I need help solving this task, if anyone had a similar problem it would help me.
Proof by mathematical induction:
$4^n>n^2$
I tried this:
$1.n=1\\4>1\\2.n=k\\4^k>k^2\\3.n=k+1\\4^{k+1}>(k+1)^2\\3k^2>2k+1$
And now, i don't know what to do next.
Thanks in advance !
If $n=1$, it is clear that $4^n>n^2$.
Now, take $n\in\Bbb N$ and suppose that $4^n>n^2$. Then\begin{align}4^{n+1}&=4\times4^n\\&>4n^2\\&\geqslant(n+1)^2,\end{align}since$$\frac{n+1}n\leqslant2\implies\frac{(n+1)^2}{n^2}\leqslant4.$$
We have
$$4^{k+1}=4 \cdot 4^k \stackrel{\text{Ind. Hyp.}}\ge 4 \cdot k^2\stackrel{?}\ge k^2+2k +1=(k+1)^2$$
and the inequality holds indeed
$$4 \cdot k^2\ge k^2+2k +1 \iff 3k^2-2k-1=(k-1)(3k+1)\ge 0$$
which is equivalent to your last inequality and true as $k\ge 1$.
