<s>[INST] Prove the function Prove 
$\forall n\in\mathbb{N}: n\ge 1 \rightarrow 2^n\le 2^{n+1}-2^{n-1}-1.$
I did that $2^n \le\ 2^{n+2} - 2^{n}$ and then $2^n < 2^{n+2} - 2^{n-1}$
but have no idea how to add $-1$ in the function and let the $2^n \le\ 2^{n+2} - 2^{n} - 1$
 [/INST] Use a geometric series to show that $1+2+\cdots +2^n = 2^{n+1}-1$.
Here is an alternative approach:
Write the equation as $2^{n+1} \ge 2^n + 2^{n-1} +1$.
For $n=1$ it is easy to check that $4 \ge 2 + 1 +1$.
Suppose the statement is true for $n$ and multiply across by $2$ to get
$2^{n+2} \ge 2^{n+1} + 2^{n} +2 \ge 2^{n+1} + 2^{n} +1$,
hence the statement is true for $n+1$.
A moment's reflection will show the connection between this approach and
my first approach above.  
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