discrete mathematics - Trouble understanding algebra in induction proof

I'm on hour 20 of studying for the discrete math midterm tomorrow, and I've got to be honest I'm a little panicked. In particular I'm having trouble with induction proofs, not because I don't understand the proofs but because of the algebra.

Here is an example of a problem/solution my professor put up:

6. Give an induction proof to prove the quantified predicate, 
(All n, n >= 1, 1/2 + 1/2^2 + ... + 1/2^n = (2^n - 1)/2^n)

Sol:
Let P(n) be the predicate 
1/2 + 1/2^2 + ... + 1/2^n = (2^n - 1)/2^n


Base Case: n=1:
P(1) <=> 1/2 = (2-1)/2 <=> 1/2 = 1/2 <=> T

Induction Step:
(All n, n >= 1, P(n) => P(n+1))

  P(n)
=> {definition of P(n)}
  1/2 + 1/2^2 + ... + 1/2^n = (2^n - 1)/2^n
=> {add 1/2^(n+1) to both sides of "="}

  1/2 + 1/2^2 + ... + 1/2^(n+1) = (2^n - 1)/2^n + 1/2^(n+1)
=> {arithmetic}
  1/2 + 1/2^2 + ... + 1/2^(n+1) = (2*(2^n - 1) + 1)/2^(n+1)
=> {arithmetic}
  1/2 + 1/2^2 + ... + 1/2^(n+1) = (2^(n+1) - 1)/2^(n+1)
=> {definition of P(n+1)}
  P(n+1)

After step two of the induction step I get confused about these transformations on the right hand side:

$$(2^n - 1)/2^n + 1/2^{n+1}$$
$$(2\cdot(2^n - 1) + 1)/2^{n+1}$$
$$(2^{n+1} - 1)/2^{n+1}$$

I'll admit I'm not the best at math, but I really need to do well in this class. Could someone explain the logic of the transformations to me? I know the point is to replace the 'n' with 'n+1', but I don't see the logic of him multiplying by two in the second transformation.

Thanks!

Answer

So we're starting with this:

$$\frac{2^n - 1}{2^n} + \frac{1}{2^{n+1}}.$$

Note that the denominators are not the same. They need to be the same in order to combine them to get to the next step. To do this we multiply the numerator and denominator of the first term by $2$. This makes the denominators the same, because $2^n \cdot 2 = 2^{n+1}$:

$$\frac{2 \cdot \left(2^n - 1\right)}{2^{n+1}} + \frac{1}{2^{n+1}}.$$

This gets us to the next expression:

$$\frac{2 \cdot \left(2^n - 1\right) + 1}{2^{n+1}}.$$

Finally, we multiply the $2$ through in the numerator, and combine terms:

$$2 \cdot \left(2^n - 1\right) + 1 = 2 \cdot 2^n - 2 + 1 = 2^{n+1}-1,$$

which gives you the last expression.

Hope that helps! Good luck on your exam.