Although $(+3)$ oxidation state is the characteristic oxidation state of lanthanoids,cerium shows $(+4)$ oxidation state also. Why?

(N/A) Cerium ($Ce$,atomic number $58$) has the electronic configuration $[Xe] \ 4f^{1} \ 5d^{1} \ 6s^{2}$.
By losing four electrons,it attains the stable noble gas configuration of Xenon $([Xe])$,which is $[Xe] \ 4f^{0} \ 5d^{0} \ 6s^{0}$.
This extra stability associated with the empty $4f$ subshell allows cerium to exhibit the $(+4)$ oxidation state.