Prove that $3^{2n-1} + 2^{n+1}$ is always a multiple of $7$. [duplicate]

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?

Solution 1:

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

$$3^{2n+1} + 2^{n+2} = \frac{7(18d - 2^{n+2})}{2} = 7(9d - 2^{n+1})$$

Solution 2:

$$2\equiv 9$$ $$2\equiv 3^2$$

$$2^{n+1}\equiv 3^{2n+2} $$

$$3^{2n-1}+3^{2n+2}=3^{2n-1}(1+27)$$

$$=4.7.3^{2n-1} $$

Done