tính giá trị A =\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2010.2011.2012}\)
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2 tháng 8 2017
S=1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 +...+ 1/2010.2011 - 1/2011.2012
S=1/1.2 - 1/2011.2012<1/2
=>S<P
Q
14 tháng 3 2016
\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{11.12}\)
\(=\frac{65}{132}\)
18 tháng 2 2016
2A=2(1/1.2.3+1/2.3.4+...+1/98.99.100)
2A=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-...+1/98.99-1/99.100
2A=1/1.2-1/99.100
2A=4949/9900
A=4949/9900:2
A=4949/19800
Vậy A=4949/198000
19 tháng 5 2016
Giải:
Ta có nhận xét:
\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{3-1}{1.2.3}=\frac{2}{1.2.3}\)
\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{4-2}{2.3.4}=\frac{2}{2.3.4}\)
=>\(\frac{1}{1.2.3}=\frac{1}{3}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)
\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)
Do đó M=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}-\frac{1}{11.12}\right)\)
=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)=\frac{1}{2}-\frac{1}{11.12}\)
=\(\frac{1}{2}.\frac{65}{132}=\frac{65}{124}\)
Vậy M=65/124
TT
4 tháng 3 2020
Ta có : \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(\Leftrightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{18.19}-\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{19.20}=\frac{189}{380}\)
\(\Rightarrow B=\frac{189}{760}\)
4 tháng 3 2020
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{380}\right)\)
\(=\frac{1}{2}.\frac{189}{380}=\frac{189}{760}\)
D
3
14 tháng 3 2017
=1+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{2}\) -\(\frac{1}{3}\) -\(\frac{1}{4}\)+\(\frac{1}{3}\) - \(\frac{1}{4}\)-\(\frac{1}{5}\)+.....+\(\frac{1}{99}\)-\(\frac{1}{100}\)-\(\frac{1}{101}\)
=1+\(\frac{1}{101}\)
=\(\frac{102}{101}\)
14 tháng 3 2017
1/1.2.3 = 1/2 .[1/1.2 - 1 / 2.3]
1/2.3.4 = 1/2[ 1/2- 1/3 ]
...................
1/99.100.101 = 1/2[ 1/99. 100 - 1/100.101]
=> A= 1/2 [ 1/1.2- 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/ 4.5 +.........+ 1/99 .100 - 1/100. 101]
A = 1/2 . [1/1.2 -1/100 .101]
A= 1/2 . 5049 /10100 = 5049 / 20200.
Mình nghĩ là vậy đó.
1 tháng 3 2017
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{98.99.100}=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{k}=\frac{1}{2}\Rightarrow k=2\)
NT
1
MS
23 tháng 4 2018
\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2010.2011.2012}\)
\(\Rightarrow\frac{1}{4}A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4\left(5-1\right)}+\frac{1}{3.4.5\left(6-2\right)}+...+\frac{1}{2010.2011.2012.\left(2013-2009\right)}\)
\(\Rightarrow\frac{1}{4}A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}-\frac{1}{1.2.3.4}+\frac{1}{3.4.5.6}-\frac{1}{2.3.4.5}+...+\frac{1}{2010.2011.2012.2013}-\frac{1}{2009.2010.2011.2012}\)
\(\Rightarrow\frac{1}{4}A=\frac{1}{2010.2011.2012.2013}\)
\(\Rightarrow A=\frac{4}{2010.2011.2012.2013}\)
\(\Rightarrow A=\frac{1}{2010.2011.503.2013}\)