Galileo's law of odd numbers 
The Galileo’s law of odd numbers states that the distances traveled are proportional to the squares of the elapsed times. In other words, in equal successive periods of time, the distances traveled by a free-falling body are proportional to the succession of odd numbers ($1, 3, 5, 7,$ etc.).

I clearly understand from kinematics equation that the distances traversed in a time interval are $-\frac{1}{2} gt^2$ so it will be proportional to squares of elapsed times. But what I don't get is how is it proportional to the succession of odd numbers?
 A: This is because $(n+1)^2-n^2=2n+1$, which is odd. Hence, as in the 1st time period, $d_1=-\frac{g} {2} $, in the 2nd time period,  $d_2=-\frac{4g} {2} $, and so on, we get $\frac{d_2 - d_1}{d_1}=\frac{4-1}{1}=3=2(1)+1$
A: According to the Galileo's law of odd numbers, "The distance covered by a falling object in successive equal time intervals is linearly proportional to the odd numbers".
Let us divide the time interval of motion of an object under free fall into  equal intervals $τ$ and find out the distances traversed during successive intervals of time. You already know that
$$y=-\frac{1}{2} gt^2$$
Using this equation, you can calculate the position of the object after different time intervals, $0, τ, 2τ, 3τ,$ etc. Alternatively, you can use $s_{n}=u+\frac{a}{2}(2 n-1)$ which gives the distance traversed in nth second (by a freely falling body, $u=0$). So the distance is directly proportional to $(2n−1)$ where $ n=1,2,3,$ etc.

This law was established by Galileo Galilei who was the first to make quantitative studies of free fall.
A: The distance traveled in the $n$-th second is
$$D = u + \frac{1}{2}a(2n-1)$$
If the body starts from rest, we have $u=0$. So
$$D = \frac{1}{2}a(2n-1)$$
Now, it is obvious that the distance traveled in the $n$-th second is $2n-1$.
