Ta thấy:
Số hạng thứ nhất:    ¹/₂ = ¹/_1.2
Số hạng thứ hai:    ¹/₆ = ¹/_2.3
Số hạng thứ ba:    ¹/₁₂ = ¹/_3.4
Số hạng thứ tư:    ¹/₂₀ = ¹/_4.5
..................................................
Số hạng thứ một trăm:    ¹/_100.101
Vậy tổng của 100 số hạng đầu của dãy là:
¹/_1.2 + ¹/_2.3 + ¹/_3.4 + ¹/_4.5 + ... + ¹/_100.101
=> 1 - ¹/₂ + ¹/₂ - ¹/₃ + ¹/₃ - ¹/₄ + ¹/₄ - ¹/₅ + ... + ¹/₁₀₀ - ¹/₁₀₁
=> 1 - ¹/₁₀₁
=> ¹⁰⁰/₁₀₁

ý b bằng 100/501

ý c bằng 100/101

Giải rõ ra chứ

Em ơi chỗ 1/100 sửa lại thành 1/110 nha

B=\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)

=\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)(Dấu . là nhân nha)

Áp dugj tổng quát \(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)ta có:

B=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

=\(\frac{1}{2}-\frac{1}{11}\)=\(\frac{9}{22}\)

H=\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\cdot\cdot\left(1-\frac{1}{2004}\right)\)

=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\cdot\cdot\frac{2003}{2004}\)

=\(\frac{1}{2004}\)

\(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...........+\frac{1}{90}\)

\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{9.10}+\frac{1}{10.10}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..........-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(H=\frac{1}{2}\times\frac{2}{3}\times..........\times\frac{2003}{2004}=\frac{1\times2\times3\times.............\times2003}{2\times3\times4\times.................\times2004}=\frac{1}{2004}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

=\(1-\frac{1}{6}\)

=\(\frac{5}{6}\)

  1/2 + 1/6 + 1/12 + 1/20 + 1/30 [mẫu chung là 60 ]

= 1x30/2x30 + 1x10/6x10 + 1x5/12x5 + 1x3/20x3 + 1x2/30x2

= 30/60 + 10/60 + 5/60 + 3/60 + 2/60

= 50 / 60 [rút gọn = 5/6 ]

`1/2+2/4+3/6+4/8+5/10+6/12`

`=1/2+1/2+1/2+1/2+1/2+1/2`

`=1/2*6=3`

`1/3+1/4+1/5+8/10+20/15+20/30`

`=(1/3+1/4)+(1/5+4/5)+(4/3+2/3)`

`=7/12+1+2`

`=7/12+3=43/12`

\(\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{6}+\dfrac{4}{8}+\dfrac{5}{10}+\dfrac{6}{12}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\)
\(=\dfrac{1}{2}\times6=3\)
\(------\)
\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{8}{10}+\dfrac{20}{15}+\dfrac{20}{30}\)
\(=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{4}{5}+\dfrac{4}{3}+\dfrac{2}{3}\)
\(=\left(\dfrac{1}{3}+\dfrac{4}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{5}+\dfrac{4}{5}\right)+\dfrac{1}{4}\)
\(=\dfrac{7}{3}+1+\dfrac{1}{4}\)
\(=\dfrac{28}{12}+\dfrac{12}{12}+\dfrac{3}{12}\)
\(=\dfrac{43}{12}\)

`=1/2+1/2+1/2+1/2+1/2+1/2= 1/2 \times 6=3`

`----`

`=1/3+1/4+1/5+4/5+4/3+2/3 =(4/3+2/3+)+(1/5+4/5)+1/3+1/4=2+1+1/3+1/4=3+1/3+1/4=43/12`

1/2+1/6+1/12+1/20+1/30+1/42

=1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+1/6*7

=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+(1/5-1/6)+(1/6-1/7)

=1-1/7

=6/7

neu co gi sai sot xin gui lai loi nhan

6/7 day minh vua lam dung ne o violympic do.

k cho minh nha cac ban.

đáp án của bài toán này là 6/7 còn cách làm thì làm như bn Hoang Ha Vy

A= 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 

=1/(1.2)+1/(2.3)+1/(3.4)+1/(4.5) +1/(5.6)+1/(6.7)+1/(7.8) +1/(8.9)+1/(9.10) 

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5.+1/5-1/6... +1/9-1/10 

=1-1/10 

=9/10

Me too ! 

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}=x-\left(2+4+..+100\right)\)

Gọi \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)là \(A\)

         \(\left(2+4+...+100\right)\)là \(B\). Ta có :  

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\)

\(A=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)

Số số hạng của \(B\)là: \(\left(100-2\right)\div2+1=50\)

Tổng của \(B\)là: \(\left(2+100\right)\times50\div2=2550\)

\(\Rightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=x-\left(2+4+..+100\right)=\frac{9}{10}=x-2550\)

\(\Rightarrow x=2550+\frac{9}{10}=2550+0,9=2550,9\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{10100}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{101-100}{100.101}\)

\(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+\frac{5}{4.5}-\frac{4}{4.5}+\frac{6}{5.6}-\frac{5}{6.5}+...+\frac{101}{100.101}-\frac{100}{100.101}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

A= 1/1.2 +1/2.3 + 1/3.4 + 1/4.5 +...............+ 1/100.101

A= 1 - 1/2 +1/2 - 1/3 + 1/3 - 1/4 +..................+1/100 - 1/101

A= 1 - 1/101

A = 100/101

sau đây là phần chữa của mình: 

\(=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)

\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

 \(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{10}\)

= \(\dfrac{3}{10}\)

= \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

= \(\dfrac{1}{2}-\dfrac{1}{10}\)

= \(\dfrac{2}{5}\)

chỗ đầu mình quên cộng thêm 1/2 rồi bạn nhớ bổ sung nha thì sẽ ra kết quả đúng nhé