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31 tháng 5 2017
A=1x2x3 + 2x3x4 +…+ 100x101x102
Nhân A với 4 ta có :
A x 4 = 1x2x3x4 + 2x3x4x 4 + 3x4x5x4 +…+100x101x102x4
A x 4 = 1x2x3x4 + 2x3x4x(5-1) + 3x4x5x(6-2) + ... + 100x101x102x(103 - 99)
A x 4 = 1x2x3x4 + 2x3x4x5 - 1x2x3x4 + 3x4x5x6 - 2x3x4x5 + ... + 100x101x102x103 - 99x100x1001x102
Sau khi cộng - trừ giản ước ta có : A x 4 = 100x101x102x103
A = 100 x101x102x103 : 4 = 26527650
VD
0
LQ
1
16 tháng 4 2018
\(A=\frac{1X2X3+2X4X6+4X8X12+8X16X24}{2X3X4+4X6X8+8X12X16+8X24X32}\)
\(A=\frac{1+1+1+1}{4+4+4+2}\)
\(A=\frac{4}{14}\)
CHÚC BẠN HỌC GIỎI !
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 = \(\frac{2010\times2011\times2012\times2013}{4}\)
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
\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{31\times32}\right)\)
\(=\frac{1}{2}\times\frac{990}{1984}\)
\(=\frac{990}{3968}=\frac{495}{1984}\)
TP
20 tháng 8 2017
Ta có:
\(F=1.2.3+2.3.4+...+98.99.100\)
\(\Rightarrow4F=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+....+98.99.100.\left(101-97\right)\)
\(\Rightarrow4F=1.2.3.4+2.3.4.5-1.2.3.4+...+98.99.100.101-97.98.99.100\)
\(\Rightarrow4F=98.99.100.101\Leftrightarrow F=\frac{98.99.100.101}{4}=24497550\)
TP
1
HT
21 tháng 9 2015
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 = \(\frac{1014552}{4058210}\)
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
A=1x2x3 + 2x3x4 +…+ 100x101x102
Nhân A với 4 ta có :
A x 4 = 1x2x3x4 + 2x3x4x 4 + 3x4x5x4 +…+100x101x102x4
A x 4 = 1x2x3x4 + 2x3x4x(5-1) + 3x4x5x(6-2) + ... + 100x101x102x(103 - 99)
A x 4 = 1x2x3x4 + 2x3x4x5 - 1x2x3x4 + 3x4x5x6 - 2x3x4x5 + ... + 100x101x102x103 - 99x100x1001x102
Sau khi cộng - trừ giản ước ta có : A x 4 = 100x101x102x103
A = 100 x101x102x103 : 4 = 26527650
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