\(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+.......+\dfrac{1}{1280}\)
\(=\dfrac{1}{5}+\dfrac{1}{5.2}+\dfrac{1}{5.4}+\dfrac{1}{5.8}+.....+\dfrac{1}{5.256}\)
\(=\dfrac{1}{5}+\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+......+\dfrac{1}{2^8}\right)\)
Đặt \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.......+\dfrac{1}{2^8}\)
\(2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^7}\)
\(2A-A=2-\dfrac{1}{2^3}\)
Thay A vào ta được :
\(\dfrac{1}{5}.A=\dfrac{1}{5}.\left(2-\dfrac{1}{2^8}\right)=\dfrac{511}{1280}\)
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