How to use an exponent that contains a variable I am trying to understand a problem that uses mathematical induction to prove the validity of a statement. This is how one section moves to another:
$$
2k + 3 = 2^{k + 1}
$$
$$
2k + 3 = (2k + 1) + 2
$$
can someone explain this to me?
Here is the proof:
Theorem: For all integers $n\geq 3$, $2n + 1 < 2^n$.
I need to show how that for all integers $k\geq3$, if $P(k)$ is true then $P(k+1)$ is also true:
$$
2n + 1 < 2^{n}
$$
$$
2k+1<2k \\ (\text{the inductive hypothesis})
$$
$$
2(k+1)+1<2^{k+1} \\ (\text{the $k+1$ term}) \\
$$
$$
$$
Body of Proof:
$$
2k + 3 = 2^{k + 1}
$$
$$
2k + 3 = (2k + 1) + 2
$$
$$
<2^{k}  + 2^{k}
$$
$$
\therefore 2k+3<2 \cdot 2^{k} = 2^{k+1}
$$
 A: So the proposition, $P(n)$ is $2n + 1 < 2^n$ for all $n\ge 3$.
So first you show the initial step that if $k = 3$ that $P(k)$ is true.
That's easy to show.  You just to do it.  $2(3) + 1 = 7 < 8 = 2^3$.  It's true.
Now the  induction step
We must show that for any $k$ so that $P(k)$ is true, (for example if $k = 3$) then it can be proven that for that particular $k$ that $P(k)$ is true.
So if $P(k)$ is true then we take it as a given that $k$ is a specific number so that $2k + 1 < 2^k$.  That's a given.
We need to prove that $2(k+1) + 1 < 2^{k+1}$.
Okay.  $2(k+1) +1 = 2k + 2 + 1 = 2k+3$.  And $2^{k+1} = 2^k*2$. 
Can we combine those to prove $2k+3 < 2^k*2$?  
We can because we were given that $2k + 1 < 2^k$
Proof:  $2(k+1) + 1 =$
$2k + 3 = $
$(2k + 1) + 2$.  We were given that $2k + 1 < 2^k$ so
$(2k + 1) + 2 < 2^k + 2$.  Now we know that as $k \ge 3 > 1$ that $2^k > 2$. so
$2^k + 2 < 2^k + 2^k$
$= 2*2^k$
$= 2^{k+1}$
So to reiterate:
$$2(k+1)+1=(2k+1) + 2< 2^k +2 < 2^k + 2^k=2*2^k = 2^{k+1}$$
And that's it we are done.
We have proven that 1) $P(k)$ is true for $k = 3$ and  2) if $P(k)$ is true for any particular $k$ (such as $k = 3$) then $P(k+1)$ is also true (so we know it is true for $4$, and therefore it is true for $4+1 = 5$, etc.).  From induction principal we can conclude $P(n)$ is true for all $n \ge 3$.
