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Question Number 141614 by ArielVyny last updated on 21/May/21

$$\mathrm{\Sigma }\frac{1}{k+1}{C}_n^k.$$

Commented by Dwaipayan Shikari last updated on 21/May/21

$$\frac{1}{k+1}\underset{n=0}{\sum ^k}{C}_n^k=\frac{1}{k+1}\underset{n=0}{\sum ^k}{C}_n^k{\left(1\right)}^{k-n}{\left(1\right)}^n=\frac{{(1+1)}^k}{k+1}=\frac{2^k}{k+1}$$

Commented by mathmax by abdo last updated on 21/May/21

$$\mathrm{not}\mathrm{correct}\mathrm{sir}...$$

Answered by mathmax by abdo last updated on 21/May/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{1}{\mathrm{k}+1}{\mathrm{C}}_{\mathrm{k}}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{k}+1}\mathrm{we}\mathrm{have}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}={(\mathrm{x}+1)}^{\mathrm{n}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\int {(\mathrm{x}+1)}^{\mathrm{n}}\mathrm{dx}+\mathrm{C}$$$$=\frac{{(\mathrm{x}+1)}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}$$$$\mathrm{f}\left(1\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}}{\mathrm{k}+1}=\frac{2^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}$$$$\mathrm{we}\mathrm{f}\left(0\right)=\frac{1}{\mathrm{n}+1}+\mathrm{C}=0\Rightarrow \mathrm{C}=-\frac{1}{\mathrm{n}+1}\Rightarrow \sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}}{\mathrm{k}+1}=\frac{2^{\mathrm{n}+1}}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+1}$$$$=\frac{2^{\mathrm{n}+1}-1}{\mathrm{n}+1}$$

Commented by ArielVyny last updated on 22/May/21

$$thanksir$$

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