If a and b are greatest values of 2nCr and 2n - 1Cr respectively, then

Let a be the greatest value of ²ⁿC_r (over r) and b the greatest value of ²ⁿ⁻¹C_r (over r).

The binomial coefficient ⁿC_r is largest at the central index. Since 2n is even, the maximum of ²ⁿC_r is the single middle term a = ²ⁿC_n.

Since 2n–1 is odd, ²ⁿ⁻¹C_r has two equal central terms, so the maximum is b = ²ⁿ⁻¹C_n (= ²ⁿ⁻¹C_{n−1}).

Now take the ratio:

a/b = [ (2n)! / (n! · n!) ] ÷ [ (2n−1)! / (n! · (n−1)!) ]

= (2n)!/(2n−1)! · (n−1)!/n! = (2n) · (1/n) = 2.

Therefore a = 2b.

Check with n = 2: greatest ⁴C_r = ⁴C_2 = 6 = a, greatest ³C_r = ³C_1 = 3 = b, and indeed 6 = 2·3. ✓