A body starts from rest with uniform acceleration. If its velocity after n second is $\text{υ}$, then its displacement in the last two seconds is

$\mathrm{v}=0+\mathrm{na}$

$\mathrm{a}=\frac{\mathrm{v}}{\mathrm{n}}$

now the distance travel in n sec

${\mathrm{s}}_{\mathrm{n}}=\frac{1}{2}{\mathrm{an}}^2$

and distance travelled in n-2 sec

${\mathrm{s}}_{\mathrm{n}-2}=\frac{1}{2}\mathrm{a}{\left(\mathrm{n}-2\right)}^2$

distance travelled in the last 2 sec

$={\mathrm{S}}_{\mathrm{n}}-{\mathrm{S}}_{\mathrm{n}-2}=\frac{1}{2}{\mathrm{an}}^2-\frac{1}{2}\mathrm{a}{\left(\mathrm{n}-2\right)}^2$

$=\frac{\mathrm{a}}{2}\left[{\mathrm{n}}^2-{\left(\mathrm{n}-2\right)}^2\right]=\frac{\mathrm{a}}{2}\left[\mathrm{n}+\left(\mathrm{n}-2\right)\right]\left[\mathrm{n}-\left(\mathrm{n}-2\right)\right]$

$=\mathrm{a}(2\mathrm{n}-2)=\frac{\mathrm{v}}{\mathrm{n}}(2\mathrm{n}-2)=\frac{2\mathrm{v}(\mathrm{n}-1)}{\mathrm{n}}$