If ∑r=0n (pr+2)⋅nCr=(25)(64) where n, p∈N, then

S=∑r=0n (pr+2)⋅nCr∴ =∑r=0n (p(n−r)+2)⋅nCr            ∵ nCr=nCn−rAdding these, we get2S=(pn+4)∑r=0n nCr⇒ S=(pn+4)×2n−1So, n = 7 and 7p + 4 = 25 or p = 3.