B = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2015.2016.2017}\)
=) 2B = \(2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2015.2016.2017}\right)\)
= \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2015.2016.2017}\)
= \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\)
= \(\frac{1}{1.2}-\frac{1}{2016.2017}=\frac{1}{2}.\left(1-\frac{1}{1008.2017}\right)=\frac{1}{2}.\left(1-\frac{1}{2033136}\right)\)
= \(\frac{1}{2}.\frac{2033135}{2033136}=\frac{1}{4066272}\)
=) B = \(\frac{1}{4066272}:2=\frac{1}{4066272}.\frac{1}{2}=\frac{1}{8132544}\)

B = 1(1/1 - 1/2 - 1/3 + 1/2 - 1/3 - 1/4 + ...+ 1/2015 - 1/2016 - 1/2017)

B = 1( 1/1 - 1/2017)

B = 1.2016

B = 2016

Mà em nói nhỏ nghe nè,đây không phải là toán lớp 9 đâu,...mà là ......toán lớp 6 thôi !

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 -_-

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

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

Áp dụng công thức trên ta có

A\(=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdot\cdot\cdot\cdot\cdot\cdot\cdot+\frac{1}{2015\cdot2016\cdot2017}\)

\(\Leftrightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2015\cdot2016\cdot2017}\)

\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{2}{3\cdot4}+....+\frac{1}{2015\cdot2016}-\frac{1}{2016\cdot2017}\)

\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2016\cdot2017}\)

\(\Rightarrow A=\left(\frac{1}{1\cdot2}-\frac{1}{2016\cdot2017}\right)\div2\approx0.25\)

Vậy A\(\approx0.25\)

* 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 !!! 

Thanks bạn Hỏa Long Natsu

=1+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{2}\) -\(\frac{1}{3}\) -\(\frac{1}{4}\)+\(\frac{1}{3}\) - \(\frac{1}{4}\)-\(\frac{1}{5}\)+.....+\(\frac{1}{99}\)-\(\frac{1}{100}\)-\(\frac{1}{101}\)

=1+\(\frac{1}{101}\)

=\(\frac{102}{101}\)

1/1.2.3 = 1/2 .[1/1.2 - 1 / 2.3]

1/2.3.4 = 1/2[ 1/2- 1/3 ] 

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

1/99.100.101 = 1/2[ 1/99. 100 - 1/100.101]

=> A= 1/2 [ 1/1.2- 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/ 4.5 +.........+ 1/99 .100 - 1/100. 101]

A = 1/2 . [1/1.2 -1/100 .101]

A= 1/2 . 5049 /10100 = 5049 / 20200.

Mình nghĩ là vậy đó.

a=100/101

\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)

\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)