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

We have (a+t)n=∑r=0n nCran−rtr Differentiating both the sides with respect to t, we get n(a+t)n−1=∑r=0n r nCran−rtr−1Multiplying both the sides by t and putting a = x and t = y, we getn(x+y)n−1y=∑r=0n r nCrxn−ryr ⇒∑r=0n r nCrxn−ryr=ny [∵ x+y=1]