\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{20\cdot21\cdot22}=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{20\cdot21\cdot22}\right)\)

                                                                     \(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{20\cdot21}-\frac{1}{21\cdot22}\right)\)

                                                                     \(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{21\cdot22}\right)\)

                                                                     \(=\frac{1}{2}\left(\frac{231}{462}-\frac{1}{462}\right)=\frac{1}{2}\cdot\frac{230}{462}=\frac{1}{2}\cdot\frac{115}{231}=\frac{115}{462}\)

\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{2018\cdot2019\cdot2020}\)

\(=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{2018\cdot2019\cdot2020}\right]\)

\(=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\right]\)

Đến đây tự tính được rồi:v

   Đặt tổng trên là A

Ta có:

\(2A=2\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{2018\cdot2019\cdot2020}\right)\)

\(=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2018\cdot2019\cdot2020}\)

\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\)

\(=\frac{1}{2}-\frac{1}{2019\cdot2020}\)

\(A=\left(\frac{1}{2}-\frac{1}{2019\cdot2020}\right)\div2\)

        *Làm tiếp*

                                          \(#Louis\)

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2018.2019.2020}\)

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2018.2019.2020}\)

Thấy : \(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

Áp dụng : 

+ Với n = 1 có : \(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)

+ Với n = 2 có : \(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\)

....

+ Với n = 2019 có : \(\frac{2}{2018.2019.2020}=\frac{1}{2018.2019}-\frac{1}{2019.2020}\)

Cộng từng vế có :

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2018.2019}-\frac{1}{2019.2020}\)

\(2A=\frac{1}{2}-\frac{1}{2019.2020}\)

   \(A=\left(\frac{1}{2}-\frac{1}{2019.2020}\right):2\)

   \(A=\left(\frac{1}{2}-\frac{1}{2019.2020}\right).\frac{1}{2}\)

   \(A=\frac{1}{4}-\frac{1}{2019.2020.2}\)

   Đến đây tắc dồi >: 

Mình không chép đề bài nhé :
Gọi biểu thức là A :
Ta có : 2A=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\)
= \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
=\(\frac{1}{1.2}-\frac{1}{49.50}\)( Rút gọn đi ta được cái này )
=1/2 - 1/2450
Vậy A = (1/2 - 1/2450):2
 

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)

\(=\frac{1}{2}.\left(\frac{741}{1482}-\frac{1}{1482}\right)\)

\(=\frac{1}{2}.\frac{740}{1482}\)

\(=\frac{185}{741}\)

Chúc bạn học tốt !!! 

Đặt 1/1.2.3 + 1/2.3.4 + ...+ 1/37.38.39 = A

Ta có : 2A = 2/1.2.3 + 2/2.3.4 +...+ 2/37.38.39

         2A = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ...+ 1/37.38 - 1/38.39

         2A = 1/1.2 - 1/38.39

         2A = 740/1482 = 370/741

           A= 370/741 . 1/2 =........

* Công thức : 

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

( VD cho dễ hiểu ) 

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}.\frac{370}{741}\)

\(=\frac{185}{741}\)

=1+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{2}\) -\(\frac{1}{3}\) -\(\frac{1}{4}\)+\(\frac{1}{3}\) - \(\frac{1}{4}\)-\(\frac{1}{5}\)+.....+\(\frac{1}{99}\)-\(\frac{1}{100}\)-\(\frac{1}{101}\)

=1+\(\frac{1}{101}\)

=\(\frac{102}{101}\)

1/1.2.3 = 1/2 .[1/1.2 - 1 / 2.3]

1/2.3.4 = 1/2[ 1/2- 1/3 ] 

...................

1/99.100.101 = 1/2[ 1/99. 100 - 1/100.101]

=> A= 1/2 [ 1/1.2- 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/ 4.5 +.........+ 1/99 .100 - 1/100. 101]

A = 1/2 . [1/1.2 -1/100 .101]

A= 1/2 . 5049 /10100 = 5049 / 20200.

Mình nghĩ là vậy đó.

a=100/101

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)

= \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)

= \(\frac{1}{1.2}-\frac{1}{49.50}\)

= \(\frac{1}{2}-\frac{1}{2450}\)

= \(\frac{612}{1225}\)

đặt

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)

\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{48.49.50}\)

\(\Rightarrow\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)

\(\Rightarrow\frac{1}{1.2}-\frac{1}{49.50}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2450}=\frac{621}{1225}\)

\(\Rightarrow A=\frac{306}{1225}\)

=>2A  = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\)

=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49-50}\)

= \(\frac{1}{1.2}-\frac{1}{49.50}\)= \(\frac{1}{2}-\frac{1}{2450}=\frac{612}{1225}\)

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)

\(2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\)

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(2A=\frac{1}{2}-\frac{1}{99.100}=\frac{49}{99.100}\Rightarrow A=\frac{49}{2.99.100}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2450}\right)\)

\(=\frac{1}{2}.\frac{612}{1225}\\ =\frac{306}{1225}\)(mà đây là toán 6 mà :V)