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Question Number 63942 by gunawan last updated on 11/Jul/19

$$\mathrm{Let}{(1+x)}^n=\underset{r=0}{\sum ^n}{C}_r{x}^r\mathrm{and}\underset{r=0}{\sum ^n}\frac{C_r}{r+1}=k,$$$$\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}k\mathrm{is}$$

Commented by mathmax by abdo last updated on 11/Jul/19

$${(1+x)}^n=\sum _{r=0}^n{C}_n^r{x}^r\Rightarrow C_r=C_n^r\Rightarrow$$$$\sum _{r=0}^n\frac{C_r}{r+1}=\sum _{r=0}^n\frac{C_n^r}{r+1}$$$$letp\left(x\right)=\sum _{r=0}^n\frac{C_n^r}{r+1}{x}^{r+1}\Rightarrow p^{{}^{\prime }}\left(x\right)=\sum _{r=0}^n{C}_n^r{x}^r={(x+1)}^n\Rightarrow$$$$p\left(x\right)=\int {(x+1)}^{n}dx+c=\frac{1}{n+1}{(x+1)}^{n+1}+c$$$$p\left(0\right)=0=\frac{1}{n+1}+c\Rightarrow c=-\frac{1}{n+1}\Rightarrow p\left(x\right)=\frac{{(x+1)}^{n+1}-1}{n+1}\Rightarrow$$$$\sum _{r=0}^{\mathrm{\infty }}\frac{C_r}{r+1}=p\left(1\right)=\frac{2^{n+1}-1}{n+1}\Rightarrow k=\frac{2^{n+1}-1}{n+1}.$$$$$$

Answered by mr W last updated on 11/Jul/19

$${(1+x)}^n=\underset{r=0}{\sum ^n}{C}_r{x}^r$$$${\int }_0^x{(1+x)}^{n}dx=\underset{r=0}{\sum ^n}{\int }_0^x{C}_r{x}^{r}dx$$$$\frac{1}{n+1}{(1+x)}^{n+1}-\frac{1}{n+1}=\underset{r=0}{\sum ^n}\frac{1}{r+1}{C}_r{x}^{r+1}$$$$letx=1:$$$$\frac{2^{n+1}-1}{n+1}=\underset{r=0}{\sum ^n}\frac{1}{r+1}{C}_r=k$$$$\Rightarrow k=\frac{2^{n+1}-1}{n+1}$$

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