Prove that $2^{n+1} \gt n^2$ for every positive integer $n$.

Question

Prove that $2^{n+1} \gt n^2$ for every positive integer $n$.

I have to prove this using mathematical induction, I know how to start the proof but I get stuck towards the end because I think I either did something wrong or do not know how to continue it.

Here are all of the steps I have done so far:
First I prove that $1 \in S$
This is obviously true because $2^{1+1} \gt 1^2$
Now assume that $a \in S$, we have $2^{a+1} \gt a^2$, now I have to prove that $a+1 \in S$, i.e. $2^{a+2} \gt (a+1)^2$.

$2^{a+2}$ becomes $2 \times 2^{a+1}$ and $(a+1)^2$ becomes $a^2+2a+1$.

We now have $2 \times 2^{a+1} \gt a^2+2a+1$

Multiply both sides of $2^{a+1} \gt a^2$ by 2 and obtain $2 \times 2^{a+1} \gt 2a^2$

to complete the proof all I have to prove is that $2a^2 \gt a^2 +2a +1$

subtracting $a^2$ from both sides I get $a^2 \gt 2a+1$, but this is only true for $a \ge 3$ and I need to prove if for all positive integers.

What did I do wrong, should I have taken a different approach or is there a way to finish what I started?

Answer 1

You need to prove more base cases.

In the case of $a=1$, we have $2^{1+1} = 4 \gt 1^2 = 1$. In the case of $a=2$, we have $2^{2+1} = 8 \gt 2^2 = 4$.

Assume that for some $a \in \mathbb N, a\ge 3,$ that $ 2^{a+1} \gt a^2$. Then we must show that the $a+1$ case is true. We have $2^{(a+1)+1)} = 2^{a+1} * 2^1 > 2 * a^2$, since by assumption $2^{a+1} \gt a^2$. Then we have $2^{a+1} * 2^1 > 2 * a^2 > (a+1)^2 = a^2 + 2a + 1$ for $a \ge 3$, and we are done.

Answer 2

Another way is using the function $$f(x)=2^{x+1}-x^2$$ whose derivative is $$f'(x)=2^{x+1}\ln 2-2x$$ One has $$2^{x+1}\ln 2-2x\gt 0\iff2^x\ln 2\gt x$$ Hence the derivative is always positive and (besides of increasing) so is the function $f(x)$ for $x\gt 0$ since $f(0)=2$. In particular for $x$ integer positive this is true.