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19 tháng 4 2018

\(\left(1+\frac{1}{1.3}\right).....\left(1+\frac{1}{2013.2015}\right)=\frac{2^2}{1.3}.....\frac{2014^2}{2013.2015}=\)\(\frac{2.3.....2014}{1.2.....2013}.\frac{2.3.....2014}{3.4.....2015}=2014.\frac{2}{2015}=\frac{4028}{2015}\)

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: \(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2013\cdot2015}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}-\dfrac{1}{2015}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2014}{2015}=\dfrac{1007}{2015}< \dfrac{1}{2}\)

21 tháng 5 2017

\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{2013.2015}+\frac{1}{2014.2016}\)

=\(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2014.2016}\right)\)

=\(\frac{1}{2}\left(1-\frac{1}{2015}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2016}\right)\)

=\(\frac{3}{4}-\left(\frac{1}{4030}+\frac{1}{4032}\right)\) < \(\frac{3}{4}\)

=> đpcm

10 tháng 4 2018

\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2013\cdot2015}\right)\)

\(=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{4056196}{2013\cdot2015}\)

\(=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2014\cdot2014\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2013\cdot2015\right)}\)

\(=\frac{\left(2\cdot3\cdot4\cdot...\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2014\right)}{\left(1\cdot2\cdot3\cdot...\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2015\right)}\)

\(=\frac{2014\cdot2}{1\cdot2015}\)

\(=\frac{4028}{2015}\)

11 tháng 12 2015

A = 1.3 + 2.4 + 3.5 + 4.6 + 5.7 + .. + 2013.2015 = [1.3 + 3.5+..+2013.2015] + [2.4 + 4.6 + .. + 2012.2014] = X + Y

X = 1.3 + 3.5 + 5.7 + .. + 2013.2015

X.6 = 1.3.﴾5 ‐ ﴾‐1﴿﴿ + 3.5.﴾7 ‐ 1﴿ + 5.7.﴾9‐3﴿ + 7.9.﴾11‐5﴿ + .. + 2011.2013.﴾2015‐2009﴿ + 2013.2015.﴾2017‐2011﴿

= ‐﴾‐1﴿.1.3 + 1.3.5 + 3.5.7 ‐ 1.3.5 + 5.7.9 ‐ 3.5.7 + .... = 1.3 + 2013.2015.2017

=> X = 1/6*﴾3 + 2013.2015.2017﴿ = 1363557553

tương tự Y = 2.4 + 4.6 + .. + 2012.2014

Y.6 = 2.4.6 + 4.6.﴾8‐2﴿ +... + 2012.2014.﴾2016‐2010﴿ = 2012.2014.2016

=> Y = 2012.2014.2016/6 = 1361528448

=> A = X + Y = 2725086001 

lâu ko làm nên  sai  hay đúng thì mình ko biết nữa

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

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