Tham khảo ở phần Câu hỏi tương tự bạn nhé :
Câu hỏi của Trịnh Thúy An - Toán lớp 5 - Học toán với OnlineMath
\(B=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\)
\(B=1\cdot\frac{1}{5}+\frac{1}{2}\cdot\frac{1}{5}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{8}\cdot\frac{1}{5}+...+\frac{1}{256}\cdot\frac{1}{5}\)
\(B=\frac{1}{5}\cdot\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\right)\)
Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)
\(\Rightarrow2A-A=2-\frac{1}{256}\)
\(A=2-\frac{1}{256}\)
Thay A vào B
có: \(B=\frac{1}{5}.\left(2-\frac{1}{256}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)
\(B=\frac{1}{5}+\frac{1}{10}+..+\frac{1}{1280}\)
\(\Rightarrow B=\frac{1}{5}+\frac{1}{5}.\frac{1}{2}+...+\frac{1}{5}.\frac{1}{256}\)
\(\Rightarrow B=\frac{1}{5}(1+\frac{1}{2}+...+\frac{1}{256})\)
ĐẶT \(A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+...+\frac{1}{128}\)
\(\Rightarrow2A-A=(2+1+...+\frac{1}{128})-(1+\frac{1}{2}+...+\frac{1}{256})\)
\(\Rightarrow A=2-\frac{1}{256}\)
\(\Rightarrow A=\frac{511}{256}\)
\(\Rightarrow B=\frac{511}{1280}\)
