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DP
2
7 tháng 8 2015
Ta có:
\(\frac{2}{3.7}+\frac{2}{7.11}+\frac{2}{11.15}+.....+\frac{2}{2015.2019}\)
\(=2.\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+....+\frac{1}{2015.2019}\right)\)
\(=\frac{2}{4}.\left(\frac{2}{3.7}+\frac{2}{7.11}+\frac{2}{11.15}+....+\frac{2}{2015.2019}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{2015}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}.\frac{224}{673}=\frac{112}{673}\)
**** ^^
SL
2
8 tháng 5 2021
Giải:
A=2/3.7+2/7.11+2/11.15+...+2/n.(n+4)
A=1/2.(4/3.7+4/7.11+4/11.15+...+4/n.(n+4)
A=1/2.(1/3-1/7+1/7-1/11+1/11-1/15+...+1/n-1/n+4)
A=1/2.(1/3-1/n+4)
A=1/6-1/2.(n+4)
⇒A<1/6
Chúc bạn học tốt!
8 tháng 5 2021
Ta có : \(A=\dfrac{2}{3.7}+\dfrac{2}{7.11}+...+\dfrac{2}{n\left(n+4\right)}\)
\(\Rightarrow4A=\dfrac{8}{3.7}+\dfrac{8}{7.11}+...+\dfrac{8}{n\left(n+4\right)}\)
\(\Rightarrow4A=\dfrac{8}{3.7}+\dfrac{8}{7.11}+...+\dfrac{8}{n\left(n+4\right)}\)\(=\dfrac{2}{3}-\dfrac{2}{7}+\dfrac{2}{7}-\dfrac{2}{11}+...+\dfrac{2}{n}-\dfrac{2}{n+4}=\dfrac{2}{3}-\dfrac{2}{n+4}\)
\(\Rightarrow A=\dfrac{1}{6}-\dfrac{1}{2\left(n+4\right)}\)
- Xét hiệu \(A-\dfrac{1}{6}=-\dfrac{1}{2\left(n+4\right)}< 0\)
Vậy A < 1/6
NL
2
HN
5 tháng 5 2016
\(A=\frac{4^2}{3.7}+\frac{4^2}{7.11}+\frac{4^2}{11.15}+...+\frac{4^2}{107.111}\)
\(A=\) \(4\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{107.111}\right)\)
\(A=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)
\(A=4\left(\frac{1}{3}-\frac{1}{111}\right)\)
\(A=4.\frac{12}{37}\)
\(A=\frac{48}{37}\)
2 tháng 8 2015
a.\(A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)
\(=\frac{1}{3}-\frac{1}{111}=\frac{37}{111}-\frac{1}{111}=\frac{36}{111}=\frac{12}{37}\)
Vậy A=\(\frac{12}{37}\)
b.\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}=\frac{2}{7}\)
Vậy \(B=\frac{2}{7}\)
c.\(C=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
\(\Rightarrow C.\frac{1}{2}=\left(\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\right).\frac{1}{2}\)
\(=\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\)
\(=\frac{1}{4}-\frac{1}{16}=\frac{4}{16}-\frac{1}{16}=\frac{3}{16}\)
Vậy \(C=\frac{3}{16}\)
HT
2 tháng 8 2015
A = \(\frac{4}{3.7}+\frac{4}{7.9}+...+\frac{4}{107.111}\)
A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{107}-\frac{1}{111}\)
A = \(\frac{1}{3}-\frac{1}{111}\)=\(\frac{12}{37}\)
2 câu sau tương tự. Mik ngại làm lắm -_-
NB
2
\(\frac{2}{3\cdot7}+\frac{2}{7\cdot11}+...+\frac{2}{71\cdot75}+\frac{2}{75\cdot79}\)
\(=\frac{2}{4}\left[\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{71\cdot75}+\frac{4}{75\cdot79}\right]\)
\(=\frac{2}{4}\left[\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{75}-\frac{1}{79}\right]\)
\(=\frac{1}{2}\left[\frac{1}{3}-\frac{1}{79}\right]=\frac{38}{237}\)
\(\frac{2}{3\cdot7}+\frac{2}{7\cdot11}+...+\frac{2}{71\cdot75}+\frac{2}{75\cdot79}\)
\(=\frac{1}{2}\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{71\cdot75}+\frac{4}{75\cdot79}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{71}-\frac{1}{75}+\frac{1}{75}-\frac{1}{79}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{79}\right)\)
\(=\frac{1}{2}\cdot\frac{76}{237}\)
\(=\frac{38}{237}\)