If x + y = 1, then value of ∑r=0n (r) nCrxn−ryr is

We have $(a + t)^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} t^{r}$

Differentiating both the sides with respect to t, we get

$n (a + t)^{n - 1} = \sum_{r = 0}^{n} r (^{n} C_{r}) a^{n - r} t^{r - 1}$

Multiplying both the sides by t and putting $a = x$ and $t = y$ , we get

$n (x + y)^{n - 1} y = \sum_{r = 0}^{n} r (^{n} C_{r}) x^{n - r} y^{r}$

$\Rightarrow \sum_{r = 0}^{n} r (^{n} C_{r}) x^{n - r} y^{r} = n y [∵ x + y = 1]$

If $x + y = 1$ , then value of $\sum_{r = 0}^{n} (r) (^{n} C_{r}) x^{n - r} y^{r}$ is