1x2x3+4x5x6+...+100x101x 102
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VK
6
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= \(\frac{1}{1.2}-\frac{1}{7.8}=\frac{27}{56}\)
TL
0
F
17 tháng 8 2017
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
TP
17 tháng 8 2017
Ta có:
\(A=1.2+2.3+3.4+...+99.100\)
\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(\Leftrightarrow3A=99.100.101\Leftrightarrow A=\frac{99.100.101}{3}=333300\)
\(B=1.2.3+2.3.4+4.5.6+...+98.99.100\)
\(\Rightarrow4B=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+4.5.6.\left(7-3\right)+...+98.99.100.\left(101-97\right)\)
\(\Rightarrow4B=1.2.3.4+2.3.4.5-1.2.3.4+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100\)
\(\Leftrightarrow4B=98.99.100.101\Leftrightarrow B=\frac{98.99.100.101}{4}=24497550\)
VN
18 tháng 6 2017
ta có : B = 1.2.3 + 2.3.4 + 3.4.5 + ...... + 2016.2017.2018
4B = 1.2.3.4 - 1.2.3.4 + 2.3.4.5 - 2.3.4.5 + ...... + 2016.2017.2018.2019
4B = 2016.2017.2018.2019
vậy B = 2016.2017.2018.2019/4
18 tháng 6 2017
Ta có : B = 1.2.3 + 2.3.4 + ...... + 2016.2017.2018
=> 4B = 1.2.3.4 - 1.2.3.4 + 2.3.4.5 - 2.3.4.5 + ...... + 2016.2017.2018.2019
=> 4B = 2016.2017.2018.2019
=> B = 2016.2017.2018.2019/4
TH
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 4
A = \(\dfrac{2}{1\times3\times5}\) + \(\dfrac{2}{3\times5\times7}\) + \(\dfrac{2}{5\times7\times9}\)+\(\dfrac{2}{7\times9\times11}\)
A = \(\dfrac{1}{2}\) x (\(\dfrac{4}{1\times3\times5}\) + \(\dfrac{4}{3\times5\times7}\) + \(\dfrac{4}{5\times7\times9}\) + \(\dfrac{4}{7\times9\times11}\))
A = \(\dfrac{1}{2}\)x (\(\dfrac{1}{1\times3}\)-\(\dfrac{1}{3\times5}\)+\(\dfrac{1}{3\times5}\)-\(\dfrac{1}{5\times7}\)+\(\dfrac{1}{5\times7}\)-\(\dfrac{1}{7\times9}\)+\(\dfrac{1}{7\times9}\)-\(\dfrac{1}{9\times11}\))
A = \(\dfrac{1}{2}\)x (\(\dfrac{1}{1\times3}\) - \(\dfrac{1}{9\times11}\))
A = \(\dfrac{1}{2}\) x (\(\dfrac{1}{3}-\dfrac{1}{99}\))
A = \(\dfrac{1}{2}\times\) \(\dfrac{32}{99}\)
A = \(\dfrac{16}{99}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 4
B = \(\dfrac{1}{1\times2\times3}\) + \(\dfrac{1}{2\times3\times4}\) + \(\dfrac{1}{3\times4\times5}\) + \(\dfrac{1}{4\times5\times6}\)
B = \(\dfrac{1}{2}\) x (\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+\dfrac{2}{4\times5\times6}\))
B = \(\dfrac{1}{2}\) x (\(\dfrac{1}{1\times2}\)-\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{2\times3}\)-\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{3\times4}\)-\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{4\times5}\)-\(\dfrac{1}{5\times6}\))
B = \(\dfrac{1}{2}\)x(\(\dfrac{1}{1\times2}\) - \(\dfrac{1}{5\times6}\))
B = \(\dfrac{1}{2}\)x (\(\dfrac{1}{2}-\dfrac{1}{30}\))
B = \(\dfrac{1}{2}\)x \(\dfrac{7}{15}\)
B = \(\dfrac{7}{30}\)
NT
0