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Ta có : \(A=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+.......+\left(1-\frac{1}{2016.2017}\right)\)
\(\Rightarrow A=\left(1+1+1+......+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{2016.2017}\right)\)
\(\Rightarrow A=2016-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(\Rightarrow A=2016-\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=2016-\frac{2016}{2017}=2015\frac{1}{2017}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{a\left(a+1\right)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{a}-\frac{1}{a+1}\)
\(=1-\frac{1}{a+1}\)
\(=\frac{a+1}{a+1}-\frac{1}{a+1}=\frac{a}{a+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có công thức :
\(\frac{1}{k\left(k+1\right)}=\frac{\left(k+1\right)-k}{k\left(k+1\right)}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}=\frac{n-1}{n}\)
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a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{a.\left(a+1\right)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{a}-\frac{1}{a+1}\)
\(=1-\frac{1}{a+1}\)
b, \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{a.\left(a+1\right).\left(a+2\right)}\)
=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{a.\left(a+1\right)}-\frac{1}{\left(a+1\right).\left(a+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{\left(a+1\right).\left(a+2\right)}\)
\(=\frac{1}{2}-\frac{1}{\left(a+1\right).\left(a+2\right)}\)
Chúc bạn học giỏi nha!!!
K cho mik vs nhé Hang Nguyen
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=>\(-B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2012}\right)\)
=\(\frac{1}{2}.\frac{2}{3}...\frac{2011}{2012}=\frac{1}{2012}\)
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a) S1 = \(-\frac{1}{1.2}-\frac{1}{2.3}-...-\frac{1}{99.100}\)
= \(-\frac{1}{1}-\frac{1}{2}-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{99}-\frac{1}{100}\)
= \(\frac{-1}{1}-\frac{1}{100}\)
= \(-\frac{101}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{a\left(a+1\right)}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{a}-\frac{1}{a+1}\)
= \(1-\frac{1}{a+1}\)