\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{94.97}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{94}-\dfrac{1}{97}\)

\(=1-\dfrac{1}{97}\)

\(=\dfrac{96}{97}\)

\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)

\(=3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\right)\)

\(=3\left(1-\dfrac{1}{97}\right)\)

\(=3.\dfrac{96}{97}=\dfrac{288}{97}\)

\(=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{37.40}\right)\)

\(=\dfrac{1}{3}.\left(3-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{3}{7}+\dfrac{3}{7}-\dfrac{3}{10}+...+\dfrac{3}{37}-\dfrac{3}{40}\right)\)

= \(\dfrac{1}{3}.\left(3-\dfrac{3}{40}\right)\)

= \(\dfrac{1}{3}.\dfrac{117}{40}\)

\(=\dfrac{39}{40}\)

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

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

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

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

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

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

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

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

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

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+........+\dfrac{3}{100.103}\)

\(\Leftrightarrow S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+.........+\dfrac{1}{100}-\dfrac{1}{103}\)

\(\Leftrightarrow S=1-\dfrac{1}{103}=\dfrac{102}{103}\)

S=\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\)

S=\(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

S=\(1-\dfrac{1}{103}=\dfrac{102}{103}\)

\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(S=1-\dfrac{1}{103}\)

\(S=\dfrac{102}{103}\)

~ chúc bn hk tốt ~

Đặt A = \(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{91.94}+\dfrac{2}{94.97}\)

\(\dfrac{3A}{2}\)= \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{91.94}+\dfrac{3}{94.97}\)

\(\dfrac{3A}{2}=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{91}-\dfrac{1}{94}+\dfrac{1}{94}-\dfrac{1}{97}\)

\(\dfrac{3A}{2}\) = \(\dfrac{1}{1}-\dfrac{1}{97}\)

\(\dfrac{3A}{2}\) = \(\dfrac{96}{97}\)

3A = \(\dfrac{96}{97}.2\)

3A = \(\dfrac{192}{97}\)

A = \(\dfrac{192}{97}:3\)

A = \(\dfrac{64}{97}\)

Vậy A = \(\dfrac{64}{97}\)

\(\dfrac{2}{1.4}\)+\(\dfrac{2}{4.7}\)+\(\dfrac{2}{7.10}\)+....+\(\dfrac{2}{91.94}\)

=2.(1/1.4+1/4.7+1/7.10+...+1/91.94)

=2.\(\dfrac{3}{3}\).(1/1.4+1/4.7+1/7.10+...+1/91.94)

=\(\dfrac{2}{3}\).(3/1.4+3/4.7+3/7.10+...+3/91.94)

=\(\dfrac{2}{3}\).(1/1-1/4+1/4-1/7+1/7-1/10+.....1/91-1/94

=\(\dfrac{2}{3}\).(1-1/94)

=\(\dfrac{2}{3}\).93/94

=31/47

\(\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}+...+\dfrac{2}{91\cdot94}+\dfrac{2}{94\cdot97}\)

\(=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{91\cdot94}+\dfrac{3}{94\cdot97}\right)\)

\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{91}-\dfrac{1}{94}+\dfrac{1}{94}-\dfrac{1}{97}\right)\)

\(=\dfrac{2}{3}\left(1-\dfrac{1}{97}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{97}{97}-\dfrac{1}{97}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{96}{97}\)

\(=\dfrac{64}{97}\)

`S_1 = 5/(1.4) + 5/(4.7) +...+ 5/(97.100)`

`S_1 = 5 (1/(1.4) + 1/(4.7) +...+ 1/(97.100))`

`S_1 = 5/3 (3/(1.4) + 3/(4.7) +...+ 3/(97.100))`

`S_1 = 5/3 (1 - 1/4 + 1/4 - 1/7 + ...+ 1/97 - 1/100)`

`S_1 = 5/3 (1 - 1/100)`

`S_1 = 5/3 . 99/100`

`S_1 = 33/20`