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

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

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

làm r sao cứ đăng hoài vậy?

S=1/1-1//2+1/2-1/3+1/3-1/4+.......=1/2014-1/2015

S=1/1-1/2015

S=2015/2015-1/2015

S=2014/2015

\(=\frac{2015}{2016}\)

2015/2016 nhé bạn.

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

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

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

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

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

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

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

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

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

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

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

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

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

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)

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

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

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

Ta có \(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+\(\frac{1}{3\cdot4}\)+...+\(\frac{1}{2015\cdot2016}\)

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

     =1-\(\frac{1}{2016}\)

     =\(\frac{2015}{2016}\)(bạn cứ nhớ công thức là làm được)

Ta thấy: \(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2};\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3};....;\frac{1}{2015.2016}=\frac{1}{2015}-\frac{1}{2016}\)

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

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

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

Ta có công thức: \(\frac{1}{n\left(n+1\right)}\) = \(\frac{1}{n}\) - \(\frac{1}{n+1}\) 

Áp dụng công thức, ta giải:

\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + ... + \(\frac{1}{2015.2016}\) = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)

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