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\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)
\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{132}\right)\)
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
\(2M=2\cdot\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+.....+\frac{1}{10\cdot11\cdot12}\right)\)
\(2M=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+.....+\frac{2}{10\cdot11\cdot12}\)
\(2M=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+.....+\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\)
\(2M=\frac{1}{1\cdot2}-\frac{1}{11\cdot12}\)
\(2M=\frac{1}{2}-\frac{1}{132}\)
\(2M=\frac{66}{132}-\frac{1}{132}\)
\(2M=\frac{65}{132}\)
\(M=\frac{65}{132}:2\)
\(M=\frac{65}{264}\)
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 nhận xét:
\(\frac{2}{n.\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
Áp dụng công thức trên vào bài tập, ta có:
B=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
Vậy \(B=\frac{185}{741}\)
Ta có nhận xét:
\(\frac{2}{n.\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
Áp dụng công thức trên vào bài tập, ta có:
\(\Rightarrow B=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+\frac{5}{3.4.5}-\frac{3}{3.4.5}+...+\frac{39}{37.38.39}-\frac{37}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{38.29}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}...+\frac{2}{37.38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)=\frac{185}{741}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{37.38.39}\right)=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1482}\right)=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}=\frac{1}{2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)
\(=\left(\frac{1}{2}-\frac{1}{38\cdot39}\right)+\left(\frac{1}{2\cdot3}-\frac{1}{2\cdot3}\right)+...+\left(\frac{1}{37\cdot38}-\frac{1}{37\cdot38}\right)=\left(\frac{741}{1482}-\frac{1}{1482}\right)+0+...+0=\frac{740}{1482}=\frac{370}{741}\)Chúc bạn học tốt!^_^
\(D=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{10\cdot11\cdot12}\)
\(D=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{10\cdot11\cdot12}\right)\)
\(D=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\right)\)
\(D=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{132}\right)=...\)
\(D=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)
\(D=\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{10.11.12}\right).\frac{1}{2}\)
\(D=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{10.11}-\frac{1}{11.12}\right).\frac{1}{2}\)
\(D=\left(\frac{1}{1.2}-\frac{1}{11.12}\right).\frac{1}{2}\)
\(D=\frac{65}{132}.\frac{1}{2}\)
\(D=\frac{65}{264}\)