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a: \(A=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2022\cdot2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{1011}{2024}=\dfrac{1011}{4848}< \dfrac{1}{4}\)
b: \(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2013\cdot2015}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2014}{2015}=\dfrac{1007}{2015}< \dfrac{1}{2}\)
=1-1/3-1/2+1/4+1/3-1/5-1/4+1/6+...+1/97-1/99-1/98+1/100
=1-1/2-1/99-1/98=2327/4851
\(\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)
\(=1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}-\frac{1}{98}+\frac{1}{100}\)
\(=1-\frac{1}{2}-\frac{1}{99}-\frac{1}{98}\)
\(=\frac{2327}{4851}\)
Đặt A=1/1.3 - 1/2.4 +1/3.5 -1/4.6 +.....+1/97.99 -1/98.100
4A= 4/1.3 -4/2.4 +4/3.5 -4/4.6 +.....+4/97.99 -4/98.100
=(4/1.3 +4/3.5 +...+4/97.99) - (4/2.4 +4/4.6 +...+4/98.100)
=(1/1 -1/3+1/3-1/5+...+1/97-1/99)-(1/2 -1/4 -....1/98-1/100)
=(1/1-1/99)-(1/2-1/100)
4A=98/99 - 99/100
A= (98/99-99/100) :4
\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)
\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)
\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)
\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)
\(B=\dfrac{4.9.16.100}{3.8.15.99}\)
\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)
\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)
\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}=\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\left(\frac{58}{45}\right)\)
\(S=\frac{29}{45}\)
\(A=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)\left(1+\frac{1}{4\cdot6}\right)...\left(1+\frac{1}{2020\cdot2022}\right)\)
\(A=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot\frac{25}{4\cdot6}\cdot....\cdot\frac{4084441}{2021\cdot2022}\)
\(A=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\frac{4\cdot4}{3\cdot5}\cdot\frac{5\cdot5}{4\cdot6}\cdot...\cdot\frac{2021\cdot2021}{2020\cdot2022}\)
\(A\frac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot2021\cdot2021}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot....\cdot2020\cdot2022}\)
\(A=\frac{\left(2\cdot3\cdot4\cdot5\cdot...\cdot2021\right)\left(2\cdot3\cdot4\cdot5\cdot...\cdot2021\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot2020\right)\left(3\cdot4\cdot5\cdot6\cdot...\cdot2022\right)}\)
\(A=\frac{2021\cdot2}{2022}\)
\(A=\frac{4042}{2022}\)
\(A=\frac{2021}{1011}\left(đpcm\right)\)