A freely falling body traveled x m in n^th second distance travelled in ${(n-1)}^{th}$ second is

u=0

$\begin{array}{l}{s}_n=g(n-\frac{1}{2})x=g\left(\frac{2n-1}{2}\right)\\ 2x=g(2n-1)\to 1\\ d=s_{n-1}-s_{n-2}=\frac{1}{2}{(n-1)}^2-\frac{1}{2}g{(n-2)}^2\\ =\frac{g}{2}(n^2+1-2n-n^2-4+4n)\end{array}$

$=\frac{g}{2}(2n-3)=\frac{g}{2}(2n-1-2)=\frac{1}{2}\left(g(2n-1)\right)$ $-\frac{g}{2}\left(2\right)$

$=\frac{1}{2}\left(2x\right)-g=x-g$