Đặt S=A

Ta có: \(\frac{3n+2}{n\left(n+1\right)\left(n+2\right)}\)

\(=\frac{2n+2+n}{n\left(n+1\right)\left(n+2\right)}=\frac{2\left(n+1\right)}{n\left(n+1\right)\left(n+2\right)}+\frac{n}{n\left(n+1\right)\left(n+2\right)}\)

\(=\frac{2}{n\left(n+2\right)}+\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}+\frac{1}{n+1}-\frac{1}{n+2}\)

\(=\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)

Do đó, ta có: \(\frac{5}{1\cdot2\cdot3}=\frac{3\cdot1+2}{1\cdot2\cdot3}=\frac11+\frac{1}{1+1}-\frac{2}{1+2}=1+\frac12-\frac23\)

\(\frac{8}{2\cdot3\cdot4}=\frac{3\cdot2+2}{2\cdot3\cdot4}=\frac12+\frac13-\frac24\)

...

Do đó, ta có: \(S=1+\frac12-\frac23+\frac12+\frac13-\frac24+\frac13+\frac14-\frac25+\ldots+\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)

\(=1+\left(\frac12+\frac12\right)+\left(-\frac23+\frac13+\frac13\right)+\left(-\frac24+\frac14+\frac14\right)+\cdots+\left(-\frac{2}{n}+\frac{1}{n}+\frac{1}{n}\right)-\frac{2}{n+1}+\frac{1}{n+1}-\frac{2}{n+2}\)

\(=1+1-\frac{1}{n+1}-\frac{2}{n+2}<2\)

=>\(S_{2018}<2\)

Ta có: \(\frac{3n+2}{n\left(n+1\right)\left(n+2\right)}\)

\(=\frac{2n+2+n}{n\left(n+1\right)\left(n+2\right)}=\frac{2\left(n+1\right)}{n\left(n+1\right)\left(n+2\right)}+\frac{n}{n\left(n+1\right)\left(n+2\right)}\)

\(=\frac{2}{n\left(n+2\right)}+\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}+\frac{1}{n+1}-\frac{1}{n+2}\)

\(=\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)

Do đó, ta có: \(\frac{5}{1\cdot2\cdot3}=\frac{3\cdot1+2}{1\cdot2\cdot3}=\frac11+\frac{1}{1+1}-\frac{2}{1+2}=1+\frac12-\frac23\)

\(\frac{8}{2\cdot3\cdot4}=\frac{3\cdot2+2}{2\cdot3\cdot4}=\frac12+\frac13-\frac24\)

...

Do đó, ta có: \(S=1+\frac12-\frac23+\frac12+\frac13-\frac24+\frac13+\frac14-\frac25+\ldots+\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)

\(=1+\left(\frac12+\frac12\right)+\left(-\frac23+\frac13+\frac13\right)+\left(-\frac24+\frac14+\frac14\right)+\cdots+\left(-\frac{2}{n}+\frac{1}{n}+\frac{1}{n}\right)-\frac{2}{n+1}+\frac{1}{n+1}-\frac{2}{n+2}\)

\(=1+1-\frac{1}{n+1}-\frac{2}{n+2}<2\)

=>\(S_{2022}=\frac{5}{1\cdot2\cdot3}+\frac{8}{2\cdot3\cdot4}+\cdots+\frac{6068}{2022\cdot2023\cdot2024}<2\)

A = 1 - 1/2 - 1/3 + 1/2 - 1/3 - 1/4 + ... + 1/2014 - 1/2015 - 1/2016

A = 1- 1/2016

A = 2015/2016

A > 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: \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{2014.2015.2016}\)

\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{2014.2015.2016}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2015.2016}\)

\(\Rightarrow A=\left(\frac{1}{2}-\frac{1}{2015.2016}\right):2\)

\(\Rightarrow A=\frac{1}{4}-\frac{1}{2015.2016.2}\)

\(\Rightarrow A< \frac{1}{4}\)

Ta có: 

A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)

2A = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\)

2A = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)

2A = \(\frac{1}{1.2}-\frac{1}{2015.2016}\)

=> 2A < \(\frac{1}{1.2}=\frac{1}{2}\)

=> A < \(\frac{1}{4}\)

Vậy A < \(\frac{1}{4}\)

A=4949/19800 và 1/4

4949/19800 < 1/4

Xong! t*** mik đê!!!

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)

\(A=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}\right)\)

\(A=0,2499998...<0,25\)

\(\Rightarrow A<\frac{1}{4}\)

bạn viết gì vậy? Mình ko thấy?

\(A=\frac{3}{1.2.3}+\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{2015.2016.2017}\)

\(A=\frac{3}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+\frac{3}{2}.\left(\frac{1}{3.4}-\frac{1}{4.5}\right)+...+\frac{3}{2}.\left(\frac{1}{2015.2016}-\frac{1}{2016.2017}\right)\)

\(A=\frac{3}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\right)\)

\(A=\frac{3}{2}.\left(\frac{1}{1.2}-\frac{1}{2016.2017}\right)\)

\(A=\frac{3}{4}-\frac{3}{2.2016.2017}< 1\)