\(P=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+...+\dfrac{1}{2021.2023}\)
Ta sẽ "tách" P làm 2 phần:
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2021.2023}\)
\(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{2020.2022}\)
Do đó \(P=A+B\)
Ta có \(A=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2023-2021}{2021.2023}\right)\)
\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)
\(A=\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)\)
\(A=\dfrac{1011}{2023}\)
Mặt khác, \(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{2020.2022}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2020.2022}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+\dfrac{8-6}{6.8}+...+\dfrac{2022-2020}{2020.2022}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(B=\dfrac{505}{2022}\)
Từ đó \(P=A+B=\dfrac{1011}{2023}+\dfrac{505}{2022}=\dfrac{3065857}{4090506}\)
