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?
 A: If
$$2 \cdot 3^{2n+1} + 9 \cdot 2^{n+2} = 7\cdot 18d$$
Then
$$2 \cdot 3^{2n+1} + 2 \cdot 2^{n+2} = 7\cdot 18d - 7 \cdot 2^{n+2}$$
And we conclude
$$3^{2n+1} + 2^{n+2} = \frac{7(18d - 2^{n+2})}{2} = 7(9d - 2^{n+1})$$
A: $3^{2n-1}+2^{n+1}=9^n3^{-1}+2^n2\equiv 2^n3^{-1}+2^n2=5\times2^n+2\times 2^n=7\times2^n\equiv0 \pmod7 $ since $3^{-1}\equiv5 \pmod7$
A: $$2\equiv 9$$
$$2\equiv 3^2$$
$$2^{n+1}\equiv 3^{2n+2} $$
$$3^{2n-1}+3^{2n+2}=3^{2n-1}(1+27)$$
$$=4.7.3^{2n-1} $$
Done
A: The idea is to extract the $n$-th case from $(n+1)$-th case. To show it for $n+1$:
$$\begin{align}&3^{2(n+1)-1}+2^{(n+1)+1}= \\
&3^{2n+1}+2^{n+2}= \\ 
&9\cdot 3^{2n-1}+2\cdot 2^{n+1}= \\
&2(3^{2n-1}+2^{n+1})+7\cdot 3^{2n-1}.\end{align}$$
A: We can proof it by induction; it is clear that $P(n)$ hold for $n=1$  and we Assume that it hold for $n$ so 
$$2^{n+1}=7k - 3^{2n-1}$$
Show it for $n+1$ 
$$ 3^{2(n+1)-1} + 2^{(n+1)+1} = 3^{2n+1} + 2.2^{n+1}=  3^{2n+1}+2(7k - 3^{2n-1})=   3^{2n+1} + 2.3^{2n-1}+14k = (9-2).3^{2n-1}+14k= 7(3^{2n-1}+2k)$$
A: 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.
