I am trying to prove the title using induction and here is my attempt:
Base Case: n = 1 : $2>1$ which is True
I.H. : $2^k > k^2$ for some $k$
Inductive Step:
$2^{k+1} = 2^k 2$
$2^{k+1} > 2k^2$ by I.H.
But I am not quite sure what to do next.
I am trying to prove the title using induction and here is my attempt:
Base Case: n = 1 : $2>1$ which is True
I.H. : $2^k > k^2$ for some $k$
Inductive Step:
$2^{k+1} = 2^k 2$
$2^{k+1} > 2k^2$ by I.H.
But I am not quite sure what to do next.
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Hint Using $k \geq 5$,
$$\begin{align} 2k^2&=k^2+k^2 \\ &\geq k^2+5k\\ &> k^2+2k+k\\ &> k^2+2k+1. \end{align}$$