2nCr(0≤r≤2n) is greatest when r is equal to

We have,  2nCr 2nCr−1=2n−r+1r            (1)For  2nCr to be greatest 2n−r+1r≥1⇒ 2n−r+1≥r⇒2r≤2n+1⇒r≤n+12                                           (2)From (1),  2nCr+1 2nCr=2n−(r+1)+1r+1=2n−rr+1.For  2nCr to be greatest 2n−rr+1≤1⇒ 2n−r≤r+1⇒2r≥2n−1⇒r≥n−12                                          (3)From (2) and (3), we get n−12≤r≤n+12⇒ r = n (since r is a positive integer)Hence,  2nCr is greatest when r = n.