\(A=\dfrac{1}{3}\left(1+\dfrac{1}{2}+...+\dfrac{1}{256}+\dfrac{1}{512}\right)\)
Đặt B=1+1/2+...+1/256+1/512
=>2B=2+1+...+1/128+1/256
=>B=2-1/512=1023/512
=>\(A=\dfrac{1}{3}\cdot\dfrac{1023}{512}=\dfrac{341}{512}\)
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