\(C=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{779}{780}=\dfrac{2.2}{3.2}.\dfrac{5.2}{6.2}.\dfrac{9.2}{10.2}...\dfrac{779.2}{780.2}\)
\(=\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{1558}{1560}=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{38.41}{39.40}\)
\(=\dfrac{1.2.3...38}{2.3.4...39}.\dfrac{4.5.6...41}{3.4.5...40}=\dfrac{1}{39}.\dfrac{41}{3}=\dfrac{41}{117}\)
\(C=\left(1-\dfrac{2}{6}\right)\left(1-\dfrac{2}{12}\right)\left(1-\dfrac{2}{20}\right)...\left(1-\dfrac{2}{1560}\right)\)
\(=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{39.40}\right)\)
Ta có: \(1-\dfrac{2}{n\left(n+1\right)}=\dfrac{n\left(n+1\right)-2}{n\left(n+1\right)}=\dfrac{n^2+n-2}{n\left(n+1\right)}=\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Do đó:
\(C=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{38.41}{39.40}\)
\(=\dfrac{1.2.3...38}{2.3.4...39}.\dfrac{4.5.6...41}{3.4.5...40}=\dfrac{1}{39}.\dfrac{41}{3}=\dfrac{41}{117}\)
