Prove that $n^2 = O(2^n)$ by mathematical induction

Question

I have been having a hard time understanding mathematical inductions and can not finalize the proof for this statement. I am currently learning about algorithms asymptotic bounding and this is a problem given in the book I am reading. I have set my $n_0$ to $4$ since during the basis step we've proven that for $n_0 \ge 4$, $n^2 \le C(2^n)$.

During my induction step, I have

$(k+1)^2 \le C \cdot 2^{k+1} $

$k^2 + 2k + 1 \le C \cdot 2^k + 2$

I thought I could then do $2k+1\le 2$ after removing $k^2$ and $2^k$ from the previous inequality due to having proven that $n^2\le 2^n$ for all $n \ge n_0$. However, doing so would cause the inequality to be false. I'm missing that one key step that will cause all of it to make sense, any help would be greatly appreciated!

Answer 1

The second inequality is wrong, you find instead :

$$(k+1)^2 \leq C \cdot 2^{k+1} \implies \frac{ k^2+2k+1}{2} \leq C \cdot 2^k $$

now find when $ \frac{ k^2+2k+1}{2} \leq k^2 $ to conclude (if you solve this since you know that $ k^2 \leq C \cdot 2^k $ then you find that $\frac{ k^2+2k+1}{2} \leq k^2 \leq C \cdot 2^k \implies (k+1)^2 \leq C \cdot 2^{k+1}$)

Answer 2

Let $C=2$. Checking $n=1-3$, $1\leq4$, $4\leq8$ and $9\leq16$ are indeed true.

Assume for $n=k$ that $k^2 \leq 2*2^k$. Then, consider $n=k+1$:

$k^2+2k+1 \leq 2*2^{k+1}$ -> $k^2+2k+1 \leq 2^k+3*2^k$

Since $k^2 \leq 2^k$(by assumption) and $2k+1 \leq k^2$ for $k\geq3$ (not very hard to prove), the above inequality for $n=k+1$ is true. Therefore, by induction, the inequality is true for all $k\geq3$. Thus, $n^2\leq2*2^n$ and so $n^2=O(2^n)$.

Update:

Here is a proof that $2n+1 \leq n^2$ for $n\geq3$.

Checking $n=3$, $7\leq9$ is indeed true.

Assume for $n=k$ that $2k+1 \leq k^2$. Then, consider $n=k+1$:

$2k+3\leq(k+1)^2$ -> $2k+3\leq k^2+2k+1$ -> $2\leq k^2$

Since $k^2\geq2$ is always true for $k\geq3$, the above inequality for $n=k+1$ is true. Therefore, by induction, the inequality is true for all $k\geq3$. Thus, $2n+1 \leq n^2$.