the problem is:
Prove by induction. In each case, n is a positive integer.
$2^{n} ≤ 2^{(n+1)} - 2^{(n-1)} - 1$
Right now I am at the following, but I got stuck and don't know where to go from here.
$2^{(n+1)} ≤ 2^{(n+2)} - 2^{n} - 1 = 2^{n}(2^{2} - 1) - 1 = 2^{n}(3) - 1 =$ $3*2^{n} - 1$
On
First, prove the base case $n = 1$:
$$2^1 \le 2^2 - 2^0 - 1 = 2$$
Now, we want to extend the equation to the $n^{th} + 1$ term. We get:
$$2^{n+1} \le 2^{n+2} - 2^n - 1$$
Working with the right side of the equation now, we can begin to simplify the expression:
$$2^{n+2} - 2^n - 1 = 2^n(4 - 1) - 1 = 3*2^n - 1$$
Dividing out $2^n$ from both sides of the expression leaves us with this:
$$2 \le 3 - \frac {1}{2^n}$$
This inequality will always hold true for $n \ge 1$, as the base case $n = 1$ leaves us with $2 \le 2$, and as $n$ increases, the amount being subtracted off of $3$ will only decrease.
Obviously you need to show the inequality holds for $n=1$ which is trivial. Then suppose you have $2^n \leq 2^{n+1}-2^{n-1}-1$ for some $n$. Multiplying everything by 2 you get $(2)2^n \leq (2)2^{n+1}-(2)2^{n-1}-2$ which equals to $2^{n+1} \leq 2^{n+2}-2^{n}-2 \leq 2^{n+2}-2^n-1$ so the inequality holds for $n+1$ which implies it holds for every natural number.