Question

Find the sum ${\sum }_{r=1}^n\frac{r\stackrel{n}{\underset{r}{C}}}{\stackrel{n}{\underset{r-1}{C}}}$

Answer 1

${\sum }_{r=1}^n\frac{r{}^n{C}_r}{{}^n{C}_{r-1}}$

$\Rightarrow {\sum }_{r=1}^n\frac{r{}^n{C}_r}{{}^n{C}_{r-1}}={\sum }_{r=1}^{n}r\frac{n-r+1}{r}$

$\Rightarrow (n+1){\sum }_{r=1}^{n}1-{\sum }_{r=1}^{n}r$

$\Rightarrow n(n+1)-\frac{n(n+1)}{2}$

$\Rightarrow \frac{n(n+1)}{2}$

Therefore, ${\sum }_{r=1}^n\frac{r{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n(n+1)}{2}$