C= 1 phần 1.2.3+1 phần 2.3.4+1 phần 3.4.5+..........+1 phần 98.99.100
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HT
2 tháng 8 2015
1/1.2.3 + 1/2.3.4 +....+1/98.99.100
= 1/2 . (3-1/1.2.3 + 4-2/2.3.4 +....+ 100-98/98.99.100)
= 1/2 . (3/1.2.3 -1/1.2.3 + 4/2.3.4 - 2/2.3.4 +.......+ 100/98.99.100 - 98/98.99.100)
= 1/2 . (1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 +......+ 1/98.99 - 1/99.100)
= 1/2 . (1/2 - 1/9900)
= 1/2 . 4949/9900
= 4949/19800
EC
1
HV
1
L
2
TN
17 tháng 9 2016
B=1/1.2.3+1/2.3.4+1/3.4.5+............+1/98.99.100
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}\cdot\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
17 tháng 9 2016
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(B=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(B=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(B=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}\)
CV
3
8 tháng 6 2017
\(\left(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\right)y=\dfrac{49}{200}\)
\(\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}+...+\dfrac{1}{98\cdot99}-\dfrac{1}{99\cdot100}\right)y=\dfrac{49}{200}\)
\(\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{99\cdot100}\right)y=\dfrac{49}{200}\)
\(\left(\dfrac{1}{4}-\dfrac{1}{19800}\right)y=\dfrac{49}{200}\)
\(\left(\dfrac{4950}{19800}-\dfrac{1}{19800}\right)y=\dfrac{49}{200}\)
\(\dfrac{4949}{19800}y=\dfrac{49}{200}\)
\(y=\dfrac{49}{200}:\dfrac{4949}{19800}\)
\(y=\dfrac{99}{101}\)
Vậy \(y=\dfrac{99}{101}\).
NT
8 tháng 6 2017
\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\right)y=\dfrac{49}{200}\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)y=\dfrac{49}{200}\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)y=\dfrac{49}{200}\\ \Rightarrow\dfrac{4949}{9900}y=\dfrac{49}{100}\\ \Rightarrow y=\dfrac{99}{101}\)
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
=> \(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{2}-\vec{\frac{1}{99.100}=\frac{4949}{99.100}}\)
\(C=\frac{4949}{2.99.100}\)