Proof that $ k^2<2^k$

Asked Dec 05 '13 at 18:07

Active Dec 05 '13 at 18:10

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I'm struggling with this problem of proof by induction:

For any natural number $k\geq 5$, prove that $k^2<2^k$.

I assumed that $k^2<2^k$

I want to show that $(k+1)^2<2^{k+1}$

The final statement is $k^2+2k+1$

Am I missing something ?

edited Dec 05 '13 at 18:10

amWhy

asked Dec 05 '13 at 18:07

user113670

3

http://math.stackexchange.com/questions/319913/proof-that-n2-2n – lab bhattacharjee Dec 05 '13 at 18:09
1

Thanks, @lab. This question has been asked more times that I can begin to count! – amWhy Dec 05 '13 at 18:10
Yes, you are missing some things: You need to consider the base case: Show the proposition is true for $k = 5$, and secondly, you haven't shown that $(k+1)^2 = k^2 + 2k + 1 \lt 2^{k+1} = 2\cdot 2^k$. – amWhy Dec 05 '13 at 18:13

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