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I am in a critical problem with the following question. Please help me.

Prove by induction: $$4^n > n^2 \text{ for }n >= 1$$

Base case: n = 1 $$4^1 > 1^2$$ 4 > 1 which is true and for some n = k > 1 we have $$4^k > k^2$$

Next, Induction step: Now setting n = k+1 we get $$4^{k+1} > (k+1)^2$$ $$4(4^k) > k^2 + 2k + 1$$ Now how will I proceed?

Mainuddin
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  • What have you tried so far? (base case, for a start? And where are you stuck in the induction step?) – Clement C. Aug 18 '15 at 18:24
  • Base case is ok, I am stuck in the induction steps. – Mainuddin Aug 18 '15 at 18:25
  • What have you assumed for the inductive step? Please put all of your work in the body of the question; this will help us determine exactly where to give a hint. – Michael Burr Aug 18 '15 at 18:26
  • To get help, you may want to show your work so far (i.e., where are you stuck in what you have tried?) – Clement C. Aug 18 '15 at 18:26
  • Basic approach: First, is it true for $n = 1$ itself? Secondly, when you increase the value of $n$ by $1$, what happens to the left side of the inequality? How much would $n$ have to increase by in order for that same change to happen to the right side? Does that ever happen when $n$ increases by $1$ (for $n \geq 1$)? – Brian Tung Aug 18 '15 at 18:28
  • 4^(k+1) > (k+1)^2 => 4*4^k > k^2 + 2k + 4 and then? – Mainuddin Aug 18 '15 at 18:28
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    $4^n+4^n+4^n\gt 2n+1$ following assumption – miniparser Aug 18 '15 at 18:28
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    Well, observe that $4^{k+1} = 4(4^k)$. And if $4^k > k^2$ then $4(4^k) > 4k^2$. And $4k^2$ is the square of what? – Brian Tung Aug 18 '15 at 18:30
  • @Mainuddin: fyi you should show your attempt in the posting that's why you got downvotes. – miniparser Aug 18 '15 at 18:36

2 Answers2

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Without induction, you can show that $2^n>n$ for all $n\in\mathbb{N}_0$, using set theory. It is known that $|A|<2^{|A|}$ for any set $A$, particularly, if $|A|=n$. Then, $4^n=\left(2^n\right)^2>n^2$.

Alternatively, you can use Bernoulli's Inequality: $2^n=(1+1)^n\geq 1+n\cdot 1>n$ for all $n\in\mathbb{N}_0$.

Batominovski
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To do induction you must start with the base case so for $n=1$ you get:

$$4^1>1^2\Rightarrow4>1$$

Wich in fact is true; now suppose the relation hold for some $n=k>1$ you have that:

$$4^k>k^2$$

Now setting $n=k+1$ leads:

$$4^{k+1}=4^k\times 4=4^k+3\times4^k$$

$$(k+1)^2=k^2+2k+1$$

Combining the two expressions you have:

$$4^k+3\times4^k>k^2+2k+1$$

Wich is actually true considering that $4^n>2n+1$ for every $n>0.$

AlienRem
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