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Ta có nhận xét: \(\frac{2}{k\left(k+1\right)\left(k+2\right)}=\frac{1}{k\left(k+1\right)}-\frac{1}{\left(k+1\right)\left(k+2\right)}\)
Áp dụng tính A ta có:
\(2.A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\)
\(\Rightarrow2.A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)
\(\Rightarrow2.A=\frac{1}{1.2}-\frac{1}{2015.2016}=\frac{2015.1008-1}{2015.2016}\)
\(\Rightarrow A=\left(\frac{2015.1008-1}{2015.2016}\right):2\)
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{2014.2015.2016}\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+.....+\left(\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2015.2016}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.2015.2016}< \dfrac{1}{4}\\ \Rightarrow A< \dfrac{1}{4}\)
Giải:
Ta có: \(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2014.2015.2016}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2014.2015.2016}\right)\)
\(=\dfrac{1}{2}\)\(\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2015.2016}\right)\)
\(=\dfrac{1}{2}.\dfrac{1}{1.2}-\dfrac{1}{2}.\dfrac{1}{2015.2016}=\dfrac{1}{4}-\) \(\dfrac{1}{2.2015.2016}\)
Mà \(\dfrac{1}{2.2015.2016}>0\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{2.2015.2016}< \dfrac{1}{4}\)
Vậy \(A< \dfrac{1}{4}\)
A = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/2014.2015.2016
=> A = 1/2.(2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ... + 2/2014.2015.2016)
=> A = 1/2.(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/2014.2015 - 1/2015.2016)
=> A = 1/2.(1/2 - 1/2015.2016)
=> A < 1/2.1/2 = 1/4
Ta có: \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{2014.2015.2016}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{2014.2015.2016}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2015.2016}\)
\(\Rightarrow A=\left(\frac{1}{2}-\frac{1}{2015.2016}\right):2\)
\(\Rightarrow A=\frac{1}{4}-\frac{1}{2015.2016.2}\)
\(\Rightarrow A< \frac{1}{4}\)