Simplification of $2^n + 2^{n-1}$

Question

I'm working through a proof right now, and have gotten to the statement:

$F(n+1) ≤ 2^n + 2^{n-1}$

This is correct, but according to my answer key, from this statement we can conclude that:

$F(n+1) ≤ 2 \cdot 2^n$

As the next step, which I don't understand. If I try to simplify this expression by factoring, I go:

$2^n + 2^{n-1}$

$2^n + 2^n2^{-1}$

$2^n + (\frac{1}{2})2^n$

$1.5 \cdot 2^n$

How on earth is the answer $2 \cdot 2^n$? I'm sure there's some small, stupid detail I'm missing but I just can't seem to get my head around it.

Answer 1

We can conclude this since : $$1.5 \cdot 2^n < 2 \cdot 2^n$$

Hence,

$$\color{blue}{F(n+1)} ≤ 2^n + 2^{n-1}=1.5 \cdot 2^n \color{blue}{< 2 \cdot 2^n}$$

Answer 2

If $a<b$ and $b<c$, then $a<c$. They basically just replaced $1.5\cdot2^n$ with $2^{n+1}$ and used the substitution in the first sentence.

Answer 3

If we have $$2^n+2^{n-1}$$ We must factor out $2^{n-1}$ to get $$2^{n-1}(2+1)$$ $$3*2^{n-1}$$ which is less than $$2*2^n$$ So, since $$F(n+1) \le 3*2^{n-1}$$ and $$3*2^{n-1} \le 2*2^n$$ then $$F(n+1) \le 2*2^n$$

Answer 4

$2^{n-1} \leq 2^n$, so $2^n + 2^{n-1} \leq 2^n + 2^n = 2 \cdot 2^n$.