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\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{20\cdot21\cdot22}=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{20\cdot21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{20\cdot21}-\frac{1}{21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{231}{462}-\frac{1}{462}\right)=\frac{1}{2}\cdot\frac{230}{462}=\frac{1}{2}\cdot\frac{115}{231}=\frac{115}{462}\)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{2018\cdot2019\cdot2020}\)
\(=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{2018\cdot2019\cdot2020}\right]\)
\(=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\right]\)
Đến đây tự tính được rồi:v
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*
\(#Louis\)
Mình không chép đề bài nhé :
Gọi biểu thức là A :
Ta có : 2A=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\)
= \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
=\(\frac{1}{1.2}-\frac{1}{49.50}\)( Rút gọn đi ta được cái này )
=1/2 - 1/2450
Vậy A = (1/2 - 1/2450):2
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)
\(=\frac{1}{2}.\left(\frac{741}{1482}-\frac{1}{1482}\right)\)
\(=\frac{1}{2}.\frac{740}{1482}\)
\(=\frac{185}{741}\)
Chúc bạn học tốt !!!
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 =........
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\frac{370}{741}\)
\(=\frac{185}{741}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)
= \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
= \(\frac{1}{1.2}-\frac{1}{49.50}\)
= \(\frac{1}{2}-\frac{1}{2450}\)
= \(\frac{612}{1225}\)
đặt
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{48.49.50}\)
\(\Rightarrow\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
\(\Rightarrow\frac{1}{1.2}-\frac{1}{49.50}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2450}=\frac{621}{1225}\)
\(\Rightarrow A=\frac{306}{1225}\)
= 1/2.(2/1.2.3+2/2.3.4+.....+2/50.51.52
=1/2.(1/1.2-1/2.3+1/2.3-1/3.4+....+1/50.51-1/51.52
=1/2.(1/1.2-1/51.52)
=1/2.(1/2-1/2652)
=1/2.1325/2652
=1325/5304
A=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/50.51-1/51.52
A=1/1.2-1/51.52
phần còn lại tự giải nhé
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)
\(2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(2A=\frac{1}{2}-\frac{1}{99.100}=\frac{49}{99.100}\Rightarrow A=\frac{49}{2.99.100}\)
\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)
\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{132}\right)\)
A = \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + ...+ \(\frac{1}{10.11.12}\)
2 x A = \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + ... + \(\frac{2}{10.11.12}\)
2A = \(\frac12\).(\(\frac{2}{1.3}\)) + \(\frac13\).(\(\frac{2}{2.4}\)) + ... + \(\frac{1}{11}\).(\(\frac{2}{10.12}\))
2A = \(\frac12\).(\(\frac11-\frac13\)) + \(\frac13\).(\(\frac12-\frac14\)) + ...+ \(\frac{1}{11}\).(\(\frac{1}{10}-\frac{1}{12}\))
2A = \(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + ...+\(\frac{1}{11.10}\) - \(\frac{1}{11.12}\)
2A = \(\frac{1}{1.2}-\frac{1}{11.12}\)
2A = \(\frac12-\frac{1}{132}\)
2A = \(\frac{66}{132}\) - \(\frac{1}{132}\)
2A = \(\frac{65}{132}\)
A = \(\frac{65}{132}\) : 2
A = \(\frac{65}{264}\)
* Công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\left(\frac{3}{6}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{20.21.22}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{20.21}-\frac{1}{21.22}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{21.22}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{462}\right)\)
\(=\frac{1}{2}.\left(\frac{231}{462}-\frac{1}{462}\right)\)
\(=\frac{1}{2}.\frac{230}{462}\)
\(=\frac{115}{462}\)
Chúc bạn học tốt !!!
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Câu hỏi tương tự
Chúc bạn học giỏi =/
tổng quát nè bn
\(\frac{1}{n\left(n+k\right)\left(n+2k\right)}+\frac{1}{\left(n+k\right)\left(n+2k\right)\left(n+3k\right)}+\frac{1}{\left(n+2k\right)\left(n+3k\right)\left(n+4k\right)}\)
\(+...+\frac{1}{\left(n+x.k\right)\left(n+x.k+k\right)\left(n+x.k+2k\right)}\)
\(=\frac{\frac{1}{n\left(n+k\right)}-\frac{1}{\left(n+x.k+k\right)\left(n+x.k+2k\right)}}{2k}\)