Prove $2^{(n+1)}>n^2$ by induction

Question

Prove $2^{(n+1)}>n^2$, for all $n \in \mathbb{N}$.

I started by verifying the condition for $1$.

$$P(1):2^{1+1}>1^2$$ $$P(1):4>1$$

That is true.

Then I supposed $P(k)$ true, for some $k \in \mathbb{N}$. Now I've to prove $P(k+1)$:

$$2^{k+1+1}>(k+1)^2$$ $$2\cdot 2^{k+1}>(k+1)^2$$

By the hypotesis one known that $2^{k+1}>k^2$ so,

$$2\cdot2^{k+1}>2k^2$$

Now one needs to prove that $2k^2\geq(k+1)^2$ to conclude that $2^{k+2}>(k+1)^2$. But I'm having some trouble to prove this inequality.

Thanks for the help.

Have you tried proving the second inequality by induction as well? It should be straightforward inductively. — Jeremy Upsal, Nov 05 '13 at 23:08
@Oria, yes the second inequality is correct, but trying with $n=1$ does not suffice, it just gives you an idea that it is correct. If you want to be rigorous prove it by induction. — Jeremy Upsal, Nov 05 '13 at 23:10

score 2 · Accepted Answer · answered Nov 05 '13 at 23:11

2

We only need to prove $2k^2 \geq (k+1)^2$ for sufficiently large $k$ (We can check the remaining cases by hand). Indeed, expanding $(k+1)^2$, we need to verify that $k^2 -2k - 1 \geq 0$. When $k \geq 3$, $k^2-2k-1 \geq 3k-2k-1 = k - 1 \geq 2 \geq 0$, and we are done.

Hence, you need to verify $P(1)$ and $P(2)$ by hand, and the rest, you induct.

answered Nov 05 '13 at 23:11

Lord Soth

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I've graphed the inequality and it's true only for $n\geq 3$. So one can say that the first inequality is true only for $n\geq 3$ – Nov 05 '13 at 23:22
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Yes, that is true. But we do not need it true for any $n$, as I have mentioned, we can check the base cases by hand. – Lord Soth Nov 05 '13 at 23:23

Prove $2^{(n+1)}>n^2$ by induction

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