<s>[INST] Simplifying fractions with exponents I'm revising for an exam, which I have the solutions to. One of the questions asks me to prove that a sequence is a Cauchy sequence, sequence is written as:
$ a_n = \frac{2^{n+2}+1}{2^n},$
and then in the solutions, it has:
$\left | a_n - a_m \right | = \left | \frac{2^{n+2}+1}{2^n} - \frac{2^{m+2}+1}{2^m} \right | = \left | \frac {1}{2^n} - \frac {1}{2^m} \right |$
I am extremely rusty on my math skills, but could someone explain how the $2^{nd}$ and $3^{rd}$ absolute values are equal?
 [/INST] Notice that 
$$
\frac{2^{n+2}+1}{2^n} = \frac{2^{n+2}}{2^n} + \frac{1}{2^n} = 2^2 + \frac{1}{2^n} = 4 + \frac{1}{2^n}.
$$
Doing the same with the second term (the one involving $m$), we find that
$$
\left| \frac{2^{n+2}+1}{2^n} - \frac{2^{m+2}+1}{2^m} \right|
= \left| 4 + \frac{1}{2^n} - \left( 4 + \frac{1}{2^m} \right) \right|
= \left| \frac{1}{2^n} - \frac{1}{2^m} \right|.
$$
</s>