\(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)

=1/2-1/380

=190/380-1/380

=189/380

Gọi biểu thức trên là S. Ta có :

\(S=\dfrac{1}{1\times2\times3}+\dfrac{1}{2\times3\times4}+\dfrac{1}{3\times4\times5}+...+\dfrac{1}{18\times19\times20}\)

\(=\dfrac{1}{2}\times\left(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{18\times19\times20}\right)\)

Trước tiên, ta áp dụng : \(\dfrac{2}{a\left(a+1\right)\left(a+2\right)}=\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}\)

Ta sẽ có : 

\(S=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{18\times19}-\dfrac{1}{19\times20}\right)\)

\(=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{19\times20}\right)\)

\(=\dfrac{1}{2}\times\dfrac{1}{1\times2}-\dfrac{1}{2}\times\dfrac{1}{19\times20}\)

\(=\dfrac{1}{4}-\dfrac{1}{760}=\dfrac{189}{760}\)

`=1/2(1/1×2 - 1/2×3 + 1/2×3 - 1/3×4 + 1/3×4 - 1/4×5 + ... + 1/18×19 - 1/19×20)`
`=1/2(1/2 - 1/19×20)`
`=1/2×189/380 `
`=189/760`

Để ngày mai mik hỏi cô giáo jup cho

\(A=1\cdot2\cdot3+2\cdot3\cdot4+...+7\cdot8\cdot9+8\cdot9\cdot10\)

\(4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+7\cdot8\cdot9\cdot4+8\cdot9\cdot10\cdot4\)

\(4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+7\cdot8\cdot9\cdot\left(10-6\right)+8\cdot9\cdot10\cdot\left(11-7\right)\)

\(4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+7\cdot8\cdot9\cdot10-6\cdot7\cdot8\cdot9+8\cdot9\cdot10\cdot11-7\cdot8\cdot9\cdot10\)

\(4A=8\cdot9\cdot10\cdot11\)

\(A=\frac{8\cdot9\cdot10\cdot11}{4}=1980\)

A=1·2·3+2·3·4+...+7·8·9+8·9·10

4A=1·2·3·4+2·3·4·4+...+7·8·9·4+8·9·10·4

4A=1·2·3·4+2·3·4·(5−1)+...+7·8·9·(10−6)+8·9·10·(11−7)

4A=1·2·3·4+2·3·4·5−1·2·3·4+...+7·8·9·10−6·7·8·9+8·9·10·11−7·8·9·10

4A=8·9·10·11

Đặt C = \(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+...+\frac{1}{30\times31\times32}\)

\(2C=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+...+\frac{2}{30\times31\times32}\)

        \(=\left(\frac{1}{1\times2}-\frac{1}{2\times3}\right)+\left(\frac{1}{2\times3}-\frac{1}{3\times4}\right)+...+\left(\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)

          \(=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\)

            \(=\frac{1}{1\times2}-\frac{1}{31\times32}\)

              \(=\frac{1}{2}-\frac{1}{992}=\frac{495}{992}\)

\(\Rightarrow C=\frac{495}{992}\div2=\frac{495}{1984}\)

Vậy ...

\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+.....+\frac{1}{30\times31\times32}\)

\(=\frac{1}{2}\times\left(\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+.....+\frac{2}{30\times31\times32}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{30.31}-\frac{1}{31.32}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{31.32}\right)=\frac{1}{2}.\frac{990}{1984}=\frac{990}{3968}\)

=1+1/32

=33/32

\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{6.7.8}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{6.7}-\frac{1}{7.8}\)

\(=\frac{1}{1.2}-\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{56}\)

\(=\frac{28}{56}-\frac{1}{56}=\frac{27}{56}\)

Dấu . là nhân nha

\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)

\(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\)

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

\(\frac{2}{6.7.8}=\frac{1}{6.7}-\frac{1}{7.8}\)

S= \(\frac{1}{1.2}-\frac{1}{7.8}=\frac{27}{56}\)

\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{6.7.8}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{6.7}-\frac{1}{7.8}\)

\(=\frac{1}{1.2}-\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{56}\)

\(=\frac{28}{56}-\frac{1}{56}=\frac{27}{56}\)

ta có:
4s=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+.........+k(k+1)(k+2)((k+3)-(k-1))
4s=1.2.3.4-1.2.3.0+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+........+k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2)
4s=k(k+1)(k+2)(k+3)
ta biết rằng tích 4 số tự nhiên liên tiếp khi cộng thêm 1 luôn là 1 số chính phương
=>4s+1 là 1 số chính phương

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

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{49.50}-\frac{1}{50.51}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{50.51}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2550}\right)\)

\(=\frac{1}{2}.\left(\frac{1275}{2550}-\frac{1}{2550}\right)\)

\(=\frac{1}{2}.\frac{1274}{2550}\)

\(=\frac{637}{2550}\)

Lưu ý : Dấu \("."\)là dấu \("\)x  \("\)

( dấu nhân ) 

Chúc bạn học giỏi !!! 

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 ( dễ hiểu ) 

\(\frac{637}{2550}\)

Chúc bạn học tốt

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

S = \(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+....+\frac{2015-2013}{2013.2014.2015}\right)\)

S = \(\frac{1}{2}.\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+...+\frac{2015}{2013.2014.2015}-\frac{2013}{2013.2014.2015}\right)\)

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

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

S = \(\frac{1014552}{4058210}\)