i need to proof on induction this questions:
$n^2<4^n$ $n \in \mathbb N$
i start with the basic $n=1$
$1^2<4^1$
then put $n=k$
$k^2<4^k$
then if $p(k)=T$ so $p(k+1)=T$
$(k+1)^2<4^{k+1}$
$k^2+2k+1<4^k\cdot 4$
by the indcution $k^2 < 4^k$
$4^k+2k+1<4^k\cdot 4$
and then I dont know what to do or even my way is right
we assume that $$4^k>k^2$$ and we want to prove that $$4^{k+1}>(k+1)^2$$ multiplying the first inequality by $4$ we get $$4^{k+1}>4k^2$$ and now we have $$4k^2\geq (k+1)^2=k^2+2k+1$$ and this is true if $$3k^2-2k-1\geq 0$$ for $k\geq 1$
Your redaction is not good.
Induction theorem says :
Let $\wp(n)$ a property.
1. There is $n_0\in \mathbb N$ s.t. $\wp(n_0)$ true.
2. For all $n\geq n_0$, $\wp(n)$ true implies $\wp(n+1)$ true.
Therefore, $\wp(n)$ true for all $n\geq n_0$.
So in your case, $n_0=0$ work. Now suppose your property is true for an unspecified $n\geq 1$. Let prove that $\wp(n+1)$ true (i.e. $(n+1)^2<4^{n+1}$).
$$(n+1)^2=n^2+2n+1\underset{hyp.}{\leq} 4^n+2n+1.$$ You also have that $n\leq n^2<4^n$ where the last inequality is by your induction hypothesis, and thus $2n<2\cdot 4^n$, and finally, $1<4^n$. Therefore, $$(n+1)^2<4^n(1+2+1)=4\cdot 4^{n}=4^{n+1}.$$ Therefore $\wp(n+1)$ is true, and thus, by the theorem, $$n^2<4^n,$$ for all $n\geq 0$.
You are trying to show that $4^{k+1}>(k+1)^2$, assuming that $4^{k}>k^2$. In order to solve this take a slightly different approach:
$$ 4^{k+1}=4\cdot4^k>4k^2=(2k)^2>(k+1)^2 $$
Note that the last step holds since $2k>k+1$ iff $k>1$, which you may assume in this example once you have checked the inequality for small values of $k$.
