Tính giá trị của biểu thức :\(\frac{1}{2\cdot4\cdot6}+\frac{1}{4\cdot6\cdot8}+\frac{1}{6\cdot8\cdot10}+...+\frac{1}{96\cdot98\cdot100}\)
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\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}.\frac{49}{100}=\frac{49}{200}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
=>\(S=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
=>\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
=>\(S=\frac{1}{2}.\left(1-\frac{1}{9}\right)-\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{10}\right)\)
=>\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
=>\(S=\frac{4}{9}-\frac{1}{5}\)
=>\(S=\frac{11}{45}\)
Tách phần lử trên ra sao cho có thể rút gọn với phần ơn dưới
\(\frac{3}{2.4}+\frac{3}{4.6}+....+\frac{3}{98.100}\)
\(=\frac{3}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{3}{2}.\frac{49}{100}=\frac{147}{200}\)
\(\frac{3}{2.4}+\frac{3}{4.6}+\frac{3}{6.8}+...+\frac{3}{98.100}\)
\(=\frac{3}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{98.100}\right)\)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{3}{2}.\frac{49}{100}=\frac{147}{200}\)
\(A=\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(A=\frac{1.2.\left(1+2^2+3^2+4^2+5^2\right)}{3.4.\left(1+2^2+3^2+4^2+5^2\right)}\)
\(A=\frac{1.2}{3.4}\)
\(A=\frac{1}{6}\)
Ta thấy : \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)
Vậy B > A
Theo đề bài, ta có:
\(A=\frac{1\times2+2\times4+3\times6+4\times8+5\times10}{3\times4+6\times8+9\times12+12\times16+15\times20}\)
\(A=\frac{1\times2\times\left(1+2^2+3^2+4^2+5^2\right)}{3\times4\times\left(1+2^2+3^2+4^2+5^2\right)}\)
\(A=\frac{1\times2}{3\times4}\)
\(A=\frac{1}{6}\)
Ta thấy rằng: \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)
Vậy \(B>A\)