If a and b are the greatest values of  2nCr and 2n−1Cr respectively. Then

We know that the greatest value of  nCr is given by nCn/2     if n is even  nCn−12 or  nCn+12,     if n is odd ∴ a= greatest value of  2nCr=2nCn and, b= Greatest value of  2n−1Cr=2n−1Cn−1⇒ab= 2nCn 2n−1Cn−1=2nn2n−1Cn−12n−1Cn−1=2⇒a=2b