\(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\)
\(=\frac{1}{5}\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
\(=\frac{\frac{1}{5}\left(1-\frac{1}{2^9}\right)}{\left(1-\frac{1}{2}\right)}\)
\(=\frac{2}{5}\left(1-\frac{1}{2^9}\right)\)
\(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1280}\)
\(=\frac{1}{5}\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)
Đặt \(A=1+\frac{1}{2}+...+\frac{1}{2^8}\)
\(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)
\(2A=2+1+...+\frac{1}{2^7}\)
\(2A-A=\left(2+1+...+\frac{1}{2^7}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)
\(A=2-\frac{1}{2^8}\).Thay A vào đc: \(\frac{1}{5}\cdot\left(2-\frac{1}{256}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)
( các bạn chỉ làm 1 câu thôi cũng được )