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.

Answer 1

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

Answer 2

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.