C=1x2x3+2x3x4+...+48x49x50
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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
TN
2
TM
2 tháng 3 2023
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 :
\(S=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{18\times19}-\dfrac{1}{19\times20}\right)\)
\(=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{19\times20}\right)\)
\(=\dfrac{1}{2}\times\dfrac{1}{1\times2}-\dfrac{1}{2}\times\dfrac{1}{19\times20}\)
\(=\dfrac{1}{4}-\dfrac{1}{760}=\dfrac{189}{760}\)
GN
3
18 tháng 3 2018
\(A=1\cdot2\cdot3+2\cdot3\cdot4+...+7\cdot8\cdot9+8\cdot9\cdot10\)
\(4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+7\cdot8\cdot9\cdot4+8\cdot9\cdot10\cdot4\)
\(4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+7\cdot8\cdot9\cdot\left(10-6\right)+8\cdot9\cdot10\cdot\left(11-7\right)\)
\(4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+7\cdot8\cdot9\cdot10-6\cdot7\cdot8\cdot9+8\cdot9\cdot10\cdot11-7\cdot8\cdot9\cdot10\)
\(4A=8\cdot9\cdot10\cdot11\)
\(A=\frac{8\cdot9\cdot10\cdot11}{4}=1980\)
VK
5
5 tháng 8 2016
\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{6.7.8}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{6.7}-\frac{1}{7.8}\)
\(=\frac{1}{1.2}-\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{56}\)
\(=\frac{28}{56}-\frac{1}{56}=\frac{27}{56}\)
Dấu . là nhân nha
5 tháng 8 2016
\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)
\(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\)
.......................................
\(\frac{2}{6.7.8}=\frac{1}{6.7}-\frac{1}{7.8}\)
S= \(\frac{1}{1.2}-\frac{1}{7.8}=\frac{27}{56}\)
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
\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{31\times32}\right)\)
\(=\frac{1}{2}\times\frac{990}{1984}\)
\(=\frac{990}{3968}=\frac{495}{1984}\)
13 tháng 8 2016
\(A=\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{36\times37\times38}+\frac{1}{37\times38\times39}\)
\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+...+\frac{2}{36\times37\times38}+\frac{2}{37\times38\times39}\)
\(2A=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{37\times38}-\frac{1}{38\times39}\)
\(2A=\frac{1}{1\times2}-\frac{1}{38\times39}\)
\(2A=\frac{741}{1482}-\frac{1}{1482}\)
\(2A=\frac{370}{741}\)
\(A=\frac{370}{741}:2=\frac{185}{741}\)
ND
4
17 tháng 8 2018
A =1x2x3 + 2x3x4 +3x4x5+....+ 2010 x2011 x 2012
4A =1x2x3x4 + 2x3x4x4 +3x4x5x4+....+ 2010 x2011 x 2012x4
4A =1x2x3x4 + 2x3x4x(5+1) +3x4x5x(6-2)+....+ 2010 x2011 x 2012x(2013-2009)
4A =1x2x3x4 + 2x3x4x5-1x2x3x4+3x4x5x6-2x3x4x5+....+ 2010 x2011 x 2012x2013-2009x2010x2011x2012
4A = 2010 x2011 x 2012x2013
A = \(\frac{2010\times2011\times2012\times2013}{4}\)
\(C=1.2.3+2.3.4+........+48.49.50\)
\(\Rightarrow4C=1.2.3.4+2.3.4.4+........+48.49.50.4\)
\(=1.2.3.4+2.3.4.\left(5-1\right)+.........+48.49.50.\left(51-47\right)\)
\(=1.2.3.4+2.3.4.5-1.2.3.4+........+48.49.50.51-47.48.49.50\)
\(=48.49.50.51\)
\(\Rightarrow C=\frac{48.49.50.51}{4}=1499400\)
Ta có C = 1 x 2 x 3 + 2 x 3 x 4 + ... + 48 x 49 x 50
=> 4C = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 4 + .... + 48 x 49 x 50 x 4
4C = 1 x 2 x 3 x 4 + 2 x 3 x 4 x (5 - 1)+ ... + 48 x 49 x 50 x (51 - 47)
4C = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + .... + 48 x 49 x 50 x 51 - 47 x 48 x 49 x 50
4C = 48 x 49 x 50 x 51
4C = 5997600
C = 5997600 : 4
C = 1499400
Vậy C = 1499400