\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(A=1-\frac{1}{2^9}\)
A= 1/2+1/2^2+1/2^3+....+1/2^9(1)
=> 2A = 1+1/2+1/2^2+...+1/2^8(2)
Lấy (2) - (1) ta có ;
=> A = 1-1/2^19
Vậy.................
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^9}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^8}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+.........+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..........+\frac{1}{2^9}\right)\)
\(\Rightarrow A=1-\frac{1}{2^9}\)