Mathematical Induction Hint

Asked Apr 14 '21 at 02:04

Active Apr 14 '21 at 02:04

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Trying to show that the inequality holds for $$n^2 \le 2^n$$

So far I have: $$\text{Assume }k \text{ is arbitrary. Our inductive hypothesis is: } k^2 \le 2^k \text{. Show this is true for } k+1\\ (k+1)^2 = k^2+2k+1 \\ k^2 < k^2+2k+1$$

Stuck here and wondering if I could get some hints.

asked Apr 14 '21 at 02:04

Peter P.

3

Hint: You know by induction that $k^2 \leqslant 2^k$. That is, $k^2+2k+1 \leqslant 2^k + 2k + 1$. Can you bound more from here? – While I Am Apr 14 '21 at 02:08
Could I say, $2^k +2k+1 \le 2^k \cdot 2^k$? Induction with inequalities isn't my strong suit. – Peter P. Apr 14 '21 at 02:12
Does this answer your question? Proof by Induction: Prove that $2^n > n^2$, for all natural numbers greater than or equal to $5$. Another very similar question here is Proof by induction that $2^n - 1 > n^2$. – John Omielan Apr 14 '21 at 02:17
You can say anything... if you can justify it. To get to $2^k + 2k + 1 \leqslant (2^k)(2^k)$, it suffices to show that $2k+1 \leqslant 2^k$ (can you see why this is?). Can you show that? – While I Am Apr 14 '21 at 02:18
Note @William and Peter P. that proving the right side of an inequality with $2^k \cdot 2^k = 2^{2k}$ doesn't help much since the right side of the inequality to prove for $n = k + 1$ is actually $2^{k + 1} = 2(2^k)$ instead. – John Omielan Apr 14 '21 at 02:33
Oh, my bad. I meant $2^k + 2^k$. – While I Am Apr 14 '21 at 12:10

Mathematical Induction Hint

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