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\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A=1-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
Cho A = 1.2 + 2.3 + ...+ 99.100
=> 3A = 1.2 .3 + 2.3.3 + ...+ 99.100.3
3A = 1.2.( 3-0) + 2.3.(4-1) + ....+ 99.100.( 101 - 98)
3A = 1.2.3 + 2.3.4 - 1.2.3 + ...+ 99.100.101 - 98.99.100
3A = ( 1.2.3 + 2.3.4 + 99.100.101) - ( 1.2.3 + ....+ 98.99.100)
3A = 99.100.101
=> A = 99.100.101 . 1/3
thay A vào B
\(B=(\frac{99.100.101.\frac{1}{3}}{99.100.101}):\frac{1}{3}\)
\(B=\frac{1}{3}:\frac{1}{3}\)
\(B=1\)
\(B=\left(\frac{1.2+2.3+...+99.100}{99.100.101}\right)\div\frac{1}{3}\)
\(\text{Đặt}:C=1.2+2.3+...+99.100\)
\(\Rightarrow3C=1.2.3+2.3.3+...+99.100.3\)
\(\Rightarrow3C=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)
\(\Rightarrow3C=1.2.3+2.3.4+...+99.100.101\)
\(\Rightarrow3C=\left(1.2.3+2.3.4+...+99.100.101\right)\)\(-\)\(\left(1.2.3+2.3.4+....+98.99.100\right)\)
\(\Rightarrow3C=99.100.101\)
\(\Rightarrow C=\frac{99.100.101}{3}\)
Thay C vào biểu thức B ta được :
\(B=\left(\frac{\frac{99.100.101}{3}}{99.100.101}\right)\div\frac{1}{3}=\frac{1}{3}\div\frac{1}{3}=1\)
Vậy B= \(1\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{98.99.100}=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{k}=\frac{1}{2}\Rightarrow k=2\)