\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{19.20.21}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{20.21}\right)\)

\(=\frac{1}{2}.\frac{209}{420}\)

\(=\frac{209}{840}\)

\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{19\cdot20\cdot21}\)

\(=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{19\cdot20\cdot21}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}-\frac{1}{20\cdot21}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot21}\right)\)

bn tự lm tp

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{19.20.21}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{19.20.21}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{19.20}-\frac{1}{20.21}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{20.21}\right)=\frac{1}{2}.\frac{209}{420}=\frac{209}{840}\)

A= \(\frac{1}{1.2.3}\)+ \(\frac{1}{2.3.4}\)+ ... + \(\frac{1}{19.20.21}\)< \(\frac{1}{4}\)

  = 1 - \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{2}\)-  \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ... + \(\frac{1}{19}-\frac{1}{20}-\frac{1}{21}\)

  = 1 - ( \(\frac{1}{2}-\frac{1}{3}\)+ \(\frac{1}{2}-\frac{1}{3}\)) + ... + ( \(\frac{1}{19}-\frac{1}{20}+\frac{1}{19}-\frac{1}{20}\))  - \(\frac{1}{21}\)

  = 1 - \(\frac{1}{21}\)

  =  \(\frac{20}{21}\)<  \(\frac{1}{4}\)

=> Đề bài có sai ko bạn?

\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right).x=5\)

\(\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{21-19}{19.20.21}\right).x=5\)

 \(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right).x=5\)

 \(\left(\frac{1}{1.2}-\frac{1}{20.21}\right).x=5\)

 \(\frac{209}{420}.x=5\)

\(\Rightarrow x=5\div\frac{209}{420}=\frac{2100}{209}\)

\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right).x=5\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{19.20.21}\right).2.x=5\)

\(\left(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{19.20}-\frac{1}{20.21}\right)\right).x.2=5\)

\(\left(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\right).x=5\div2\)

\(\left(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{20.21}\right)\right).x=2,5\)

\(\left(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{420}\right)\right).x=2,5\)

\(\left(\frac{1}{2}\times\frac{209}{420}\right)\times x=2,5\)

\(\frac{209}{840}\times x=2,5\)

\(x=2,5\div\frac{209}{840}=10\frac{10}{209}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{4.5.6}+....+\frac{1}{98.99.100}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}\)

\(=\frac{99}{100}\)

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

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(=\frac{1}{2}.\frac{4949}{9900}\)

\(=\frac{1}{19800}\)

Nhầm , kết quả bằng :

\(=\frac{4949}{19800}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)

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

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(=\frac{1}{2}.\frac{4949}{9900}\)

\(=\frac{4949}{19800}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{49.50.51}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.5}+...+\frac{1}{49.50}-\frac{1}{50.51}\)

\(=\frac{1}{2}-\frac{1}{50.51}\)

\(=\frac{1}{2}-\frac{1}{2550}=\frac{637}{1275}\)

Gọi A là tổng dãy phân số trên

Ta có :

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{49.50.51}\)

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{49.50.51}\)

Ta thấy:

\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3};\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4};...;\frac{2}{49.50.51}=\frac{2}{49.50}-\frac{2}{50.51}\text{​​}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{49.50}-\frac{1}{50.51}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{50.51}\)

\(\Rightarrow2A=\frac{1}{2}-\frac{1}{2550}\)

\(\Rightarrow2A=\frac{1275}{2550}-\frac{1}{2550}\)

\(\Rightarrow2A=\frac{637}{1275}\Rightarrow A=\frac{637}{1275}:2=\frac{637}{2550}\)

Vậy tổng dãy phân số trên là :\(\frac{637}{2550}\)

Chúc bạn học tốt !!! :D

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2004.2005.2006}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2004.2005}-\frac{1}{2005.2006}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2005.2006}\right)\)

\(=\frac{1}{4}-\frac{1}{2.2005.2006}\)

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

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

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

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 >: 

A=1/2 *(1/1*2-1/2*3+1/2*3-1/3*4+........+1/98*99-1/99*100)

=1/2*(1/2-1/99*100)

=1/2*(4950-1/9900)

=4950/19800

\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)

\(A=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right]\)

\(A=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right]\)

\(A=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{99\cdot100}\right]=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)

A=1/2.(2/1.2.3+2/2.3.4+...+2/98.99.100

=1/2.(1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100