Prove $2^{3^n}+1 $ is a multiple of $3^{n+1}$ by induction

Question

I am using induction and I understand that when $n=1$ it is true. The induction hypothesis is when $n=k$ so $2^{3^n}+1$ is a multiple of $3^{n+1}$. So for the induction step we have $n=k$ so $2^{3^{k+1}}+1=3^{k+1} m$, $m$ is integer. For the next step we have $n=k+1$ so $2^{3^{n+1+1}}+1$. Next, as I understand it, you need to take a step in which the number $2^{3^{n+1+1}}+1$ is expanded to the terms that are divided by $3^{k+2}$ and the easiest way to do this is to cancel out $2^{3^{k+1}}+1$, but I do not know how to do this - please help.

Answer 1

Prove $2^{3^{n}}+1$ is a multiple of $3^{n+1}$ by induction

For $n=1$: $2^{3^1}+1=9=3^2 \ \ \checkmark$

Assume for $n=k$: $2^{3^k}+1=3^{k+1}m$.

Prove it for $n=k+1$: $$\begin{align}2^{3^{k+1}}+1=(2^{3^{k}})^3+1&=(3^{k+1}m-1)^3+1=\\ &=3^{3k+3}m^3-3^{2k+3}m^2+3^{k+2}m-1+1=\\ &=3^{k+2}(3^{2k+1}m^3-3^{k+1}m^2+m)=\\ &=3^{k+2}t.\end{align}$$