`#3107.101107`

1.

a)

`1/(1*4) + 1/(4*7) + 1/(7*10) + ... + 1/(100*103)`

`= 1/3 * (3/(1*4) + 3/(4*7) + 3/(7*10) + ... + 3/(100*103) )`

`= 1/3 * (1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)`

`= 1/3* (1 - 1/103)`

`= 1/3*102/103`

`= 34/103`

b)

`-1/3 + (-1/15) + (-1/35) + (-1/63) + ... + (-1/9999)`

`= - 1/3 - 1/15 - 1/35 - 1/63 - ... - 1/9999`

`= - (1/3 + 1/15 + 1/35 + ... + 1/9999)`

`= - (1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/99*101)`

`= - 1/2 * (2/(1*3) + 2/(3*5) + 2/(5*7) + ... + 2/99*101)`

`= - 1/2* (1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)`

`= -1/2 * (1 - 1/101)`

`= -1/2*100/101`

`= -50/101`

2.

`3/(1*4) + 3/(4*7) + ... + 3/(94*97) + 3/(97*100)`

`= 1 - 1/4 + 1/4 - 1/7 + ... + 1/94 - 1/97 + 1/97 - 1/100`

`= 1-1/100`

`= 99/100`

\(A=\frac{9}{1.4}+\frac{9}{4.7}+\frac{9}{7.10}+...+\frac{9}{97.100}\)

\(A=9\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)

\(A=9.\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...-\frac{1}{100}\right)\)

\(A=\frac{9}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(A=3\left(\frac{99}{100}\right)=\frac{297}{100}\)

`3/1.4+3/4.7+3/7.10+...+3/94.97`

`=1/1-1/4+1/4-1/7+1/7-1/10+...+1/94-1/97`

`=1-1/97`

`=96/97`

\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\\ =1-\dfrac{1}{97}=\dfrac{96}{97}\)

tớ ko biết

S = 3 - \(\frac{3}{100}\)= \(\frac{300}{100}-\frac{3}{100}=\frac{297}{100}\)

S=3/1.4+3/4.7+3/7.10+.....+3/97.100

S=1/1-1/4+1/4-1/7+1/7-1/10+.....+1/97-1/100

S=1-1/100

S=99/100

3/1.4+3/4.7+....+3/97.100

= 1-1/4+1/4-1/7+....+1/97-1/100

=1-1/100

=99/100

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)

\(\Rightarrow A=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(\Rightarrow A=3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(\Rightarrow A=3\left(1-\frac{1}{100}\right)\)

\(\Rightarrow A=3.\frac{99}{100}\)

\(\Rightarrow A=3.\frac{99}{100}\)

\(\Rightarrow A=\frac{297}{100}\)

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

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

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

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

mà \(\frac{96}{97}< 1\)

\(\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.07}< 1\)

vậy..................

\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{91\cdot94}+\frac{3}{94\cdot97}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\)

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

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

\(\Rightarrow\frac{96}{97}< 1\)

\(\Rightarrow\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)

Vậy \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+....+\frac{3^2}{97.100}\)

\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)

\(A=3.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=3.\left(\frac{1}{1}-\frac{1}{100}\right)=3-\frac{3}{100}=\frac{297}{100}\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+\frac{3^2}{13.16}+...+\frac{3^2}{97.100}\)

\(A=\frac{3}{1}-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+\frac{3}{10}-\frac{3}{13}+\frac{3}{13}-\frac{3}{16}+...+\frac{3}{97}-\frac{3}{100}\)

\(A=\frac{3}{1}-\frac{3}{100}\)

\(A=\frac{297}{100}\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)

\(\Rightarrow A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)

\(A=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=3\left(1-\frac{1}{100}\right)\)

\(=3.\frac{99}{100}\)

\(=\frac{297}{100}\)

Chào bạn, bạn hãy theo dõi bài giải của mình nhé!

\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\)

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

\(=1-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

Chúc bạn học tốt!