It was written in the presentation that the volume bound inequality

assuming minimum distance between 2 codewords id d=3, is:

(2^k)*(n+1) <= 2^n

I thought about the case that d>3, and i came up with this inequality:

(2^k)*((n^A)+1) <= 2^n

A = floor((n-1)/2)

Do you agree with that claim?