Help with algebraic manipulation for proof: $\sum_{k=0}^{n} \frac{1}{2^{k}}=2-\frac{1}{2^{n}}$

Question

I'm having trouble finding the algebraic manipulation process for the following inductive proof I am doing:

$$\sum_{k=0}^{n} \frac{1}{2^{k}}=2-\frac{1}{2^{n}}$$

I have gotten as far as the following:

Basis: $n=0$

$\sum_{k=0}^{0} \frac{1}{2^{0}}=2-\frac{1}{2^{0}}$

$=\sum_{k=0}^{0} \frac{1}{2^{0}}=2-\frac{1}{2^{0}}$

$=\sum_{k=0}^{0} 1=2-1$

$=\sum_{k=0}^{0}1=1$

Thus, the basis holds.

Inductive step: Assume $n\geq 0$ and prove that if $\sum_{k=0}^{n} \frac{1}{2^{k}}=2-\frac{1}{2^{n}}$ holds, then $\sum_{k=0}^{n+1} \frac{1}{2^{k}}=2-\frac{1}{2^{n+1}}$ also holds.

$S(n)=\sum_{k=0}^{n} \frac{1}{2^{k}}=2-\frac{1}{2^{n}}$

$S(n+1)=\sum_{k=0}^{n+1} \frac{1}{2^{k}}=2-\frac{1}{2^{n+1}}$

Expanding & replacing hypothesis:

$\sum_{k=0}^{n+1} \frac{1}{2^{k}}$

$=[S(n)]+\frac{1}{2^{n+1}}$

$=[\sum_{k=0}^{n} \frac{1}{2^{k}}]+\frac{1}{2^{n+1}}$

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

This is exactly where I run into trouble. I cannot figure out the steps needed to algebraically transform $2-\frac{1}{2^{n}}+\frac{1}{2^{n+1}}$ into $2-\frac{1}{2^{n+1}}$. I've tried messing around with applying exponent rules in a different order and so on, but many sheets of paper later and I'm still unable to figure it out. I feel like it must be quite obvious and I'm just not seeing something.

Any help regarding this final step of my proof would be very appreciated.

Answer 1

$$\frac{1}{2^n} - \frac{1}{2^{n+1}} = \frac{1}{2^n} - \frac 1 2 \cdot \frac{1}{2^{n}} = \frac 1 {2^n} (1 - \frac1 2) = \frac 1 {2^{n+1}}$$

Basic factoring is where you had the bottleneck, now go finish your proof :)

Answer 2

Note that: $$2-\frac{1}{2^n}+\frac{1}{2^{n+1}}=2-\frac{1}{2^n}\left(1-\frac{1}{2}\right)=2-\frac{1}{2^{n+1}}=S(n+1) $$