\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{10.11}-\frac{1}{11.12}\)

\(M=\frac{1}{2}-\frac{1}{11.12}=\frac{65}{132}\)

Ta có \(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)

Áp dụng:

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{10\cdot11\cdot12}\\ =\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{10\cdot11}-\dfrac{1}{11\cdot12}\\ =\dfrac{1}{2}-\dfrac{1}{11\cdot12}=\dfrac{1}{2}-\dfrac{1}{132}=\dfrac{65}{132}\)

Ta có \(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)

Áp dụng

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{10\cdot11\cdot12}\\ =\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{10\cdot11\cdot12}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+..+\dfrac{1}{10\cdot11}-\dfrac{1}{11\cdot12}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{11\cdot12}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{132}\right)=\dfrac{1}{2}\cdot\dfrac{65}{132}=\dfrac{65}{264}\)

Ta có: \(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)

Đặt \(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)

   \(\Leftrightarrow2A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\)

             \(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\)

             \(=\dfrac{1}{2}-\dfrac{1}{11.12}=\dfrac{65}{132}\)

  \(\Rightarrow A=\dfrac{65}{132}:2=\dfrac{65}{264}\)

\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)

\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)

Nhận xét rằng:
2/[(n - 1)n(n +1)] = 1/[(n-1).n] - 1/[n(n+1)]
Do đó
2M = 2/(1.2.3) + 2/(2.3.4) + 2/(3.4.5) + ... + 2(10.11.12)
= 1/(1.2) - 1/(2.3) + 1/(2.3) - 1/(3.4) + 1/(3.4) - 1/(4.5) + .... + 1/(10.11) - 1/(11.12)
= 1/(1.2) - 1/(11.12) = 65/132
=> M = 65/264

Ta có nhận xét: \(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{3-1}{1.2.3}=\dfrac{2}{1.2.3}\),

\(\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{4-2}{2.3.4}=\dfrac{2}{2.3.4};...\)

\(\Rightarrow\dfrac{1}{1.2.3}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)\);

\(\dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\);...

Do đó \(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{1.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)

/n.(n+1)(n+2)=1/n*(1/(n+1)-1/(n+2)) = 1/n(n+1) - 1/n(n+2) = 1/n - 1/(n+1) - 1/2n + 1/2(n+2)
. . . ...... ... . . . = 1/2n - 1/(n+1) + 1/2(n+2)
như vậy ta có
1/1.2.3 = 1/2 - 1/2 + 1/6
1/2.3.4 = 1/4 - 1/3 + 1/8
1/3.4.5 = 1/6 - 1/4 + 1/10
1/4.5.6 = 1/8 - 1/5 + 1/12
.........................................
1/(n-1)n(n+1)= 1/2(n-1) - 1/n + 1/2(n+1)
1/n.(n+1)(n+2)= 1/2n - 1/(n+1) + 1/2(n+2)
=> tong = 1/4 - 1/2(n+1) + 1/2(n+2)

\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.....+\frac{1}{10.11.12}\)

\(M=\frac{1}{2}-\frac{1}{11.12}\)

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

Ngắn gọn , xúc tích !!! :))

\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)

\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)

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

\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)

\(=\frac{1}{2}-\frac{1}{11.12}\)

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

Ta có nhận xét: 1/1.2 - 1/2.3 = 3-1/1.2.3 = 2/1.2.3          

                       1/2.3 - 1/3.4 = 4-2/2.3.4 = 2/2.3.4      

Suy ra:             1/1.2.3 = 1/2(1/1.2 - 1/2.3)    

                         1/2.3.4 = 1/2(1/2.3 -1/3.4)  

Do đó:              M = 1/2(1/1.2-1/2.3 + 1/2.3 -1/3.4 + ... + 1/10.11 -1/11.12)

                           = 1/2(1/1.2 - 1/11.12) = 1/2(1/2-11/12 )      

                           = 1/2.65/132 = 65/264

Phức tạp lắm

Bạn ơi nhưng 1/(1.2.3) không bằng 1/(1.2) -1/(2.3)?

\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)

\(=\frac{1}{2}-\frac{1}{11.12}\)

 \(=\frac{65}{132}\)

65/132

65/132

\(D=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{10\cdot11\cdot12}\)

\(D=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{10\cdot11\cdot12}\right)\)

\(D=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\right)\)

\(D=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{132}\right)=...\)

\(D=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)

\(D=\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{10.11.12}\right).\frac{1}{2}\)

\(D=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{10.11}-\frac{1}{11.12}\right).\frac{1}{2}\)

\(D=\left(\frac{1}{1.2}-\frac{1}{11.12}\right).\frac{1}{2}\)

\(D=\frac{65}{132}.\frac{1}{2}\)

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