$T(n) = T(n-2) + n^{^{2}}$
$T(n-2) = T(n-4) + (n-2 )^{^{2}}$
$T(n-4) = T(n-6) + (n-4 )^{^{2}}$
so this goes on till n goes to zero
$n-2x = 0$
$\rightarrow x= \frac{n}{2}$
so finally
$T(n) = 1+n^{^{2}} + (n-2)^{^{2}}+(n-4)^{^{2}}..........$ $\frac{n}{2} times$
Each contain a $n^{2}$ term and they are $\frac{n}{2}$ of those so it will be $O(n^{3})$