\(\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+\cdot\cdot\cdot+\frac{1}{96\times98}+\frac{1}{98\times100}\)= ?
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NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 4 2019
\(1+\frac{2}{n\left(n+3\right)}=\frac{n^2+3n+2}{n\left(n+3\right)}=\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+3\right)}\)
\(\Rightarrow A=\frac{2.3}{1.4}.\frac{3.4}{2.5}.\frac{4.5}{3.6}...\frac{2020.2021}{2019.2022}\)
\(\Rightarrow A=\frac{2.3.4...2020}{1.2.3...2019}.\frac{3.4.5...2021}{4.5.6...2022}=\frac{2020}{1}.\frac{3}{2022}=\frac{1010}{337}\)
1 tháng 5 2015
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
30 tháng 4 2019
\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{7}+\frac{1}{7}\cdot\frac{1}{8}+\frac{1}{8}\cdot\frac{1}{9}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
BD
30 tháng 4 2019
\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{8}\cdot\frac{1}{9}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}\)
* LÀM NỐT *
#Louis
DD
6 tháng 8 2015
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{99}\right)=\frac{3}{2}.\frac{4}{3}...\frac{100}{99}=\frac{100}{2}=50\)
6 tháng 8 2015
= \(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\cdot\cdot\cdot\frac{99}{98}\cdot\frac{100}{99}=\frac{3.4.5....99.100}{2.3.4....98.99}=\frac{100}{2}=50\)
\(S.2=\frac{2}{2.4}+\frac{2}{4.6}+...........+\frac{2}{98.100}\)
\(S.2=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{98}-\frac{1}{100}\)
\(S.2=\frac{1}{2}-\frac{1}{100}\)
\(S.2=\frac{49}{100}\)
\(S=\frac{49}{100}:2\)
\(S=\frac{49}{200}\)