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Bài 1
Nhân 2 vào biểu thức
Rút gọn và trừ đi 1 lần nó
còn lại \(\frac{1}{2}_{ }-\frac{1}{2^{10}}\)
\(A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(A=\frac{1}{2}-\frac{1}{2^{10}}\)
#)Giải :
\(A=1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{4950}\)
\(2A=2+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)
\(2A=2+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(2A=2+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(2A=2+\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\Leftrightarrow A=1+\left(1-\frac{1}{50}\right)\)
\(\Leftrightarrow A=\frac{99}{50}\)
\(A=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{4851}+\frac{1}{4950}\)
\(=2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9702}+\frac{1}{9900}\right)\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{1000}\right)\)
\(=2.\left(1-\frac{1}{100}\right)\)
\(=2.\frac{99}{100}\)
\(=\frac{99}{50}\)
\(-1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-....-\frac{1}{1225}\)
\(=-2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{2450}\right)\)
\(=-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\right)\)
\(=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=-2\left(1-\frac{1}{50}\right)=-2\cdot\frac{49}{50}=-\frac{49}{25}\)
Đặt : \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1225}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{2450}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{49.50}\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{49}-\frac{1}{50}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(A=2.\frac{12}{25}\)
\(A=\frac{24}{25}\)