Proving by induction that $2^n \le 2^{n+1}-2^{n-1} - 1$ . Does my proof make sense?

I'm not sure if this proof is valid or not and could use some feedback.

Let P be the supposition that $2^n \le 2^{n+1}-2^{n-1} - 1$ for all positive integers n.

Base Case: Let n=1

$2^1 \le 2^2-2^0-1$

$2\le2$

Suppose that P holds true for some positive integer n=k

$$2^{k+1}=2*2^k$$

By inductive hypothesis:

$$2*2^k \le 2(2^{k+1}-2^{k-1}-1)$$ $$2^{k+1} \le 2^{k+2}-2^{k}-2$$

Since we know $2^{k+1} \le 2^{k+2}-2^{k}-2$, we also know $2^{k+1} \le 2^{k+2}-2^{k}-1$ because the former is always smaller.

This proves that: $$2^k \le 2^{k+1}-2^{k-1} - 1\Rightarrow 2^{k+1} \le2^{k+2}-2^{k}-1 $$

Solution 1:

It looks fine!

There's really no particular need for induction, though, if you know that $x\mapsto 2^x$ is an increasing function. Instead, show that $$2^{n+1}=2^n+2^n=2^n+2^{n-1}+2^{n-1},$$ so since $n-1\ge0$ for all positive integers $n,$ we have $2^{n-1}\ge 2^0=1,$ so that $$2^{n+1}\ge 2^n+2^{n-1}+1,$$ from which the result follows.

More generally, we can use this approach to show that, for real numbers $x,$ we have $2^x\le2^{x+1}-2^{x-1}-1$ if and only if $x\ge1.$