How is it that $(2^k+1)3 \gt (2^k+1)2\gt 2^{k+1}+2\gt 2^{k+1}+1$?

Question

I am working on a mathematical induction worksheet, and my professor gave us the key. I have run across something that makes zero sense to me, so please explain if you can.

Additional info: $k \ge2$

Here is what she has:

What needs to be shown: $$ 3^{k+1} \gt 1 +2^{k+1} \tag0$$

$$3^{k+1} = 3^k(3) \gt (2^k+1)3 \tag1$$ This is where I am confused.. $$ \gt (2^k+1)2 \tag2$$ $$ \gt 2^{k+1}+2 \tag3$$ $$ \gt 2^{k+1}+1 \tag4$$

Answer 1

$\text{As is it was mentioned in the comments we have:}$ $(2^k+1) \cdot 3 >(2^k+1) \cdot 2 \Rightarrow (2^k+1) \cdot 3>2^k \cdot 2+2 \Rightarrow (2^k+1) \cdot 3>2^{k+1}\cdot 2>2^{k+1}+1$ $\text{Finally:}$ $(2^k+1) \cdot 3 >2^{k+1}+1.$
$\text{A general formoula would be:}$ $\ (2^k+1) \cdot 3 >(2^k+1) \cdot c \text{ with c<3}$