A = 1/1.3 +1/3.5 + 1/5.7 + ... +1/199.201
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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)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\frac{2010}{2011}\)
\(=\frac{1005}{2011}\)
26 tháng 12 2018
a) Đặt B= 1/1.3 + 1/3.5 + 1/5.7 + .....+ 1/19.21
Ta có: 2B= 2/1.3 + 2/3.5 + 2/5.7 + ....+ 2/19.21
= 1- 1/3 + 1/3-1/5 + 1/5-1/7 +....+ 1/19-1/21
= 1-1/21 = 20/21
=> B= 20/21 : 2 => B= 10/21
b) Như trên, ta có: 2A= 1- (1/2n + 1) => A=( 1-1/2n+1).1/2
=> A= 1/2- 1/2n+1
=> A< 1/2 ( đpcm )
10 tháng 8 2015
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}:2=\frac{100}{101}\times\frac{1}{2}=\frac{50}{101}\)
PT
4
NH
17 tháng 6 2019
\(D=\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{199.201}\)
\(D=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{199.201}\right)\)
\(D=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{199}-\frac{1}{201}\right)\)
\(D=\frac{3}{2}\left(1-\frac{1}{201}\right)\)
\(D=\frac{3}{2}.\frac{200}{201}\)
\(D=\frac{100}{67}\)
T
17 tháng 6 2019
#)Giải :
\(D=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{199.201}\)
\(D=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{199.201}\right)\)
\(D=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\right)\)
\(D=\frac{3}{2}\left(1-\frac{1}{201}\right)\)
\(D=\frac{3}{2}\times\frac{200}{201}\)
\(D=\frac{100}{67}\)
TA
2
HP
12 tháng 5 2016
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{2013.2015}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{2013.2015}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2014}{2015}=\frac{1007}{2015}\)
Vậy A=1007/2015
TN
12 tháng 5 2016
\(2A=2\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\)
\(2A=1-\frac{1}{2015}\)
\(A=\frac{2014}{2015}:2\)
\(A=\frac{1007}{2015}\)
12 tháng 5 2016
1/1.3+1/3.5+...+1/2013.2015
=1/2.(1/1-1/3+1/3-1/5+...+1/2013-1/2015)
=1/2.(1/1-1/2015)
=1/2.2014/2015
=1007/2015
12 tháng 5 2016
A=1/1.3+1/3.5+1/5.7+...+1/2013.2015
2A=2.(1/1.3+1/3.5+1/5.7+...+1/2013.2015)
=2/1.3+2/3.5+2/5.7+...+2/2013.2015
=1-1/3+1/5-1/7+1/7-1/9+...+1/2013-1/2015
=1-1/2015
=2014/2015
=>2A=2014/2015=>A=1007/2015
PM
6
11 tháng 5 2019
A = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/2011.2013
A = 1/2.(2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2011.2013)
A = 1/2.(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2011 - 1/2013)
A = 1/2.(1 - 1/2013)
A = 1/2.2012/2013
A = 1006/2013
11 tháng 5 2019
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2011.2013}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\)
\(2A=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{2011}-\frac{1}{2011}\right)-\frac{1}{2013}\)
\(2A=1-\frac{1}{2013}\)
\(2A=\frac{2012}{2013}\)
\(A=\frac{2012}{2013}:2\)
\(A=\frac{1006}{2013}\)
~ Hok tốt ~
T
25 tháng 8 2023
\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2009\cdot2011}\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2009\cdot2011}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2010}{2011}=\dfrac{1005}{2011}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{199.201}\).
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{199.201}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\)
\(2A=\frac{1}{1}-\frac{1}{201}\)
\(2A=\frac{201-1}{201}\)
\(2A=\frac{200}{201}\)
\(A=\frac{200}{201}:2\)
\(A=\frac{200}{402}\)
Đáp số là 100/201