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\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
Ta có:
\(A=\dfrac{1}{2}.\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{2016^2}{2015.2017}\)
\(A=\dfrac{1}{2}.\dfrac{2^2}{3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{2016^2}{2015.2017}\)
\(A=\left(\dfrac{2.3.4...2016}{2.3.4.5...2015}\right).\left(\dfrac{2.3.4...2016}{2.3.4.5...2017}\right)\)
\(A=2016.\dfrac{1}{2017}=\dfrac{2016}{2017}\)
S = \(\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.....\dfrac{4068289}{2016.2018}\)
\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{2017^2}{2016.2018}\\ =\dfrac{2.2017}{2018}\\ =\dfrac{2017}{1009}.\)