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_

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=1/1.2.3+1/2.3.4+1/3.4.5+...+1/2014.2015.2016

A=1/2.(1/2.3+1/3.4+1/4.5+...+1/2014.2015+1/2015.2016)

A=1/2.(1/2.3-1/2015.2016)

A=1/2.(1/2-1/2015.2016)

A=1/4-A1/2.2015.2016<1/4

suy ra A<1/4

A=1/4-1/2015.1008

=)A<4 (ĐPCM)
Nhớ k nha

A= 1 - 1/2 - 1/3 + 1/2 - 1/3 - 1/4 + 1/3 - 1/4 - 1/5 + ....... + 1/2014 - 1/2015 - 1/2016 

Rồi đoạn sau tự tính tiếp nhé :)) Đến đôạn này chắc trừ được

conan mà không biết làm mấy bài này conan lởm

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

3.

Ta có :

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

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

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2105}-\frac{1}{2015.2016}\right)\)

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

\(A=\frac{1}{4}-\frac{1}{2.2015.2016}< \frac{1}{4}\)

mấy câu này dễ

Lời giải:

$2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2014-2012}{2012.2013.2014}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2012.2013}-\frac{1}{2013.2014}$

$=\frac{1}{1.2}-\frac{1}{2013.2014}=\frac{1}{2}-\frac{1}{2013.2014}<\frac{1}{2}$

$\Rightarrow A< \frac{1}{4}$

\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2013.2014.2015}\)

\(S=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2014.2015}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4058210}\right)\)

\(S=\frac{1}{2}.\left(\frac{2029105}{4058210}-\frac{1}{4058210}\right)\)

\(S=\frac{1}{2}.\frac{2029104}{4058210}\)

\(S=\frac{1014552}{4058210}\)

Chúc bạn học tốt !!! 

Công thức : 

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

có cái gì sai sai đúng ko bạn 

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

A>1/4

A<1/4 BN NHÉ

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

Từ kết quả của các bài trên,ta nhận thấy :

\(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}=\frac{3-1}{1\cdot2\cdot3}=\frac{2}{1\cdot2\cdot3}\)

\(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}=\frac{4-2}{2\cdot3\cdot4}=\frac{2}{2\cdot3\cdot4}\)

Vậy : \(\frac{1}{1\cdot2\cdot3}=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right];\frac{1}{2\cdot3\cdot4}=\frac{1}{2}\left[\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right];...\)

\(\frac{1}{2015\cdot2016\cdot2017}=\frac{1}{2}\left[\frac{1}{2015\cdot2016}-\frac{1}{2016\cdot2017}\right]\)

Cộng các số hạng của vế trái và các số hạng của vế phải ta được :

\(A=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right]+\frac{1}{2}\left[\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right]+...+\frac{1}{2}\left[\frac{1}{2015\cdot2016}-\frac{1}{2016\cdot2017}\right]\)

\(=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}-\frac{1}{2016\cdot2017}\right]\)

\(=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2016\cdot2017}\right]=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{2016\cdot2017}\right]=...\)

Tính nốt và so sánh -_-