Prove that $3\cdot 5^{2n+1} +2^{3n+1}$ is divisible by $17$ for all $n ∈ \mathbb{N}$ 
Use mathematical induction to prove that $3\cdot 5^{2n+1} +2^{3n+1}$
  is divisible by $17$ for all $n ∈ \mathbb{N}$.

I've tried to do it as follow. 
If $n = 1$ then  $392/17 = 23$.
Assume it is true when $n = p$. Therefore $3\cdot 5^{2p+1} +2^{3p+1} = 17k $ where $k ∈ \mathbb{N} $. Consider now $n=p+1$. Then 
\begin{align}
&3\cdot 5^{2(p+1)+1} +2^{3(p+1)+1}=\\
&3\cdot 5^{2p+1+2} + 2^{3p+1+3}=\\
&3\cdot5^{2p+1}\cdot  5^{2} + 2^{3p+1}\cdot 2^{3}.
\end{align}
I reached a dead end from here. If someone could help me in the direction of the next step it would be really helpful. Thanks in advance.

I'll start with the inductive step
$$3\cdot5^{2(n+1)+1}+2^{3(n+1)+1}$$
$$3\cdot5^{2n+1+2}+2^{3n+1+3}$$
$$3\cdot25\cdot5^{2(n+1)}+8\cdot2^{3n+1}$$
We know that $$3\cdot5^{2n+1}+2^{3n+1}=17k$$
Therefore $$3\cdot5^{2n+1}=17k-2^{3n+1}$$
Plug this in: 
$$25\cdot(17k-2^{3n+1})+8\cdot2^{3n+1}$$
$$425k\cdot-25\cdot2^{3n+1}+8\cdot2^{3n+1}$$
$$425k\cdot-17\cdot2^{3n+1}$$
$$17(25-2^{3n+1})$$
