<s>[INST] How does $\left(2 - \frac{1}{2^{k}}\right) + \frac{1}{2^{k + 1}}$ become $2 - \frac{1}{2^{k+1}}$? I'm studying induction here and throughout the proof we got from here:
$$ \begin{align}
&\left(2 - \frac{1}{2^{k}}\right) + \frac{1}{2^{k + 1}} \tag1\\[0.75em]
=\; &2 - \frac{1}{2^{k+1}} ( 2-1) \tag2\\[0.75em]
=\; &2 - \frac{1}{2^{k+1}} \tag3
\end{align}$$
How was that done?
 [/INST] They did this by adding up the two fractions:
$\frac{1}{2^k}$ - $\frac{1}{2^{k+1}}$
When dealing with the denominators, you can write $2^{k+1}$ as ${2^k}$$\cdot$${2^1}$
Then the $2^k$ cancels out with the first fraction's denominator and you are left with $2$$\cdot$$1$ which totals to 2. In the second fraction you are just left with 1.
Then this all adds upto: $\frac{2}{2^{k+1}}$ - $\frac{1}{2^{k+1}}$.  You can pull out $2^{k+1}$ which equals $\frac{1}{2^{k+1}}$$(2-1)$ = $\frac{1}{2^{k+1}}$.
With the complete expression this equals to what we needed, $2$ - $\frac{1}{2^{k+1}}$
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