If an = Σ1/nCr , then Σr/nCr equals where summation is from r=0 to r=n.

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Please find below the solution to the asked query:

$$\begin{gathered} {\mathrm{a}}_{\mathrm{n}}=\sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{1}{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}} \\ =\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_0}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_1}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_2}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_3}+.....+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-3}}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-2}}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-1}}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}} \\ =1+\left\{\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_1}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_2}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_3}+.....+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_3}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_2}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_1}\right\}+1 \left[\mathrm{As}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-\mathrm{r}} \mathrm{and}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{=}^{\mathrm{n}}{\mathrm{C}}_0=1\right] \\ {\mathrm{a}}_{\mathrm{n}}=2+2\left\{\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_1}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_2}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_3}+\right\} \\ \Rightarrow \frac{{\mathrm{a}}_{\mathrm{n}}-2}{2}=\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_1}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_2}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_3}+...\left(\mathrm{i}\right) \\ \mathrm{S}=\sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{\mathrm{r}}{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}} \\ =0+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_1}+\frac{2}{{}^{\mathrm{n}}{\mathrm{C}}_2}+\frac{3}{{}^{\mathrm{n}}{\mathrm{C}}_3}+.....+\frac{\mathrm{n}-3}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-3}}+\frac{\mathrm{n}-2}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-2}}+\frac{\mathrm{n}-1}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-1}}+\frac{\mathrm{n}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}} \\ =\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_1}+\frac{2}{{}^{\mathrm{n}}{\mathrm{C}}_2}+\frac{3}{{}^{\mathrm{n}}{\mathrm{C}}_3}+.....+\frac{\mathrm{n}-3}{{}^{\mathrm{n}}{\mathrm{C}}_3}+\frac{\mathrm{n}-2}{{}^{\mathrm{n}}{\mathrm{C}}_2}+\frac{\mathrm{n}-1}{{}^{\mathrm{n}}{\mathrm{C}}_1}+\frac{\mathrm{n}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}} \left[\mathrm{As}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-\mathrm{r}}\right] \\ =\frac{1+\mathrm{n}-1}{{}^{\mathrm{n}}{\mathrm{C}}_1}+\frac{2+\mathrm{n}-2}{{}^{\mathrm{n}}{\mathrm{C}}_2}+\frac{3+\mathrm{n}-3}{{}^{\mathrm{n}}{\mathrm{C}}_3}+... \frac{\mathrm{n}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}} \\ =\frac{\mathrm{n}}{{}^{\mathrm{n}}{\mathrm{C}}_1}+\frac{\mathrm{n}}{{}^{\mathrm{n}}{\mathrm{C}}_2}+\frac{\mathrm{n}}{{}^{\mathrm{n}}{\mathrm{C}}_3}+... \frac{\mathrm{n}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}} \\ =\mathrm{n}\left\{\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_1}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_2}+\frac{1}{{}^{\mathrm{n}}{\mathrm{C}}_3}+..\right\}+... \frac{\mathrm{n}}{1} \\ =\mathrm{n}\left\{\frac{{\mathrm{a}}_{\mathrm{n}}-2}{2}\right\}+\mathrm{n} \left[\mathrm{Using} \left(\mathrm{i}\right)\right] \\ =\mathrm{n}\left\{\frac{{\mathrm{a}}_{\mathrm{n}}-2}{2}+1\right\} \\ =\mathrm{n}\left\{\frac{{\mathrm{a}}_{\mathrm{n}}-2+2}{2}\right\} \\ =\frac{\mathrm{n} {\mathrm{a}}_{\mathrm{n}}}{2} \\ \mathbf{\Rightarrow }\mathbf{\sum }_{\mathbf{r}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathbf{ }\frac{\mathbf{r}}{{\mathbf{ }}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}}}\mathbf{=}\frac{\mathbf{n}\mathbf{ }{\mathbf{a}}_{\mathbf{n}}}{\mathbf{2}} \end{gathered}$$

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