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29 tháng 4 2018

* Công thức :  \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\left(\frac{3}{6}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)

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

\(\Rightarrow A=3.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2015.2016.2017}\right)\)

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

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

\(\Rightarrow A=3.\left(\frac{1}{2}-\frac{1}{4066272}\right)\)

\(\Rightarrow A=3.\left(\frac{2033136}{4066272}-\frac{1}{4066272}\right)\)

\(\Rightarrow A=3.\frac{2033135}{4066272}>3.\frac{1355424}{4066272}\)

\(\Rightarrow A>3.\frac{1}{3}\)

\(\Rightarrow A>1\)

Chúc bạn học tốt !!! 

29 tháng 4 2018

Thanks bạn Hỏa Long Natsu

5 tháng 5 2017

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

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

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

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

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

16 tháng 5 2018

Ta có : 

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

\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)

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

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

\(\Rightarrow A=\frac{1}{4}-\frac{1}{2015.2016}\)

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

Vậy A < \(\frac{1}{4}\)

_Chúc bạn học tốt_

16 tháng 5 2018

Ta có:

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{2014+2015+2016}\)

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

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

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

\(\Rightarrow2A< \frac{1}{1.2}=\frac{1}{2}\)

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

Vậy .... 

A=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2015.2016.2017}\)

\(\Leftrightarrow\)A=\(\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}-\frac{2}{2017}\)

\(\Leftrightarrow\)A=\(\frac{1}{1}-\frac{1}{2017}\)

\(\Leftrightarrow\)A=\(\frac{2016}{2017}\)

mk quên:Có \(\frac{2016}{2017}< \frac{1}{4}\) \(\Rightarrow\)S<\(\frac{1}{4}\)

28 tháng 4 2019

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)

\(A=\frac{1}{2}.\frac{370}{741}\)

\(A=\frac{185}{741}\)

28 tháng 4 2019

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)

Tự tính tiếp nha =)) mỏi tay quá

15 tháng 4 2019

\(M=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{100.101.102}\right)\)

\(M=\frac{1}{2}.\left(1-\frac{1}{102}\right)\)

\(M=\frac{101}{204}< 1\left(đpcm\right)\)

 

Ta có: M=11.2.3  +12.3.4  +13.4.5  +...+1100.101.102  

         M=2.(11.2.3  +12.3.4  +13.4.5  +...+1100.101.102  ).12 

          M=(21.2.3  +22.3.4  +23.4.5  +...+2100.101.102  ).12 

          M=(11.2  -12.3  +12.3  -13.4  +13.4  -14.5 +...+1100.101 1101.102  ).12 

          M=( 11.2 1101.102 ).12 

          Mà 11.2 1101.102 <1

         Và 12 <1 

        =>  (11.2 1101.102  ) .12  <1

        => M <1

nhớ 9 k đó
15 tháng 5 2016

2A=\(\frac{2}{1\cdot2\cdot3}\)+\(\frac{2}{2\cdot3\cdot4}\)+\(\frac{2}{3\cdot4\cdot5}\)+...+\(\frac{2}{2014\cdot2015\cdot2016}\)

2A=\(\frac{1}{1\cdot2}\)-\(\frac{1}{2\cdot3}\)+\(\frac{1}{2\cdot3}\)-\(\frac{1}{3\cdot4}\)+\(\frac{1}{3\cdot4}\)-\(\frac{1}{4\cdot5}\)+...+\(\frac{1}{2014\cdot2015}\)-\(\frac{1}{2015\cdot2016}\)

2A=\(\frac{1}{2}\)-\(\frac{1}{2015\cdot2016}\)

A=(\(\frac{1}{2}\)-\(\frac{1}{2015\cdot2016}\)):2

A=\(\frac{1}{2}\):2-\(\frac{1}{2015\cdot2016}\):2

A=\(\frac{1}{4}\)-\(\frac{1}{2015\cdot2016\cdot2}\)<\(\frac{1}{4}\)

Vậy A<\(\frac{1}{4}\)