Question

\(A=\frac{1}{1x2x3}+\frac{1}{2x3x4}+\frac{1}{3x4x5}+.....+\frac{1}{2014x2015x2016}\)

So sánh A với \(\frac{1}{4}\)

 

Answer 1

A = \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2014.2015.2016}\right)=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)=\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2015.2016}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4062240}\right)=\frac{1}{4}-\frac{1}{8124480}

Answer 2

Nhận xét: \(\frac{2}{1.2.3}=\frac{3-1}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)

\(\frac{2}{2.3.4}=\frac{4-2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\)

........................

\(\frac{2}{2014.2015.2016}=\frac{2016-2014}{2014.2015.2016}=\frac{1}{2014.2015}-\frac{1}{2015.2016}\)

=> \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2014.2015.2016}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)

=> 2.A = \(2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2014.2015.2016}\right)=\frac{1}{1.2}-\frac{1}{2015.2016}

Answer 3

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

\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\right)\)

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

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

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4062240}\right)\)

\(A=\frac{1}{4}-\frac{1}{8124480}\)

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

~Học tốt~

Answer 4

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

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

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

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

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

 MÀ    __________________________  < \(\frac{1}{4}\). đpcm