Đặt:
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
=>\(\frac{A}{2}=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{11}}\)
=>\(A-\frac{A}{2}=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{11}}\right)\)
=>\(\frac{A}{2}=\frac{1}{2}-\frac{1}{2^{11}}=\frac{1023}{2048}\Rightarrow A=\frac{1023}{2048}.2=\frac{1023}{1024}\)