The distance travelled by a particle in $n^{th}$second is $S_n=u+\frac{a}{2}(2n-1)$ where u is the velocity and a is the acceleration. The equation is

$S_n=S_1-S_2$

$S_n=un+\frac{1}{2}a{n}^2-\left(u(n-1\right)+\frac{1}{2}a{(n-1)}^2)$$S_n=un+\frac{1}{2}a{n}^2-\left(un-u+\frac{1}{2}a(n^2+1.1-2n(1\left)\right)\right)$$S_n=un+\frac{1}{2}a{n}^2-un+u.1-\frac{1}{2}a{n}^2-\frac{1}{2}a.1.1+\frac{1}{2}a.2.n\left(1\right)$$S_n=u.1+\frac{a}{2}(2n.1-1.1) Here n and 1 are the magnitudes of time and measured in seconds.$$D.F of L.H.S\left[S_n\right]=\left[L\right]$

$R.H.S D.F of u.1=\left[L{T}^{-1}.T\right]=\left[L\right]$$D.F of \frac{a}{2}(2n.1-1.1)=\left[L{T}^{-2}\right]\left[T.T-T.T\right]=\left[L{T}^{-2}\right]\left[T^2\right]=\left[L\right]$$\therefore R.H.S=\left[L+L\right]=\left[L\right]$

L.H.S = R.H.S

  Hence the formula is dimensionally correct.