Can someone please look at my work below? I want to make sure everything I'm doing to my inequality in my inductive step is allowable/if I need to justify along the way (for instance when I subtract $-1$ from the LHS in second to last step). Also, any advice/other tips on how I constructed this proof welcome.

$S_n:$ For every $n\in\mathbb{N}$, it follows that $2^n+1\le3^n$.

$Proof.$ We will prove this using mathematical induction.

Base Case. Let $n=1$. Then $2^n+1=2^{(1)}+1=3$ and $3^n=3^{(1)}=3$. Thus our inequality holds for $n=1$ since $3\le3$.

Inductive Step. Let the natural number $n=k\ge1$. We need to show that $S_k\Rightarrow S_{k+1}$. We use direct proof. Suppose $2^k+1\le 3^k$. Observe that

$$\begin{align*} 2^k+1 &\le 3^k \\ 2(2^k+1) &\le 3(3^k)\enspace (since\enspace (a\le b)\land( c < d) \Rightarrow ac\le bd)\\ 2^{k+1}+2&\le3^{k+1}\\ 2^{k+1}+2-1&\le3^{k+1}\\ 2^{k+1}+1&\le3^{k+1} \end{align*}$$

Thus $S_k\Rightarrow S_{k+1}$.

It follows by mathematical induction that $S_n$ is true.


1 Answer 1


Yes it is correct. But of course you don't need to use induction.

$$3^n=3*3^{n-1} = 2*3^{n-1} + 3^{n-1} \ge 2*3^{n-1} + 1 \ge 2*2^{n-1} + 1 = 2^n+1$$


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