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S=1/1x2x3+1/2x3x4+1/3x4x5+.............+1/2013x2014x2015
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TP
0
MT
1
31 tháng 1 2021
Đặt \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)
Ta có: \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)
\(\Leftrightarrow2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{98\cdot99\cdot100}\)
\(\Leftrightarrow2A=-\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}-\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}-\dfrac{1}{4\cdot5}+...-\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)
\(\Leftrightarrow2A=-\dfrac{1}{2}+\dfrac{1}{99\cdot100}\)
\(\Leftrightarrow2A=\dfrac{-1}{2}+\dfrac{1}{9900}\)
\(\Leftrightarrow2A=\dfrac{-4950}{9900}+\dfrac{1}{9900}=\dfrac{-4949}{9900}\)
hay \(A=\dfrac{-4949}{19800}\)
TT
3
A
17 tháng 11 2018
Đặt \(A=\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\)
\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+...+\frac{2}{30\times31\times32}\)
\(=\left(\frac{1}{1\times2}-\frac{1}{2\times3}\right)+\left(\frac{1}{2\times3}-\frac{1}{3\times4}\right)+\left(\frac{1}{3\times4}-\frac{1}{4\times5}\right)+...+\left(\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)
\(=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\)
\(=\frac{1}{1\times2}-\frac{1}{31\times32}\)
\(=\frac{1}{2}-\frac{1}{992}\)
\(=\frac{495}{992}\)
\(\Rightarrow A=\frac{495}{992}\div2=\frac{495}{1984}\)
D
17 tháng 11 2018
\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{31\times32}\right)\)
\(=\frac{1}{2}\times\frac{990}{1984}\)
\(=\frac{990}{3968}=\frac{495}{1984}\)
LD
3
11 tháng 11 2018
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 ...
11 tháng 11 2018
\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+.....+\frac{1}{30\times31\times32}\)
\(=\frac{1}{2}\times\left(\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+.....+\frac{2}{30\times31\times32}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{30.31}-\frac{1}{31.32}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{31.32}\right)=\frac{1}{2}.\frac{990}{1984}=\frac{990}{3968}\)
TT
1
7 tháng 4 2015
=1/2-1/3-1/4+1/3-1/4-1/5+1/5-1/6-1/7+...+1/35-1/36-1/37
giao hoán, kết hợp là ra nha
AD
2
2
17 tháng 5 2022
`A=1/[1xx2xx3]+1/[2xx3xx4]+1/[3xx4xx5]+....+1/[98xx99xx100]`
`A=1/2xx(2/[1xx2xx3]+2/[2xx3xx4]+2/[3xx4xx5]+....+2/[98xx99xx100])`
`A=1/2xx(1/[1xx2]-1/[2xx3]+1/[2xx3]-1/[3xx4]+1/[3xx4]-1/[4xx5]+....+1/[98xx99]-1/[99xx100])`
`A=1/2xx(1/[1xx2]-1/[99xx100])`
`A=1/2xx(1/2-1/9900)`
`A=1/2xx(4950/9900-1/9900)`
`A=1/2xx4949/9900`
`A=4949/19800`
17 tháng 5 2022
\(A=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}\)
\(A=\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right):2\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{12}-\dfrac{1}{20}+...+\dfrac{1}{9702}-\dfrac{1}{990}\right):2\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{990}\right):2\)
\(A=\dfrac{4949}{9900}:2\)
\(A=\dfrac{4949}{19800}\)
TN
1
3 tháng 3 2023
\(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)
=1/2-1/380
=190/380-1/380
=189/380
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 = \(\frac{1014552}{4058210}\)