Đặt \(A=\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\) có:
\(2A=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\)
\(\Rightarrow2A-A=\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\right)-\left(\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{1}{2}-\left(\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)=\dfrac{1}{2}-\left(\dfrac{1}{2}-\dfrac{1}{1024}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{1024}=\dfrac{1}{1024}\)
Vậy...
Cách của Tuấn Anh Phan Nguyễn đây.
\(=\dfrac{1}{2}-\left[\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{512}+\dfrac{1}{1024}\right]\)
\(=\dfrac{1}{2}-\left[\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{8}\right)+\left(\dfrac{1}{8}-\dfrac{1}{16}\right)+...+\left(\dfrac{1}{512}-\dfrac{1}{1024}\right)\right]\)\(=\dfrac{1}{2}-\left(\dfrac{1}{2}-\dfrac{1}{1024}\right)=\dfrac{1}{1024}.\)
Tách nó ra thành 21;22;23..........đến 210đó bạn
Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.........+\frac{1}{1024} (1)\)
Ta có: \(2A=2+1+\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{512} (2)\)
Từ (1) và (2)
\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...........+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+..........+\frac{1}{1024}\right)\)
\(\Rightarrow A=2-\frac{1}{1024}\)
\(\Rightarrow A=\frac{2047}{1024}\)
Sửa lại tí :P
\(\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{1024}\)
\(=\left(1-\dfrac{1}{2}\right)-\left(\dfrac{1}{2}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{512}-\dfrac{1}{1024}\right)\)
\(=1-\dfrac{1}{2014}\)
\(=\dfrac{1023}{1024}\)