Bài giải:

Ta có:
$\begin{align}& \frac{1}{2}=1-\frac{1}{2} \\& \frac{1}{4}=\frac{1}{2}-\frac{1}{4} \\& \frac{1}{8}=\frac{1}{4}-\frac{1}{8} \\& ... \\& \frac{1}{2048}=\frac{1}{2024}-\frac{1}{2048} \\\end{align}$

Suy ra:
$\begin{align}& \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2048} \\& =1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...-\frac{1}{2024}+\frac{1}{2024}-\frac{1}{2048} \\& =1-\frac{1}{2048} \\& =\frac{2047}{2048} \\\end{align}$