If $\mathrm{x}+\mathrm{y}=1$, then $\sum_{\mathrm{r}=0}^{\mathrm{n}} \mathrm{r}^{\mathrm{n}} \mathrm

Question

If $\mathrm{x}+\mathrm{y}=1$, then $\sum_{\mathrm{r}=0}^{\mathrm{n}} \mathrm{r}^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}} \mathrm{y}^{\mathrm{n}-\mathrm{r}}$ equals

Professor PC Thomas Classes · Answer

$ \begin{aligned} & \sum_{r=0}^n r \cdot{ }^n C_r x^r y^{n-r}=\sum_{r=1}^n n \cdot{ }^{n-1} C_{r-1} x^r y^{n-r} \ & =n x \sum_{r=1}^n{ }^{n-1} C_{r-1} x^{r-1} y^{(n-1)-(r-1)} \ & =n x \end{aligned} $