# Prove $2^n > 10n^2$ for all sufficiently large integers n.

How do I prove $2^n > 10n^2$ inductively? I know you can prove this to be true using calculus (i.e. taking derivatives). But how would I do it inductively?

Hints: $2^{10} =1024$ and $2\cdot 10n^2 > 10(n+1)^2$.
You can't prove inductively that there is some integer $n$ such that $2^n>10n^2$, but you can prove that if there exist such an $n$ which is "big enough", then all integers greater than $n$ satisfy the relation, by induction.
If $n$ satisfies the relation, then $$2^{n+1}=2\cdot2^n>2\cdot10n^2\ge 10n^2 + 20n+100$$ if $10n^2\ge 20n + 100$. This last relation is certainly true if $n\ge10$.
$\blacksquare$
It is satisfied for $n=10$ since $1024=2^{10}>10.10^2=1000$. We need to prove $2^{n+1}>10(n+1)^2$ for $n\geq 10$. Let us see that it is indeed true:
$2^{n+1}>2.10n^2=10 (n^2+n^2)>10(n+1)^2$ since $n^2>2n+1$ for $n\geq10$.