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.

Here are two alternative routes.
Using the binomial theorem:
$$
3\cdot 5^{2n+1} +2^{3n+1}
= 15\cdot 25^{n} +2\cdot 8^{n}
= 15\cdot (17+8)^{n} +2\cdot 8^{n}\\
= 15\cdot (17a+8^{n}) +2\cdot 8^{n}
= 17b+15\cdot 8^{n} +2\cdot 8^{n}
= 17b+17\cdot 8^{n}
$$
Using recurrences:
Let
$
x_n=3\cdot 5^{2n+1} +2^{3n+1}
$,
Since, as above,
$
x_n= 15\cdot 25^{n} +2\cdot 8^{n}
$,
we have the recurrence $x_{n}=33x_{n-1}-200x_{n-2}$, for $n\ge 2$ (*).
The result follows by induction because $x_0$ and $x_1$ are multiples of $17$.
(*) In general, if $x_n=a\alpha^{n} +b\beta^{n}$, then $x_{n}=(\alpha+\beta)x_{n-1}-(\alpha\beta)x_{n-2}$, for $n\ge 2$. This follows because $\alpha$ and $\beta$ are the roots of the quadratic $(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+(\alpha\beta)$.
