$13\mid4^{2n+1}+3^{n+2}$ How can I prove that $4^{2n+1}+3^{n+2}$ is always divisible by 13?

$$4^{2n+1}+3^{n+2}=16^n\cdot 4+3^n\cdot 9\\16^n\cdot4+3^n\cdot 9\equiv3^n\cdot4+3^n\cdot 9\pmod {13}\\3^n(4+9)\equiv3^n\cdot13\equiv0\pmod{13}$$
This can also be solved with induction,for $n=0$
$$4+3^2=13$$
Assume it holds for $n=k$
$$4^{2k+1}+3^{k+2}$$
Prove it holds for $n=k+1$
$$4^{2k+3}+3^{k+3}=16\cdot4^{2k+1}+3\cdot3^{k+2}=16\cdot4^{2k+1}+16\cdot3^{k+2}-13\cdot3^{k+2}=16(4^{2k+1}+3^{k+2})-13\cdot3^{k+2}$$
by the inductive hypothesis $4^{2k+1}+3^{k+2}$ is divisible by $13$ because of that the whole expression is
