\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{1}{3}\cdot\frac{99}{202}=\frac{33}{202}\)
sửa lại đề 1 chút :
\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{98\cdot101}\)
\(=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{98\cdot101}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{1}{3}\cdot\frac{99}{202}\)
\(=\frac{33}{202}\)
Giải:
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)
\(\Leftrightarrow\frac{3}{3}\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\right)\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\right)\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(\Leftrightarrow\frac{1}{3}.\frac{99}{202}\)
\(\Leftrightarrow\frac{33}{202}\)
\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+ \frac{1}{98\cdot101}\)
\(=\frac{1}{3}\cdot\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{98\cdot101}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{1}{3}\cdot\frac{99}{202}\)
\(=\frac{33}{202}\)
