Theo bài ra, ta có: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2017.2018.2019}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2017.2018.2019}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2018.2019}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2018.2019}\right)\)
Giải thích:
\(\frac{2}{1.2.3}=\frac{3}{1.2.3}-\frac{1}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)
\(\frac{2}{2.3.4}=\frac{4}{2.3.4}-\frac{2}{2.3.4}=\frac{1}{1.2}-\frac{1}{3.4}\)
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\(\frac{2}{2017.2018.2019}=\frac{2019}{2017.2018.2019}-\frac{2017}{2017.2018.2019}=\frac{1}{2017.2018}-\frac{1}{2018.2019}\)