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

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

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

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

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

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

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

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

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

bn tự lm tp

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

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

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

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

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

\(\Rightarrow2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

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

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

\(\Rightarrow C=\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\div2\)

Giải :

\(\text{S}=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{998\cdot999\cdot1000}\)

\(\text{S}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{998}-\frac{1}{999}+\frac{1}{999}-\frac{1}{1000}\)

\(\text{S}=1-\frac{1}{1000}=\frac{999}{1000}\)

\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{998.999.1000}\)

  \(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.1000}\right)\)

  \(=\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{1000-998}{998.999.1000}\right)\)

 \(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{998.999}-\frac{1}{999.1000}\right)\)

 \(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{999.1000}\right)\)

 \(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{999000}\right)\)

 \(=\frac{1}{2}.\frac{499499}{999000}\)

 \(=\frac{499499}{1998000}\)

Study well ! >_<

Trả lời:

\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{998.999.1000}\)

\(S=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.1000}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{998.999}-\frac{1}{999.1000}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{999.1000}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{999000}\right)\)

\(S=\frac{1}{2}.\frac{499499}{999000}\)

\(S=\frac{499499}{1998000}\)

Đặt

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{37.38.39}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}=\frac{1}{1.2}-\frac{1}{38.39}=\frac{1}{2}-\frac{1}{1428}\Rightarrow A=\left(\frac{1}{2}-\frac{1}{1428}\right):2=\frac{713}{1428}.\frac{1}{2}\)

=>S=\(\frac{713}{1428}.\frac{1}{2}.1428+185.8=\frac{713}{2}+185.8=\frac{713}{2}+1480=356+\frac{1}{2}+1480=1836\frac{1}{2}\)

S=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)+\(\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)+..............+\(\frac{1}{2}\left(\frac{1}{37.38}-\frac{1}{38.39}\right)\).1428+185.8

S=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+........\frac{1}{37.38}-\frac{1}{38.39}\right).\)1428+185.8

S=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\).1428+185.8

S=\(\frac{185}{741}\).1428+1480

S=1836,518219

Ta có công thức:

\(\frac{a}{c.\left[c+1\right].\left[c+2\right]}=\frac{a}{2}\left[\frac{1}{c.\left[c+1\right]}-\frac{1}{\left[c+1\right].\left[c+2\right]}\right]\)

vậy

\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{11.12}-\frac{1}{12.13}\right]\)

\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{12.13}\right]\)

\(C=\frac{1}{2}.\frac{77}{156}=\frac{77}{312}\)

mình làm đầu tiên đó, 

Chúc bạn học tốt !

\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{11.12.13}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{11.12}-\frac{1}{12.13}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{156}\right)\)

\(=\frac{1}{2}\cdot\frac{77}{156}\)

\(=\frac{77}{312}\)

T/c:A=1/1*2*3+1/2*3*4+1/3*4*5+1/4*5*6+...+1/97*98*99+1/98*99*100

2A=2/1*2*3+2/2*3*4+2/3*4*5+2/4*5*6+...+2/97*98*99+1/98*99*100

2A=(1/1*2-1/2*3)+(1/2*3-1/3*4)+(1/3*4-1/4*5)+.....+(1/97*98-1/98*99)+(1/98*99-1/99*100)

2A=1/2+1/99*100

A=tự tính nha

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

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

A=-101/198/2

A=-101/396

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

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

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

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

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

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

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

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

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

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

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 >: