Prove that for every natural number n, \(2^n+1 \leq 3^n \)

We will prove that for \(n \in \mathbb{N}, 2^n+1 \leq 3^n \) by induction. First we will set up a base case at \[ \begin{align} n&=1\\ 2^1+1&=3\\ 3^1&=3\\ so \ 2^1+1 &\leq 3^1 \end{align} \] Since this base case holds we will move on to the induction step and assume that for some \( n \in \mathbb{N}, \ 2^n+1 \leq 3^n \), and will show that this implies that \(2^{n+1}+1 \leq 3^{n+1}\) \[ \begin{align} 2^{n+1}+1 &= (2*2^n)+1\\ &= 2(2^n+1)-1\\ &=((2^n+1)+(2^n+1))-1\\ &\leq 3^n+3^n-1\\ &=3^{n+1}-1-(3^n)\\ &\leq3^{n+1} \end{align} \] So, by induction, for every \(n \in \mathbb{N}, 2^n+1 \leq 3^n\).