Question

[๖²⁴ʱ ₡ảM↭₦ắ₦G ︵²ᵏ⁶]

• 1/2 nhân 4 + 1/6 nhân 8 +...+ 1/96 nhân 98 + 1/98 nhân 100.tìm tổng

Answer 1

[Huỳnh Quang Sang]

\(\frac{1}{2\cdot4}+\frac{1}{6\cdot8}+...+\frac{1}{96\cdot98}+\frac{1}{98\cdot100}\)

\(=\frac{1}{2}\left[\frac{2}{2\cdot4}+\frac{2}{6\cdot8}+...+\frac{2}{96\cdot98}+\frac{2}{98\cdot100}\right]\)

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

\(=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{100}\right]=\frac{1}{2}\left[\frac{50}{100}-\frac{1}{100}\right]=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)

Answer 2

[_Shadow_]

\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}...+\frac{1}{96.98}+\frac{1}{98.100}\)

\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{96.98}+\frac{2}{98.100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\right)\)

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

\(=\frac{1}{2}.\frac{49}{100}\)

\(=\frac{49}{200}\)

~Học tốt~