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\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.....+\frac{1}{10.11.12}\)
\(M=\frac{1}{2}-\frac{1}{11.12}\)
\(M=\frac{65}{132}\)
Ngắn gọn , xúc tích !!! :))
\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{11.12}\)
\(=\frac{65}{132}\)
Ta có nhận xét: 1/1.2 - 1/2.3 = 3-1/1.2.3 = 2/1.2.3
1/2.3 - 1/3.4 = 4-2/2.3.4 = 2/2.3.4
Suy ra: 1/1.2.3 = 1/2(1/1.2 - 1/2.3)
1/2.3.4 = 1/2(1/2.3 -1/3.4)
Do đó: M = 1/2(1/1.2-1/2.3 + 1/2.3 -1/3.4 + ... + 1/10.11 -1/11.12)
= 1/2(1/1.2 - 1/11.12) = 1/2(1/2-11/12 )
= 1/2.65/132 = 65/264
Phức tạp lắm
\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{11.12}\)
\(=\frac{65}{132}\)
Giải:
Ta có nhận xét:
\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{3-1}{1.2.3}=\frac{2}{1.2.3}\)
\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{4-2}{2.3.4}=\frac{2}{2.3.4}\)
=>\(\frac{1}{1.2.3}=\frac{1}{3}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)
\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)
Do đó M=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}-\frac{1}{11.12}\right)\)
=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)=\frac{1}{2}-\frac{1}{11.12}\)
=\(\frac{1}{2}.\frac{65}{132}=\frac{65}{124}\)
Vậy M=65/124
Ta có :
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{3}{1.2.3}-\dfrac{1}{1.2.3}=\dfrac{2}{1.2.3}\)
\(\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{4}{2.3.4}-\dfrac{2}{2.3.4}=\dfrac{2}{2.3.4}\)
...
Do đó :
\(\dfrac{1}{1.2.3}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)\)
\(\dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\)
Vậy :
\(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
Nhận xét rằng:
2/[(n - 1)n(n +1)] = 1/[(n-1).n] - 1/[n(n+1)]
Do đó
2M = 2/(1.2.3) + 2/(2.3.4) + 2/(3.4.5) + ... + 2(10.11.12)
= 1/(1.2) - 1/(2.3) + 1/(2.3) - 1/(3.4) + 1/(3.4) - 1/(4.5) + .... + 1/(10.11) - 1/(11.12)
= 1/(1.2) - 1/(11.12) = 65/132
=> M = 65/264
Ta có nhận xét: \(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{3-1}{1.2.3}=\dfrac{2}{1.2.3}\),
\(\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{4-2}{2.3.4}=\dfrac{2}{2.3.4};...\)
\(\Rightarrow\dfrac{1}{1.2.3}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)\);
\(\dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\);...
Do đó \(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{1.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
M = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/10.11.12
M = 1/2.(2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ... + 2/10.11.12)
M = 1/2.(1/1.2 - 1/2.3 + 1/2.3- 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/10.11 - 1/11.12)
M = 1/2.(1/1.2 - 1/11.12)
M = 1/4 - 1/264
M = 65/264
Ta có \(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)
Áp dụng:
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{10\cdot11\cdot12}\\ =\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{10\cdot11}-\dfrac{1}{11\cdot12}\\ =\dfrac{1}{2}-\dfrac{1}{11\cdot12}=\dfrac{1}{2}-\dfrac{1}{132}=\dfrac{65}{132}\)
sai rồi kìa
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3}\) mà
4N = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 2015.2016.2017.(2018-2014)
4N = 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 2015.2016.2017.2018 - 2014.2015.2016.2017
4N = (1.2.3.4 + 2.3.4.5 + 3.4.5.6 + ... + 2015.2016.2017.2018) - (0.1.2.3 + 1.2.3.4 + 2.3.4.5 + ... + 2014.2015.2016.2017)
4N = 2015.2016.2017.2018 - 0.1.2.3
4N = 2015.2016.2017.2018
N = 2015.2016.504.2018 (kq hơi to nên bn tự tính nhé)