Prove that $3^n>n^2$ for any positive integer n>2
Let $n=1$ $$3^1>1^2$$ Let $n=2$ $$3^2>2^2$$ Assume P holds for n=k $$3^k>k^2$$ Let $n=k+1$ $$ \begin{align} 3^{k+1}>(k+1)^2 \\3\times3^k>(k+1)^2 \end{align}$$ From here I can not find where to go to finish the proof.
The main part of the question is the proof, however; I would like to also know if using $n=k+1$ is always the way to go? I have only done a few proofs by induction and so far to my understanding is that the whole point is to prove the function, series or statement for all positive integers. Is there a special way to go about these types of problems when only given an inequality or a single statement? In comparison to being give a sequence and told what that sequence as a function is, i.e $1+2+3+...+n=\frac{n(n+1)}{2}$ as an example as an easy sequence to prove with $n=k+1$
You proved $n=1,2$.
So we do $$3^{k+1}=3 \times 3^{k} >3k^2 $$ From the assumption. If $k \ge 2$, it follows that $k^2\ge2k$, $k^2>1$ so,$$3k^2=k^2+k^2+k^2>k^2+2k+1=(k+1)^2$$ So $$3^{k+1}>3k^2>(k+1)^2$$Thus, $P$ holds is $n=k+1$. We are done!
As for your second question, most induction does use $$n=k \rightarrow n=k+1$$ However, there are several different kinds of induction, such as using $$n=k,k+1 \rightarrow n=k+2$$ or $$n=1,2,3,4,\dots,k \rightarrow k+1$$ The last is called Strong Induction.