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I've been trying to prove this statement by induction; however, in following the steps I normally take I end up utterly stuck. I know that I must be missing something, but I have been stuck on this for a while now and can't seem to figure it out. Does anyone know how I can get to a sum on the RHS of my proof statements? Any other advice on how to find this?

Prove $**2^n > n^2**$ by induction using a basis > 4: Basis: $**n = 5** **2^n > n^2**$ 32 > 25 Assume: $**2^n > n^2**$

Prove: $**2^(n+1) > (n+1) ^2**$

$**2^{n+1} = 2^n * 2** $ LHS Prove = LHS Assume + Fix
$**2^{n+1} > (n^2) *2** $ Utilize assumption
$**2^{n+1} > 2(n^2)** $ Could utilize assumption again...
$**2^{n+1} > 1(n^2)** $ ..doesn't seem to help
$**2^{n+1} > (n * 1)^2** $ Basically just$ n^2$

usukidoll
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  • can you use $(n+1)^2 = n^2 +2n + 1$ ? – Brad S. Mar 20 '14 at 02:29
  • Although this is a perfect question as far as 1) work shown, 2) problem description, 3) tags, and 4) title, I am voting to close this as a "duplicate" for organization's sake--please don't take it personally. I really like your question, though, so please do come back and ask more--we need more people who ask questions like you do. :) – apnorton Mar 20 '14 at 03:14

3 Answers3

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For sure this question has been asked so many times here...

As this is your first question it would be kind of rude if i vote to close this..

Coming to the Question :

You have $2^n>n^2$

You have to prove $2^{n+1}>(n+1)^2$

So, you could consider (as you did)

$2^{n+1}=2^n.2>2n^2$

Suppose you prove $2n^2>(n+1)^2$ then you are done..

I guess you know what is $(n+1)^2$!!

Compare $(n+1)^2$ (just by expanding it) with $2n^2$

Then you are done!

  • Oh! Thanks, but then how do I prove that n^2 + n^2 > n^2 + 2n + 1? –  Mar 20 '14 at 03:41
  • I guess you can cancel $n^2$ in both sides... what would you be left with? –  Mar 20 '14 at 03:42
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You are on the right track. Under the inductive hypothesis, you've shown that $2^{n+1} > 2n^2$.

If you could show that $2n^2 \ge (n+1)^2$, you would be done.

This is equivalent to $n^2 - 2n - 1 \ge 0$, or $(n-1)^2 \ge 1$. That's definitely true!

user11977
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You can also show this directly without using induction.

Define $f:(0,\infty)\rightarrow (0,\infty)$ by $f(x)=x^{\frac1x}$. Then it is easy to compute $f'(x)=x^{\frac1x-2}(1-\log x)$, so that $f$ is strictly decreasing for $x>e$. This means that $5^{\frac15}>n^{\frac1n}$ for $n=6,7,8,\ldots$; and since one can easily compute $2^{\frac12}\approx 1.41 >1.38\approx 5^{\frac15}$, we have (for integral $n\geq 5$) $$2^{\frac12}>n^{{\frac1n}}$$ $$\left(2^{\frac12}\right)^{2n}>\left(n^{{\frac1n}}\right)^{2n}$$ $$\boxed{2^n>n^2}$$ for integral $n\geq 5$ as desired.

MPW
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