How to prove that $2^{n+2}+3^{2n+1}$ is divisible by 7 using induction? I want to prove that $2^{n+2}+3^{2n+1}$ is divisible by 7 using induction. My first step is replace $n$ with $1$.


*

*$2^{1+2}+3^{2(1)+1}$

*$2^3+3^3$

*$8+27$

*$35 = 7\times 5$


The next step is assume that $2^{n+2}+3^{2n+1}$ is divisible by 7. And the last step is to prove that $2^{(n+1)+2}+3^{2(n+1)+1}$.


*

*$2^{(n+1)+2}+3^{2(n+1)+1}$

*$2^{n+3}+3^{2n+3}$

*I got stuck here.
How can I prove it using induction?

Let $$f(n)=2^{n+2}+3^{2n+1}$$ so that $$f(n+1)=2^{n+3}+3^{2n+3}$$
Then
$$ \begin{align} f(n+1)-f(n) &= 2^{n+3}+3^{2n+3}-2^{n+2}-3^{2n+1} \\
&= 2^{n+2} \left(2-1 \right)+3^{2n+1}\left(3^2 - 1 \right) \\
&= 2^{n+2} + 8\left(3^{2n+1}\right) \\
&= 2^{n+2} + 3^{2n+1} + 7\left(3^{2n+1}\right) \end{align} $$
We assumed that $2^{n+2} + 3^{2n+1} ~|~ 7$ and $7\left(3^{2n+1}\right)$ is obviously divisble by 7. So we're done.
