\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{43.46}\\ S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\\ S=1-\dfrac{1}{46}< 1\)

Vậy S < 1 (đpcm)

\(S=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{43\cdot46}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\)

\(S=1-\dfrac{1}{46}< 1\)

S= \(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{40\cdot43}+\dfrac{3}{43\cdot46}\)

S= \(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{42}-\dfrac{1}{46}\)

S= \(1-\dfrac{1}{46}\)

S= \(\dfrac{45}{46}\)

Mà \(\dfrac{45}{46}< 1\)

\(\Rightarrow S< 1\)

Vậy S < 1

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...................+\dfrac{3}{n\left(n+1\right)}\)

\(\Rightarrow S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.............+\dfrac{1}{n}-\dfrac{1}{n+1}\)

\(\Rightarrow S=1-\dfrac{1}{n+1}< 1\)

\(\Rightarrow S< 1\rightarrowđpcm\)

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n.\left(n+1\right)}\)

\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...-\dfrac{1}{n+1}\)

\(S=1-\dfrac{1}{n+1}\)\(< 1\)

\(\Leftrightarrow S< 1\)

tik cho mik nhé

\(S=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{40\cdot43}+\dfrac{3}{43\cdot46}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\)

\(=1-\dfrac{1}{46}< 1\)

Vậy S < 1 ( ĐPCM )

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n\left(n+3\right)}\)

\(\Rightarrow S=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{\left(n+3\right)-n}{n\left(n+3\right)}\)

\(\Rightarrow S=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{n+3}{n\left(n+3\right)}-\dfrac{n}{n\left(n+3\right)}\)

\(\Rightarrow S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{n}-\dfrac{1}{n+3}\)

\(\Rightarrow S=1-\dfrac{1}{n+3}< 1\Rightarrow S< 1\)

Vậy S < 1

\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}\\ =1-\dfrac{1}{43}\\ =\dfrac{42}{43}\)

e) 3/1.4 + 3/4.7 + 3/7.10+ ... + 3/40.43
= 1-1/4 + 1/4 -1/7 + 1/7-1/10+...+1/40-1/43
= 1-1/43
= 42/43

\(=-\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{61}-\dfrac{1}{64}\right)=-\dfrac{1}{63}\)

\(S=\) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}\)

\(=\dfrac{99}{100}\)

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{97.100}\)

\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+....+\dfrac{1}{97}-\dfrac{1}{100}\)

(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với mọi \(a\in N\)*)

\(S=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Vậy \(S=\dfrac{99}{100}\)

Chúc bạn học tốt!!!

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{97.100}\)

\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.....+\dfrac{1}{97}-\dfrac{1}{100}\)

\(S=1-\dfrac{1}{100}\)

\(S=\dfrac{99}{100}\)

\(B=1-\dfrac{3}{1\cdot4}-\dfrac{3}{4\cdot7}-...-\dfrac{3}{2020\cdot2023}\\ =1-\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2020\cdot2023}\right)\\ =1-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2020}-\dfrac{1}{2023}\right)\\ =1-\left(1-\dfrac{1}{2023}\right)\\ =1-\dfrac{2022}{2023}=\dfrac{1}{2023}\)

`B=1-3/(1.4)-3/(4.7)-3/(7.10)-....-3/(2020.2023)`

`B=1-(3/(1.4)+3/(4.7)+.....+3/(2020.2023))`

`B=1-(1-1/4+1/4-1/7+.....+1/2020-1/2023)`

`B=1-(1-1/2023)`

`B=1-1+1/2023=1/2023`

A = \(-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2020}\right)=-1+\dfrac{1}{2020}=\dfrac{-2019}{2020}\)