\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{19.20.21}\)
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G6
3
KT
19 tháng 8 2018
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{19.20.21}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{20.21}\right)\)
\(=\frac{1}{2}.\frac{209}{420}\)
\(=\frac{209}{840}\)
U
19 tháng 8 2018
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{19\cdot20\cdot21}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{19\cdot20\cdot21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}-\frac{1}{20\cdot21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot21}\right)\)
bn tự lm tp
27 tháng 4 2017
A= \(\frac{1}{1.2.3}\)+ \(\frac{1}{2.3.4}\)+ ... + \(\frac{1}{19.20.21}\)< \(\frac{1}{4}\)
= 1 - \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ... + \(\frac{1}{19}-\frac{1}{20}-\frac{1}{21}\)
= 1 - ( \(\frac{1}{2}-\frac{1}{3}\)+ \(\frac{1}{2}-\frac{1}{3}\)) + ... + ( \(\frac{1}{19}-\frac{1}{20}+\frac{1}{19}-\frac{1}{20}\)) - \(\frac{1}{21}\)
= 1 - \(\frac{1}{21}\)
= \(\frac{20}{21}\)< \(\frac{1}{4}\)
=> Đề bài có sai ko bạn?
U
2
DP
24 tháng 6 2017
\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right).x=5\)
\(\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{21-19}{19.20.21}\right).x=5\)
\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right).x=5\)
\(\left(\frac{1}{1.2}-\frac{1}{20.21}\right).x=5\)
\(\frac{209}{420}.x=5\)
\(\Rightarrow x=5\div\frac{209}{420}=\frac{2100}{209}\)
24 tháng 6 2017
\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right).x=5\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{19.20.21}\right).2.x=5\)
\(\left(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{19.20}-\frac{1}{20.21}\right)\right).x.2=5\)
\(\left(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\right).x=5\div2\)
\(\left(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{20.21}\right)\right).x=2,5\)
\(\left(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{420}\right)\right).x=2,5\)
\(\left(\frac{1}{2}\times\frac{209}{420}\right)\times x=2,5\)
\(\frac{209}{840}\times x=2,5\)
\(x=2,5\div\frac{209}{840}=10\frac{10}{209}\)
EC
3
TV
6 tháng 7 2016
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{4.5.6}+....+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{99}{100}\)
6 tháng 7 2016
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{1}{19800}\)
VA
1
11 tháng 4 2019
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
NN
2
6 tháng 9 2019
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{49.50.51}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.5}+...+\frac{1}{49.50}-\frac{1}{50.51}\)
\(=\frac{1}{2}-\frac{1}{50.51}\)
\(=\frac{1}{2}-\frac{1}{2550}=\frac{637}{1275}\)
DT
6 tháng 9 2019
Gọi A là tổng dãy phân số trên
Ta có :
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{49.50.51}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{49.50.51}\)
Ta thấy:
\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3};\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4};...;\frac{2}{49.50.51}=\frac{2}{49.50}-\frac{2}{50.51}\text{}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{49.50}-\frac{1}{50.51}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{50.51}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{2550}\)
\(\Rightarrow2A=\frac{1275}{2550}-\frac{1}{2550}\)
\(\Rightarrow2A=\frac{637}{1275}\Rightarrow A=\frac{637}{1275}:2=\frac{637}{2550}\)
Vậy tổng dãy phân số trên là :\(\frac{637}{2550}\)
Chúc bạn học tốt !!! :D
MC
1
14 tháng 7 2018
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2004.2005.2006}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2004.2005}-\frac{1}{2005.2006}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2005.2006}\right)\)
\(=\frac{1}{4}-\frac{1}{2.2005.2006}\)
PH
3
15 tháng 5 2019
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{2018\cdot2019\cdot2020}\)
\(=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{2018\cdot2019\cdot2020}\right]\)
\(=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\right]\)
Đến đây tự tính được rồi:v
BD
15 tháng 5 2019
Đặt tổng trên là A
Ta có:
\(2A=2\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{2018\cdot2019\cdot2020}\right)\)
\(=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2018\cdot2019\cdot2020}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\)
\(=\frac{1}{2}-\frac{1}{2019\cdot2020}\)
\(A=\left(\frac{1}{2}-\frac{1}{2019\cdot2020}\right)\div2\)
*Làm tiếp*
\(#Louis\)
PT
4
14 tháng 4 2019
A=1/2 *(1/1*2-1/2*3+1/2*3-1/3*4+........+1/98*99-1/99*100)
=1/2*(1/2-1/99*100)
=1/2*(4950-1/9900)
=4950/19800
14 tháng 4 2019
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(A=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{99\cdot100}\right]=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{19.20.21}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{420}\right)=\frac{1}{2}.\frac{209}{420}=\frac{209}{840}\)
=\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right)\)
=\(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{21-19}{19.20.21}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{420}\right)=\frac{1}{2}.\frac{209}{420}=\frac{209}{840}\)