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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 !!!
QA
2
16 tháng 11 2018
A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 20.21.22
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + ... + 20.21.22.4
4A = 1.2.3.4 + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) + ... + 20.21.22.(23 - 19)
4A = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 20.21.22.23 - 19.20.21.22
4A = 20.21.22.23
A = 20.21.22.23 : 4
A = 53130
NV
16 tháng 11 2018
Đặt A = 1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 +...+ 20.21.22
\(\Rightarrow4A=1.2.3.4+2.3.4.4+3.4.5.4+4.5.6.4+...+20.21.22.4\)
\(=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+4.5.6.\left(7-3\right)+...+20.21.22.\left(23-19\right)\)
= 1.2.3.4 + 2.3.4.5 -1.2.3.4 + 3.4.5.6 - 2.3.4.5 + 4.5.6.7 - 3.4.5.6 +...+ 20.21.22.23 - 19.20.21.22
= 20.21.22.23
= 212520
_Hok tốt_
!!!
4 tháng 5 2018
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{20\cdot21\cdot22}=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{20\cdot21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{20\cdot21}-\frac{1}{21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{231}{462}-\frac{1}{462}\right)=\frac{1}{2}\cdot\frac{230}{462}=\frac{1}{2}\cdot\frac{115}{231}=\frac{115}{462}\)
DG
0
DT
7
AK
11 tháng 4 2018
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)
\(=\frac{1}{2}.\left(\frac{741}{1482}-\frac{1}{1482}\right)\)
\(=\frac{1}{2}.\frac{740}{1482}\)
\(=\frac{185}{741}\)
Chúc bạn học tốt !!!
11 tháng 4 2018
Đặt 1/1.2.3 + 1/2.3.4 + ...+ 1/37.38.39 = A
Ta có : 2A = 2/1.2.3 + 2/2.3.4 +...+ 2/37.38.39
2A = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ...+ 1/37.38 - 1/38.39
2A = 1/1.2 - 1/38.39
2A = 740/1482 = 370/741
A= 370/741 . 1/2 =........
TH
13 tháng 2 2018
A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)
3A-A= \(1-\frac{1}{3^{2008}}\)
AQ
0
NG
2
19 tháng 4 2019
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{20\cdot21\cdot22}\)
\(A=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{20\cdot21\cdot22}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{20\cdot21}-\frac{1}{21\cdot22}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{21\cdot22}\right]=\frac{1}{2}\cdot\frac{115}{231}=\frac{230}{231}>\frac{57}{231}(đpcm)\)
19 tháng 4 2019
Sửa dùm chút :v
Chỗ \(\frac{1}{2}\cdot\frac{115}{231}=\frac{115}{462}>\frac{57}{231}(đpcm)\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{20.21.22}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{20.21.22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{20.21}-\frac{1}{21.22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{462}\right)=\frac{1}{2}.\frac{115}{231}=\frac{115}{462}\)