Prove that $3^{2n-1} + 2^{n+1}$ is always a multiple of $7$. I'm trying to prove the following statement:
$P(n) = 3^{2n-1} + 2^{n+1}$ is always a multiple of $7$  $\forall n\geq1$.
I want to use induction, so the base case is $P(1) = 7$ so that's okay.
Now I need to prove that if $P(n)$ is true then $P(n+1)$ is true.
So there exists a $d \in \mathbb{N}$ such that 
$$ 3^{2n-1} + 2^{n+1} = 7d $$
From this I need to say that there exists a $k \in \mathbb{N}$ such that:
$$ 3^{2n+1} + 2^{n+2} = 7k $$
With a little algebraic manipulation, I have managed to say:
$$ 2 \cdot 3^{2n+1} + 9 \cdot 2^{n+2} = 7\cdot(18d) $$
But now I am stuck. How should I keep going?

Alternatively, suppose $3^{2n-1}+2^{n+1}= 7x$.
Then, $x=\frac17\left(3^{2n-1}+2^{n+1}\right)=\frac{1}{21}\left(3^{2n}+6(2^{n})\right)=\frac{1}{21}\left(9^{n}+6(2^{n})\right)\equiv 0 \pmod3$
Now, $9\equiv 2 \pmod7\Rightarrow9^n\equiv 2^n \pmod3$, $6\equiv -1 \pmod7$ and hence, $6\cdot2^n\equiv (-1)\cdot2^n\pmod7$
Therefore, $9^{n}+6(2^{n})\equiv 0 \pmod7$, because $9^{n}\equiv 2^n \pmod7$ and $6\cdot2^{n}\equiv -2^n \pmod7$
Therefore, since $9^{n}+6(2^{n})\equiv 0 \pmod3$, $9^{n}+6(2^{n})\equiv 0 \pmod{21}$, hence $\frac{1}{21}\left(9^{n}+6(2^{n})\right)=\frac{1}{7}\left(3^{2n-1}+2^{n+1}\right)$ is an integer.
