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5 tháng 5 2019
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=\(\frac{2016}{2017}\)
5 tháng 5 2019
mk quên:Có \(\frac{2016}{2017}< \frac{1}{4}\) \(\Rightarrow\)S<\(\frac{1}{4}\)
21 tháng 1 2017
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\(\approx0.25\)
AK
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 !!!
NT
1
MS
23 tháng 4 2018
\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)
AQ
0
PH
3
15 tháng 5 2019
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{2018\cdot2019\cdot2020}\)
\(=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{2018\cdot2019\cdot2020}\right]\)
\(=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\right]\)
Đến đây tự tính được rồi:v
BD
15 tháng 5 2019
Đặt tổng trên là A
Ta có:
\(2A=2\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{2018\cdot2019\cdot2020}\right)\)
\(=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2018\cdot2019\cdot2020}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\)
\(=\frac{1}{2}-\frac{1}{2019\cdot2020}\)
\(A=\left(\frac{1}{2}-\frac{1}{2019\cdot2020}\right)\div2\)
*Làm tiếp*
\(#Louis\)
BQ
2
24 tháng 5 2018
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2450}\right)\)
\(A=\frac{1}{2}.\frac{612}{1225}=\frac{306}{1225}\)
9 tháng 8 2019
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{1.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}\right)-\frac{1}{3.4}+...\frac{1}{2}\left(\frac{1}{48.49}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(A=\left(\frac{1}{2}.\frac{1}{49.50}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{2}.\frac{1}{1225}=\frac{306}{1225}\)