Proving $2^n -1 > \frac {n(17n-89)}{2}$ for all positive $n$ without induction If $n$ is any positive integer then $2^n -1 > \frac {n(17n-89)}{2}$ is true . How to prove this inequality without induction?
 A: I'll hide the induction in claims about 


*

*how many ways there are to choose $k$ element subsets out of an $n$ element set

*how many subsets an $n$-element set has.


If $n>m+2$, then the number of ways to select $2$ out of $n$ objects and paint them blue, and afterwards select an arbitrary subset of the first $m$ non-blue objects and paint them red, is
$$ {n\choose 2}\cdot2^m.$$
This gives us one of $2^n-1$ nonempty sets of painted objects (actually, the set has $2\le k\le m+2$ elements) and each such subsets occurs with at most $m+2\choose 2$ different red/blue patterns. Hence
$$ 2^m\cdot{n\choose 2}\le {m+2\choose 2}\cdot (2^n-1)$$
or $$\frac{2^{m+1}}{(m+2)(m+1)}\cdot{n\choose 2}\le 2^n-1.$$
Since
$$\frac{n(17n-89)}2<\frac{n(17n-17)}2 =17\cdot\frac{n(n-1)}2$$
we find that
$$ \frac{n(17n-89)}2<2^n-1$$
if we can choose $m=11$ (as $\frac{2^{12}}{13\cdot12}=\frac{1024}{39}>17$).
So this solves the problem for all $n>13$, while the rest can be checked manually.
A: You can try considering $$\lim_{n\rightarrow\infty}\frac{n^2}{2^n}=0$$
A: Fact $1$: $2^n$ moves way faster than $n^2$, for $n \ge 4$ 
Fact $2$: $2^n$ moves way faster than $n^3$, for $n \ge 10$ 
