\(P=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+...+\dfrac{1}{2021.2023}\)

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 \(P=A+B=\dfrac{1011}{2023}+\dfrac{505}{2022}=\dfrac{3065857}{4090506}\)

\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{2016}{2017}\)

\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{2016}{2017}\)

\(A=\dfrac{1}{2}\left(\dfrac{2.2}{1.3}\right).\left(\dfrac{3.3}{2.4}\right)...\left(\dfrac{2020.2020}{2019.2021}\right)\)

\(=\dfrac{1.2.2.3.3...2020.2020}{1.2.2.3.3.4.4...2019.2021}\)

\(=\dfrac{1}{2021}\)

\(A=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)...\left(1+\dfrac{1}{2019\cdot2021}\right)\)

\(A=\dfrac{1}{2}\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\left(1+\dfrac{1}{4^2-1}\right)...\left(1+\dfrac{1}{2020^2-1}\right)\)

\(A=\dfrac{1}{2}\cdot\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\cdot\left(3+1\right)}...\left(\dfrac{2020^2}{\left(2020-1\right)\cdot\left(2020+1\right)}\right)\)

\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{2}{3}\cdot\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2020}{2021}\)

\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}\)

\(A=\dfrac{1}{2}\cdot2020\cdot\dfrac{2}{2021}=\dfrac{2020}{2021}\)

Bài này nếu dùng hằng đẳng thức lớp 8 thì hay hơn.Thôi cứ làm vầy nhé:

1+1/1.3=2^2/1.3  ;   1+1/2.4=3^2/2.4  ;   1+1/3.5=4^2/3.5  ;  ......;1+1/2007.2009=2008^2/2007.2009    Thấy quy luật rồi chứ!

ta được A=(2^2.3^2.4^2.....2008^2)/1.3.2.4.3.5.4.6.5.7...2007.2009    Chú ý quan sát và sắp xếp 

để giản ước hết.

=(2^2.3^2.4^2....2008^2)/(1.2.3.4.5.6....2007.3.4.5.6....2007.2008.2009)    Chuẩn đó

=(2^2.3^2.4^2....2008^2)/(2.3^2.4^2.5^2....2007^2.2008.2009)   Viết ra nháp thì dễ nhìn hơn

=2^2.2008^2/2.2008.2009=2.2008/2009   Tựbấm máy và nhớ TICK đó.