Prove by induction: $2^{n+2} +3^{2n+1}$ is divisible by $7$ for all $n \in \mathbb{N}$ I want to prove that $2^{n+2} +3^{2n+1}$ is divisible by $7$ for all $n \in \mathbb{N}$ using proof by induction.
Attempt
Prove true for $n = 1$
$2^{1+2} + 3^{2(1) +1} = 35$
35 is divisible by 7 so true for $n =1$
Induction step: Assume true for $n = k$ and prove true for $n = k+1$
$n = k$
$2^{k+2} + 3^{2k +1} = 2^k \cdot 2^2 + 3^{2k} \cdot 3^1 = (4) 2^k + (3) 3^{2k}$
$n = k+1$
$2^{k+3} +3^{2k+3} = (8) 2^k + (27) 3^{2k}$
Thoughts:
I know I need to use the expression for $n = k$ to prove that it is true for $n = k+1$ but I am not sure where to start. I think that using modular arithmetic would be too complex.  Any ideas would be greatly appreciated!
Update
Thanks for all of the useful answers/suggestions. I understand the approach to take now!
 A: \begin{eqnarray*}
2^{k+3}+3^{2k+3} =(9-7)2^{k+2} +9 \times 3^{2k+1}= 9(\color{blue}{2^{k+2}+3^{2k+1}})-\color{blue}{7} \times2^{k+2}
\end{eqnarray*}
A: Hint: Write the equation for $n=k$ equal to $7(c)$ where $c$ is some constant. Now, look hard at the equation where $n=k+1$.
A: You're almost there. What is $2^{k+3} + 3^{2k + 3} - 2(2^{k+2} +3^{2k+1})$, according to your rewritings?
A: HINT.-$x+y\equiv0\pmod7$ this implies $2x+9y\equiv2x+2y\equiv2(x+y)\equiv0\pmod7$
A: Yet another view:
Take your $n=k$ version and double it.  Then add $21(3^{2k})$ and you have your $k+1$ expression.  Both things you're adding are divisible by $7$
A: You mentioned that modular arithmetic would be to complex, but that's the approach I'd take and I actually found it fairly straightforward, so here's an alternative solution. We're working modulo 7, so there can't be any more than 7 different answers for $k^n$ that we care about. That's small enough for an exhaustive search;
\begin{align}
n=0 \quad\quad& 2^{2} \cong 4 \quad\quad& 3^{1} \cong 3 \quad\quad& 4+3 \cong 0                                        \\
n=1 \quad\quad& 2^{3} \cong 1 \quad\quad& 3^{3} \cong 6 \quad\quad& 1+6 \cong 0                                        \\
n=2 \quad\quad& 2^{4} \cong 2 \quad\quad& 3^{5} \cong 5 \quad\quad& 2+5 \cong 0                                        \\
n=3 \quad\quad& 2^{5} \cong 4 \quad\quad& 3^{7} \cong 3 \quad\quad                                       
\end{align}
And lucky us, we can stop early! Our residues for $n=3$ are the same as for $n=0$, and we generate each new row by multiplying both of the values in the two middle columns by 2 (since $3^2 \cong 2$).
So the pairs of residues can only ever by $(4,3), (1,6)$ or $(2,5)$. Adding any of these pairs yields a multiple of 7, QED.
