\(M=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{100\cdot101\cdot102}\\ M=\frac{1}{2}\cdot\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{100\cdot101\cdot102}\right)\\ M=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{100\cdot101}-\frac{1}{101\cdot102}\right)\\ M=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{101\cdot102}\right)\\ M=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{10302}\right)\\ M=\frac{1}{2}\cdot\left(\frac{5151}{10302}-\frac{1}{10302}\right)\\ M=\frac{1}{2}\cdot\frac{25}{51}\\ M=\frac{25}{102}\\ \Rightarrow M< 1\)

Vậy M < 1

M<1 ok?

Ta có: M=\(\frac{1}{1.2.3}\) +\(\frac{1}{2.3.4}\) +\(\frac{1}{3.4.5}\) +...+\(\frac{1}{100.101.102}\) 

         M=2.(\(\frac{1}{1.2.3}\) +\(\frac{1}{2.3.4}\) +\(\frac{1}{3.4.5}\) +...+\(\frac{1}{100.101.102}\) ).\(\frac{1}{2}\)

          M=(\(\frac{2}{1.2.3}\) +\(\frac{2}{2.3.4}\) +\(\frac{2}{3.4.5}\) +...+\(\frac{2}{100.101.102}\) ).\(\frac{1}{2}\)

          M=(\(\frac{1}{1.2}\) -\(\frac{1}{2.3}\) +\(\frac{1}{2.3}\) -\(\frac{1}{3.4}\) +\(\frac{1}{3.4}\) -\(\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\) ).\(\frac{1}{2}\)

          M=( \(\frac{1}{1.2}-\frac{1}{101.102}\)).\(\frac{1}{2}\)

          Mà \(\frac{1}{1.2}-\frac{1}{101.102}<1\)

         Và \(\frac{1}{2}<1\) 

        \(=>\)  (\(\frac{1}{1.2}-\frac{1}{101.102}\) ) .\(\frac{1}{2}\) \(<1\)

        \(=>\) M <1

Ta có : 

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)

\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2015.2016}\)

\(\Rightarrow A=\left(\frac{1}{2}-\frac{1}{2015.2016}\right):2\)

\(\Rightarrow A=\frac{1}{4}-\frac{1}{2015.2016}\)

\(\Rightarrow A< \frac{1}{4}\)

Vậy A < \(\frac{1}{4}\)

_Chúc bạn học tốt_

Ta có:

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{2014+2015+2016}\)

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+.....+\frac{2}{2014.2015.2016}\)

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)

\(2A=\frac{1}{1.2}-\frac{1}{2015.2016}\)

\(\Rightarrow2A< \frac{1}{1.2}=\frac{1}{2}\)

\(\Rightarrow A< \frac{1}{4}\)

Vậy .... 

 A=1/1.2.3+1/2.3.4+1/3.4.5+...+1/2014.2015.2016

A=1/2.(1/2.3+1/3.4+1/4.5+...+1/2014.2015+1/2015.2016)

A=1/2.(1/2.3-1/2015.2016)

A=1/2.(1/2-1/2015.2016)

A=1/4-A1/2.2015.2016<1/4

suy ra A<1/4

\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2013.2014.2015}\)

\(S=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2014.2015}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4058210}\right)\)

\(S=\frac{1}{2}.\left(\frac{2029105}{4058210}-\frac{1}{4058210}\right)\)

\(S=\frac{1}{2}.\frac{2029104}{4058210}\)

\(S=\frac{1014552}{4058210}\)

Chúc bạn học tốt !!! 

Công thức : 

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\left(\frac{3}{6}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)

có cái gì sai sai đúng ko bạn 

\(2M=2\cdot\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+.....+\frac{1}{10\cdot11\cdot12}\right)\)

\(2M=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+.....+\frac{2}{10\cdot11\cdot12}\)

\(2M=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+.....+\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\)

\(2M=\frac{1}{1\cdot2}-\frac{1}{11\cdot12}\)

\(2M=\frac{1}{2}-\frac{1}{132}\)

\(2M=\frac{66}{132}-\frac{1}{132}\)

\(2M=\frac{65}{132}\)

\(M=\frac{65}{132}:2\)

\(M=\frac{65}{264}\)

giong nhu dap an minh viet khi nay do 

nho k cho minh voi nha

A=1/2(2/1.2.3+2/2.3.4+...+2/2014.2015.2016)~A=1/2(1/1.2-1/2.3+1/2.3-1/3.4+...+1/2014.2015-1/2015.2016)~~A=1/2(1/1.2-1/2015.2016)~A=1/2(1/2-1/4062240)~A=1/2.2031119/4062240~A=203119/8124480.                     Dấu/= dấu gạch ps còn ~ là dấu xuống dòng.   Còn bài này thì ko biết dung hay sai nua

L mi này đụ mẹ