3
$\begingroup$

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 $n=k+1$

  • 0
    similar to http://math.stackexchange.com/questions/497092/proof-by-induction-2n-n2-for-all-integer-n-greater-than-4, same as http://math.stackexchange.com/questions/1236667/how-to-prove-with-mathematical-induction-3n-n2 and http://math.stackexchange.com/questions/2128203/is-my-solution-for-proving-3n-n2-using-induction-correct – 2017-02-18

2 Answers 2

5

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.

  • 0
    hmm where does the $(k+1)^2$ come from? – 2017-02-18
  • 2
    @NickPavini Our aim , which is to prove $P$ holds if $n=k+1$. In order for this, we have to prove $$3^{k+1}>(k+1)^2$$ – 2017-02-18
  • 0
    Is $3k^2 = (k+1)^2$ in this case? – 2017-02-18
  • 1
    @NickPavini No. $3k^2>(k+1)^2$. – 2017-02-18
  • 0
    @NickPavini Do you understand? – 2017-02-18
  • 0
    I'm really trying to wrap my head around how the $3k^2>(k+1)^2$ I see how we get $3^{k+1}>3k^2$ Just having a little trouble with the click that $3k^2>(k+1)^2$ – 2017-02-18
  • 1
    @NickPavini I will elaborate; $$3k^2>(k+1)^2 \Leftrightarrow 2k^2-2k-1>0$$ This is true if $k \ge 2$ as $k^2-k$ is an increasing function if $k \ge 2$. – 2017-02-18
  • 0
    ok I am seeing that a little bit more now – 2017-02-18
  • 1
    May this can help: We know that by hypothesis, $k>2$ and so $k^2>2k$ as well as $k^2>1$ and so $$3k^2=k^2+k^2+k^2>k^2+2k+1=(k+1)^2.$$ – 2017-02-18
  • 0
    @ΘΣΦGenSan That actually helped me a lot! Because I proved for $k=1$ and $k=2$, we know that $k>2$ and thats the comparison I wasn't seeing. Thank you. – 2017-02-18
  • 1
    @NickPavini I have aded the edit to make it a bit easier. :) (I would appreciate it if you were to accept my answer; no pressure though). – 2017-02-18
0

$$(k+1)^2 = k^2 + 2k + 1 < k^2 + k^2 + k^2 = 3k^2 < 3\cdot 3^k = 3^{k+1}$$