Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\)
\(\dfrac{1}{4}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\\ \Leftrightarrow A-\dfrac{1}{4}A=\dfrac{1}{2}-\dfrac{1}{2^{10}}=\dfrac{2^9-1}{2^{10}}\\ \Leftrightarrow A=\dfrac{2^9-1}{2^{10}}:\dfrac{3}{4}=\dfrac{2^{11}-2^2}{2^{10}\cdot3}=\dfrac{2^9-1}{2^8\cdot3}=\dfrac{511}{768}\)
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\)
\(2A=2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\right)\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{256}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{256}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\right)\)
\(A=1-\dfrac{1}{512}\)
\(=\dfrac{511}{512}\)
