A freely falling body travels a distance X in the n^th second.  In the next second if it travels a distance, Y.  Then

Relation between distances travelled in nth and (n+1)th seconds by a freely falling body

Let a freely falling body travel a distance X during the nth second, and distance Y during the (n+1)th second.

Distance travelled in nth second: X = (1/2) g (2n - 1)

Distance travelled in (n+1)th second: Y = (1/2) g (2n + 1)

Subtracting, we get:

Y - X = (1/2) g (2n + 1) - (1/2) g (2n - 1) = g

Final relation:

Y - X = g

where g is the acceleration due to gravity.