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\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+......+\left(\frac{1}{2016}-\frac{1}{2017}\right)\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2016}-\frac{1}{2017}\)
\(A=\frac{1}{1}-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
A=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)
A=\(\frac{1}{1}-\frac{1}{2017}\)
A=\(\frac{2016}{2017}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
= \(1-\frac{1}{2017}\)
= \(\frac{2016}{2017}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{2016}+\frac{1}{2016}\right)-\frac{1}{2017}\)
\(A=1+0+0+...+0-\frac{1}{2017}\)
\(A=1-\frac{1}{2017}\)
\(A=\frac{2017}{2017}-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
Vậy: \(A=\frac{2016}{2017}\)
S=2-1/1.2 . 3-2/2.3............2016-2015/2015.2016
=1/1 - 1/2 + 1/2 - 1/3+........+1/2015 - 1/2016
=1/1 - 1/2016
=2015/2016
Ta có \(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+\(\frac{1}{3\cdot4}\)+...+\(\frac{1}{2015\cdot2016}\)
=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{2015}\)-\(\frac{1}{2016}\)
=1-\(\frac{1}{2016}\)
=\(\frac{2015}{2016}\)(bạn cứ nhớ công thức là làm được)
Ta thấy: \(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2};\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3};....;\frac{1}{2015.2016}=\frac{1}{2015}-\frac{1}{2016}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2015.2016}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{2016}\)
\(=1-\frac{1}{2016}=\frac{2015}{2016}\)
Làm bậy, mà đúng
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ … + \(\frac{1}{99.100}\)
= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)-\(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ … + \(\frac{1}{99}\)- \(\frac{1}{100}\)
= \(\frac{1}{1}\)- \(\frac{1}{100}\)
= \(\frac{99}{100}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A=1-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}\)
\(=\frac{5}{6}\)
1/1.2+1/2.3+1/3.4+1/4.5+1/5.6
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6
=1-1/6
=5/6
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}=\frac{2016}{2017}\)