Induction Problem ($n^2 ≤ 2^n$)

Asked Sep 29 '13 at 22:07

Active Sep 29 '13 at 22:32

Viewed 579 times

So I have this induction problem:

$$n^2 ≤ 2^n \;\text{ for } \; n ≥ 4$$

I know the base case is ($n=4$) which checks out.

I know the hyp is ($n=k$) giving $k^2 ≤ 2^k$.

However, I am getting stuck on the actual proof part (i.e. $n = k+1$).

I know $2^{k+1}$ can be broken apart into $2\times 2^k$ but I haven't seen where that can help.

Edits:

Expanding out the $(k+1)^2$, I get $k^2 + 2k + 1 ≤ 2^{k+1}$ Using the hyp I can do $2^k + 2k +1 ≤ (2)(2k)$

edited Sep 29 '13 at 22:32

asked Sep 29 '13 at 22:07

KyleCrowley

What happens when you expand $(k+1)^2$ and use that $k^2 \le 2^k$? – grantfgates Sep 29 '13 at 22:09
2

Tenplicate.${{}}$ – Git Gud Sep 29 '13 at 22:12
@grantfgates Updated with your comments. Still a bit lost. – KyleCrowley Sep 29 '13 at 22:33
By using condition that $k\ge4$, we know that $k^2\ge4k$, and $2k\ge2\cdot4>1$. – peterwhy Sep 29 '13 at 22:36

0 Answers0