*S=1-1/4+1/4-1/7+1/7-1/11+1/11-1/14+1/14-1/17

S=1-1/17=16/17

*M=2(1/1.2+1/2.3+...+1/15.16)

M=2(1-1/2+1/2-1/3+..+1/15-1/16)

M=2(1-1/16)

M=2.15/16

M=15/8

:w

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(S=1-\frac{1}{17}\)

\(S=\frac{16}{17}\)

\(M=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{15.16}\)

\(M=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(M=2.\left(1-\frac{1}{16}\right)\)

\(M=2.\frac{15}{16}\)

\(=\frac{30}{16}=\frac{15}{8}\)

a) 4/ 3x7 + 4/7x11+ 4/11x15+...+ 4/107x111

=1/3-1/7+ 1/7-1/11+ 1/11- 1/15+...+1/107 - 1/111

= 1/3-1/111

=12/37

\(b,\frac{3^2}{8\cdot11}+\frac{3^2}{11\cdot14}+\frac{3^2}{14\cdot17}+...+\frac{3^2}{197\cdot200}\)

\(=3\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{197\cdot200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)

\(=3\cdot\frac{3}{25}=\frac{9}{25}\)

mình nghĩ bài náy sai đề rồi

sửa lại đề \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(S=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{11-7}{7.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(S=1-\frac{1}{17}=\frac{16}{17}\)

\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{14}-\frac{1}{17}=1-\frac{1}{17}=\frac{16}{17}\)

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(\Rightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{14}-\frac{1}{17}\)

\(\Rightarrow S=1-\frac{1}{17}\)

\(\Rightarrow S=\frac{16}{17}\)

Vậy S = \(\frac{16}{17}\)

\(\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+...+\dfrac{3^2}{197.200}\)

=\(3.\left(\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{197.200}\right)\)

=\(3.\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)

=\(3.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)

=\(3.\dfrac{3}{25}=\dfrac{9}{25}\)

Ta có: \(\dfrac{3^2}{8\cdot11}+\dfrac{3^2}{11\cdot14}+\dfrac{3^2}{14\cdot17}+...+\dfrac{3^2}{197\cdot200}\)

\(=3\left(\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+...+\dfrac{3}{197\cdot200}\right)\)

\(=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)

\(=3\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)

\(=3\cdot\dfrac{24}{200}=\dfrac{72}{200}=\dfrac{9}{25}\)

\(B=3.\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{197\cdot200}\right)=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{200}\right)=3\left(\frac{25}{200}-\frac{1}{200}\right)=3\cdot\frac{24}{200}=\frac{72}{200}=\frac{9}{25}\)

mk đi thi hsg mk bik chắc chắn mk ko sai đâu k mk nhé
3/1.4 +3/4.7+....+3/14.17
=1/1-1/4+1/4-1/7+...+1/14-1/17
=1-1/17
=16/17

\(\frac{3}{1}-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+...\)\(\frac{3}{14}-\frac{3}{17}\)

\(\frac{3}{1}-\frac{3}{17}\)

= \(\frac{18}{7}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(=1-\frac{1}{17}\)

\(=\frac{16}{17}\)

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!

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