CM:A=1/1.3+1/3.5+1/5.7+...+1/2015.2016<1/2
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LD
1
23 tháng 7 2015
a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\frac{6}{7}\)
\(=\frac{3}{7}\)
b)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\frac{2010}{2011}\)
\(=\frac{1005}{2011}\)
TV
1
T
25 tháng 8 2023
\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2009\cdot2011}\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2009\cdot2011}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2010}{2011}=\dfrac{1005}{2011}\)
DB
1
22 tháng 3 2021
S= 1/1.3 + 1/3.5 + 1/5.7 +................+ 1/200.202
=>S=1/2.(2/1.3+2/3.5+2/5.7+...+2/200.202)
=>S=1/2.(3-1/1.3+5-3/3.5+...+202-200/200.202)
=>S=1/2.(1-1/3+1/3-1/5+...+1/200-1/202)
=>S=1/2.(1-1/202)
=>S=1/2.201/202
=>S=201/404
Vậy S=201/404
11 tháng 4 2023
=1/2(2/1*3+2/3*5+...+2/2017*2019)
=1/2(1-1/3+1/3-1/5+...+1/2017-1/2019)
=1/2*2018/2019
=1009/2019
11 tháng 4 2023
=1/2(2/1x3+2/3x5+...+2/2017x2019)
=1/2(1-1/3+1/3-1/5+...+1/2017-1/2019)
=1/2x2018/2019
=1008/2019
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2015.2016}=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2016}\right)=\dfrac{1}{2}-\dfrac{1}{2016.2}< \dfrac{1}{2}\left(đpcm\right)\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2015.2017}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2017}\right)\\ < \dfrac{1}{2}.1=\dfrac{1}{2}\)