\(S=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{17\cdot20}\\ =\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{17\cdot20}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\\ =\dfrac{1}{3}\cdot\dfrac{9}{20}\\ =\dfrac{3}{20}\)
Giải:
Ta có:
\(S=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{17.20}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{17.20}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\)
\(=\dfrac{1}{3}.\dfrac{9}{20}=\dfrac{3}{20}\)
Vậy \(S=\dfrac{3}{20}\)
\(S=\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{17.20}\)
\(S=\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\)
\(S=1-\dfrac{1}{20}=\dfrac{19}{20}\)
Lời giải:
\(S=\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{17.20}\)
\(\Rightarrow S=\dfrac{1.3}{2.5.3}+\dfrac{1.3}{5.8.3}+...+\dfrac{1.3}{17.20.3}\)
\(\Rightarrow S=\dfrac{1}{3}.\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{17.20}\right)\)
\(\Rightarrow S=\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)
\(\Rightarrow S=\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\)
\(\Rightarrow S=\dfrac{1}{3}.\dfrac{9}{20}\)
\(\Rightarrow S=\dfrac{3}{20}\)
