Prove $ 3^{n}> n^{2}$ for $n=2$ by induction

Question

I understand base case at $n=1$, and $n=2$. Then I do understand the inductive hypothesis of assuming $n=k$. The part that confuses me is when showing $n=k+1$. On other tutorials that are online, they multiply by 3 to both sides instead of plugging in $k+1$ for n and proving from there. I started the proof for $n=k+1$ like this:

Proving the statement holds for $n=k+1$:

$ 3^{k} > k^{2}$ = $ 3^{k+1} > (k+1)^{2}$

=> $ 3^{k}*3 > k^{2}+2k+1$ I am confused right here.

Or compare the left side of the inequality with the right side $ 3^{k}*3 > k^{2}$

Where $3^{k}*3 > k^{2}$, $ 3^{k} $ is less than $ k^{2}$ by inductive hypothesis Then I am not sure where to go from there.

You have $3^{k+1}=3\cdot 3^k>3k^2$. The rest should be easy. — Peter, May 13 '20 at 20:00
Use equals signs only for equality. You write $3^k>k^2\color{red}{=}3^{k+1}>(k+1)^2$. You mean to write "We wish to prove that $3^k>k^2$ implies that $3^{k+1}>(k+1)^2$" — JMoravitz, May 13 '20 at 20:01
Read the linked question on how to write a proper proof by induction. You should be trying to line everything up in one line if possible, or if not then at least in one line of implications. You should be able in this problem be able to write $3^{k+1}\geq \dots \geq \dots \geq \dots \geq (k+1)^2$ where the inbetween steps follow from simple algebraic manipulations, the fact that $k$ is at least as large as your base case, and the induction hypothesis. — JMoravitz, May 13 '20 at 20:03
@Theoneandonly then modify my line to say $\geq \dots \geq \dots \geq \dots >\dots \geq \dots (k+1)^2$. All that is required is a single strict inequality in the chain and the rest be non-strict. The point is that some of those in the chain can be strict equalities if one chooses. That will fall naturally from using the induction hypothesis anyways for this problem — JMoravitz, May 13 '20 at 20:10

score 0 · Answer 1 · answered May 13 '20 at 20:01

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$$ 3^{n+1} = 3^n \cdot 3 > 3 n^2 = n^2 + n^2 + n^2 > n^2 + 2n + 1 = (n+1)^2 QED$$

answered May 13 '20 at 20:01

Theoneandonly

653

why the downvote ? – Theoneandonly May 13 '20 at 20:04
Why did you multiply the $n^{2}$ by 3 though? – user1829 May 13 '20 at 20:08
induction hypothesis: $n^2 > 3^n$ and so $3 3^{n} > 3 n^2 $ – Theoneandonly May 13 '20 at 20:09
do you mean $n^{2} < 3^{n}$ – user1829 May 13 '20 at 20:11
@user1829 Recall that $3^{a+b}=3^a\times 3^b$... in particular $3^{k+1}=3^k\times 3^1$ or simplified $3^k\times 3$. Now... recall that if $a>b$ and $c>0$ then $a\times c > b\times c$... so $3^k\times 3 > k^2 \times 3$ – JMoravitz May 13 '20 at 20:11
yes which is $3^n > n^2$ – Theoneandonly May 13 '20 at 20:12
I still dont understand the downvote tho. – Theoneandonly May 13 '20 at 20:12
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It wasn't me, but for low effort questions like this a popular strategy for teaching is explicitly not to give them a full answer and holding their hands but rather to give them the tools they need to complete it themselves, asking leading questions, showing them other related but distinct examples, etc... I could see someone thinking your answer here was too complete. – JMoravitz May 13 '20 at 20:14
Obviously it was a hint and proof of this is the OP asking some questions about it. He still needs to figure out things and write it out nicely. Do you really think this is too complete? – Theoneandonly May 13 '20 at 20:15
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@JMoravitz Now the answer that has just been posted that is too complete. – Theoneandonly May 13 '20 at 20:16

score 0 · Answer 2 · answered May 13 '20 at 20:16

First of all.... don't write $3^{k+1} > (k+1)^2$. That's what you have to prove.... you don't know it yet.

DO not start with what you need to prove and work backwards.

Startt with what you know and work foreward.

What you know is $3^k > k^2$.

So $3*3^k > 3*k^2$

So $3^{k+1} > 3*k^2$.

Now what you need to prove is (scribble this on note paper to the side) $3k^2 \ge (k+1)^2= k^2 + 2k + 1$.

So $3^{k+1} > k^2 + k^2 + k^2$

Now what we need to prove (still scribbling on note paper to the side) is that $k^2 + k^2 \ge 2k + 1$.

We know that $k \ge 2$ so $k*k \ge 2*k$ so

So $3^{k+1} > k^2 + k^2 + k^2 \ge k^2 + 2k + k^2$.

Now we need to prove (scribble scribble) that $k^2 \ge 1$.

We know that $k \ge 1$ so $k*k \ge 1*k = k \ge 2 > 1$ so

So $3^{k+1} > 3k^2= k^2+k^2 + k^2 \ge k^2 + 2k + k^2 > k^2 + 2k + 1 =(k+1)^2$.

....

Now we are done.

Prove $ 3^{n}> n^{2}$ for $n=2$ by induction

2 Answers2