The distance travelled in the $n^{\text{th}}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$. For a freely falling body, $u=0$ and $a=g$. Thus, $S_n \propto (2n-1)$. For $n=1, 2, 3, 4$, the ratios are $(2(1)-1) : (2(2)-1) : (2(3)-1) : (2(4)-1) = 1 : 3 : 5 : 7$.