A=1/1x2x3 +1/2x3x4 + .......+ 1/2014x2015x2016
K
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7 tháng 5 2015
A = \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2014.2015.2016}\right)=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)=\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2015.2016}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4062240}\right)=\frac{1}{4}-\frac{1}{8124480}
7 tháng 5 2015
Nhận xét: \(\frac{2}{1.2.3}=\frac{3-1}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)
\(\frac{2}{2.3.4}=\frac{4-2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\)
........................
\(\frac{2}{2014.2015.2016}=\frac{2016-2014}{2014.2015.2016}=\frac{1}{2014.2015}-\frac{1}{2015.2016}\)
=> \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2014.2015.2016}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)
=> 2.A = \(2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2014.2015.2016}\right)=\frac{1}{1.2}-\frac{1}{2015.2016}
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 \(A=\dfrac{-4949}{19800}\)
NH
1
ND
Nguyễn Đăng Nhân
CTVHS
24 tháng 3 2022
Ta có:
\(A=\frac{1}{1\text{x}2\text{x}3}+\frac{1}{2\text{x}3\text{x}4}+\frac{1}{3\text{x}4\text{x}5}+...+\frac{1}{18\text{x}19\text{x}20}< \frac{1}{4}\)
\(A=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{20}< \frac{1}{4}\)
\(A=1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\frac{1}{20}< \frac{1}{4}\)
\(A=1+\frac{1}{20}< \frac{1}{4}\)
\(A=\frac{19}{20}< \frac{1}{4}\)
\(A=\frac{19}{20}< \frac{5}{20}\)
\(A>\frac{1}{4}\)
KS
10 tháng 5 2022
`1/(1.2.3) + 1/(2.3.4) +.....+ 1/(98.99.100)`
`2/(1.2.3) + 2/(2.3.4) + ...+ 2/(98.99.100)`
`1/(1.2) - 1/(2.3) + 1/(2.3) - 1/(3.4) + ... + 1/(98.99) - 1/(99.100)`
`1/(1.2) - 1/(99.100)`
`1/2 - 1/9900`
= `4949/9900`
XO
10 tháng 6 2020
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{1}{4}-\frac{1}{760}< \frac{1}{4}\)(ĐPCM)
NM
2
7 tháng 2 2017
\(D=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2015.2016.2017}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2016.2017}\right)=\frac{1}{2}.\left(\frac{2016.2017:2-1}{2016.2017}\right)\)
7 tháng 2 2017
D=1/1.2.3+1/2.3.4+....+1/2015.2016.2017
D=1/2(1/1.2-1/2.3+1/2.3-1/3.4+.......+1/2015.2016-1/2016.2017)
D=1/2(1/1.2-1/2016.2017)
K
18 tháng 7 2020
Trả lời:
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2018.2019.2020}+\frac{1}{2.2019.2020}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2018.2019.2020}+\frac{2}{2.2019.2020}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2018.2019}-\frac{1}{2019.2020}+\frac{1}{2019.2020}\right)\)
\(A=\frac{1}{2}.\frac{1}{1.2}\)
\(A=\frac{1}{4}\)
DQ
2
DP
16 tháng 7 2017
\(\Leftrightarrow3x-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\right)\)
\(\Leftrightarrow3x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+....+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(\Leftrightarrow3x-\left(1-\frac{1}{100}\right)=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(\Leftrightarrow3x-\frac{99}{100}=\frac{1}{2}\cdot\frac{189}{380}\)
\(\Leftrightarrow3x-\frac{99}{100}=\frac{189}{760}\)
\(\Leftrightarrow3x=\frac{189}{760}+\frac{99}{100}=\frac{4707}{3800}\)
\(\Leftrightarrow x=\frac{1569}{3800}\)
\(\text{Vậy }x=\frac{1569}{3800}\)
2A = 2/1.2.3 + 2/2.3.4 +.......+ 2/2014.2015.2016
2A = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 +.......+ 1/2014.2015 - 1/2015.2016
2A = 1/1.2 - 1/2015.2016
2A = ................Tự tính rồi tự làm nốt!