\Rightarrow I_{2n+1}=\frac{2^n(n!)}{3.5....(2n+1)}{I}_{1}but{I}_1=1$$$$S_n=\frac{2^{2n}{(n!)}^2}{(2n+1)!}.$$

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