Question

\(\frac{3}{2\cdot4}+\frac{3}{4\cdot6}+\frac{3}{6\cdot8}+.....+\frac{3}{98\cdot100}\)

 

Answer 1

\(\frac{3}{2.4}+\frac{3}{4.6}+....+\frac{3}{98.100}\)

\(=\frac{3}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=\frac{3}{2}.\frac{49}{100}=\frac{147}{200}\)

Answer 2

\(\frac{3}{2.4}+\frac{3}{4.6}+\frac{3}{6.8}+...+\frac{3}{98.100}\)

\(=\frac{3}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{98.100}\right)\)

\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=\frac{3}{2}.\frac{49}{100}=\frac{147}{200}\)

Answer 3

Đặt \(A=\frac{3}{2.4}+\frac{3}{4.6}+\frac{3}{6.8}+...+\frac{3}{98.100}\)

\(A=\frac{3}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{98.100}\right)\)

\(A=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(A=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(A=\frac{3}{2}.\frac{49}{100}\)

\(A=\frac{147}{200}\)

Vậy \(A=\frac{147}{200}\)

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Answer 4

\(\frac{3}{2.4}+\frac{3}{4.6}+\frac{3}{6.8}+...+\frac{3}{98.100}\)

= \(\frac{3}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

= \(\frac{3}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)

= \(\frac{3}{2}.\frac{49}{100}=\frac{147}{200}\)

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