mình sẽ ủng hộ bạn có câu trả lời đúng nhất nhé

= 1/2 . (1/1 - 1/3 + 1/3 - 1/5 +... + 1/2009 - 1/2011)

= 1/2 . (1/1 - 1/2011)

= 1/2 . 2010 / 2011

= 1005/2011

= 1 - 1/2011

= 2010 / 2011

   1/2 ( 2/1.3 + 2/3.5 +...+ 2 /2009.2011)

= 1/2 ( 1/1 - 1/3 + 1/3 - 1/5 +... + 1/2009 - 1/2011)

= 1/2( 1/1 - 1/2011)

= 1/2 . 2010 / 2011

=1005/2011

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+....+\frac{1}{99\cdot101}\)

\(=2\cdot\frac{1}{2}\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\)

\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2} \cdot\left(1-\frac{1}{101}\right)\)

\(=\frac{1}{2}\cdot\frac{100}{101}\)

\(=\frac{50}{101}\)

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2021.2023}\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{2021.2023}\right)\)

\(=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)

\(=\dfrac{1}{2}.\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}.\dfrac{2022}{2023}=\dfrac{1011}{2023}\)

Ta có A = \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2021\cdot2023}\)

            = \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\right)\)

            = \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}+\dfrac{1}{2023}\right)\)

            = \(\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}\cdot\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
 

A=1.31​+3.51​+5.71​+.....+2021.20231​

=12.(21.3+23.5+25.7+....+22021.2023)=21​.(1.32​+3.52​+5.72​+....+2021.20232​)

=12.(1−13+13−15+15−17+....+12021−12023)=21​.(1−31​+31​−51​+51​−71​+....+20211​−20231​)

=12.(1−12023)=12.20222023=10112023=21​.(1−20231​)=21​.20232022​=20231011​

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2020}\)\(-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}\)

\(=\dfrac{2021}{2022}\)

\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + \(\dfrac{2}{7.9}\) + ... + \(\dfrac{2}{2021.2023}\)

= 1 - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{2021}\) - \(\dfrac{1}{2023}\)

=> 1 - \(\dfrac{1}{2023}\)

= \(\dfrac{2022}{2023}\)

B : 7/2 =2/1.3+2/3.5+...+2/99.101

B:7/2=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101

B:7/2=1-1/101=100/101

B=100/101*7/2=700/202=350/101

B=7/2(2/1.3+2/3.5+ ...+2/99.101)

B=7/2(1-1/3+1/3-1/5+...+1/99-1/101)

B=7/2(1-1/101)=7/2.100/101=350/101

k nha bạn

B=1/1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101

B=1-1/101=100/101

vậy B=100/101

[(2n+1)(2n+2)(2n+3)(2n+4):12]+(n+1)

\(A=1.3+3.5+5.7+...+\left(2n+1\right)\left(2n+3\right)\)

\(6A=1.3.\left(5+1\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+\left(2n+1\right)\left(2n+3\right)\left(2n+5-2n+1\right)\)

\(6A=3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9-...-\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)\)

\(+\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)\)

\(6A=\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)+3\)

\(A=\frac{\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)+3}{6}\)