\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+........+\frac{1}{20.21.22}\)
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Những câu hỏi liên quan
TL
4
AK
9 tháng 4 2018
* Công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\left(\frac{3}{6}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
AK
9 tháng 4 2018
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{20.21.22}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{20.21}-\frac{1}{21.22}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{21.22}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{462}\right)\)
\(=\frac{1}{2}.\left(\frac{231}{462}-\frac{1}{462}\right)\)
\(=\frac{1}{2}.\frac{230}{462}\)
\(=\frac{115}{462}\)
Chúc bạn học tốt !!!
28 tháng 4 2019
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(A=\frac{1}{2}.\frac{370}{741}\)
\(A=\frac{185}{741}\)
H
28 tháng 4 2019
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
Tự tính tiếp nha =)) mỏi tay quá
8 tháng 11 2016
A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)
=> A = \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
= \(\frac{1}{2}.\frac{5049}{10100}\)
= \(\frac{5049}{20200}\)
8 tháng 11 2016
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{99.100.101}\)
Ta thấy:
\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3};\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4};...;\frac{2}{99.100.101}=\frac{1}{99.100}-\frac{1}{100.101}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{100.101}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{10100}\)
\(\Rightarrow2A=\frac{5050}{10100}-\frac{1}{10100}\)
\(\Rightarrow2A=\frac{5049}{10100}\Rightarrow A=\frac{5049}{10100}:2=\frac{5049}{20200}\)
TD
20 tháng 1 2019
Đặt biểu thức là A
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+.2018.2019\)
\(2A=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)
\(2A=\frac{1}{2}-\frac{1}{2018.2019}\)
A= 1/4 - 1/(2018.2019)
Vậy A = ... (tự ghi)
20 tháng 1 2019
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2017.2018.2019}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2017.2018.2019}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2018.2019}\right)\)
Tự làm nốt
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
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{380}\right)\)
\(=\frac{1}{2}.\frac{189}{380}=\frac{189}{760}\)
VA
1
11 tháng 4 2019
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
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
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(A=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{99\cdot100}\right]=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{20\cdot21\cdot22}=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{20\cdot21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{20\cdot21}-\frac{1}{21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{231}{462}-\frac{1}{462}\right)=\frac{1}{2}\cdot\frac{230}{462}=\frac{1}{2}\cdot\frac{115}{231}=\frac{115}{462}\)