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10 tháng 5 2016

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2015.2016}\)

\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2015}-\frac{1}{2016}\)

\(S=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)

10 tháng 5 2016

làm r sao cứ đăng hoài vậy?

4 tháng 8 2015

a) \(A=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{10200}\)

\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)

\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\)

\(2A=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{8}\right)+...+\left(\frac{1}{100}-\frac{1}{102}\right)\)

\(2A=\frac{1}{2}-\frac{1}{102}\)

\(2A=\frac{25}{51}\)

\(A=\frac{25}{51}:2\)

\(A=\frac{25}{102}\)

Vậy \(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{10200}=\frac{25}{102}\)

 

 

b) \(B=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2015.2016}\)

\(B=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right)\)

\(B=3.\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{2015}-\frac{1}{2016}\right)\right]\)

\(B=3.\left(\frac{1}{1}-\frac{1}{2016}\right)\)

\(B=3.\frac{2015}{2016}\)

\(B=\frac{2015}{672}\)

Vậy \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2015.2016}=\frac{2015}{672}\)

 

 

 

 

 

 

13 tháng 2 2022

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1008}\right)\)

\(A=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2015}+\frac{1}{2016}\)

\(\Rightarrow B-A=\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)-\left(\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}\right)\)

\(\Rightarrow B-A=\frac{1}{1008}\)

1 tháng 5 2016

B = 1/1.2 + 1/3.4 +..+1/2015.2016
B = 1-1/2 + 1/3 - 1/4 +...+ 1/2015 - 1/2016
B = 1+ 1/2 + 1/3 +..+1/2015 + 1/2016 - 2( 1/2 + 1/4 + ..1/2016)
B = 1/1009 + 1/1010 +.. + 1/2016 ( dpcm)

10 tháng 5 2017

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(=1-\frac{1}{2017}=\frac{2016}{2017}\)

16 tháng 4 2016

S=2-1/1.2  .  3-2/2.3............2016-2015/2015.2016

  =1/1 - 1/2 + 1/2 - 1/3+........+1/2015 - 1/2016

  =1/1 - 1/2016

   =2015/2016

15 tháng 5 2017

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)

\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+......+\left(\frac{1}{2016}-\frac{1}{2017}\right)\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2016}-\frac{1}{2017}\)

\(A=\frac{1}{1}-\frac{1}{2017}\)

\(A=\frac{2016}{2017}\)

15 tháng 5 2017

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2016}-\frac{1}{2017}\)

\(\Rightarrow A=1-\frac{1}{2017}\)

\(\Rightarrow A=\frac{2016}{2017}\)