Ta có: \(S=1.2+2.3+3.4+...+99.100\)

\(\Rightarrow3S=1.2.3+2.3.3+3.3.4+....+99.100.3\)

\(\Rightarrow3S=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)....99.100.\left(101-98\right)\)

\(\Rightarrow3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(\Rightarrow3S=99.100.101\)

\(\Rightarrow S=\frac{99.100.101}{3}=\frac{999900}{3}=333300\)

S=  1.2 + 2.3 +... + 99.100

=>S= \(\frac{99.100.101}{3}\)=333300

ko ghi lại đề bài 

=1/1-1/2+1/2-1.3+...+1/99-1/100

=1/1-1/100

=99/100

hc tốt

ko ghi lại đề 

=1/1-1/2+1/2-1/3+...+1/99-1/100

=1/1-1/100

=99/100

\(S=\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{98\times99}+\frac{2}{99\times100}\)

\(S=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)

\(S=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(S=2\times\left(1-\frac{1}{100}\right)\)

\(S=2\times\frac{99}{100}\)

\(S=\frac{99}{50}\)

\(S=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{98.99}+\frac{2}{99.100}\)

\(S=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(S=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}+\frac{1}{100}\right)\)

\(S=2.\left(\frac{1}{1}-\frac{1}{100}\right)\\ S=2.\left(\frac{100}{100}+\frac{-1}{100}\right)\\ S=2.\frac{99}{100}\\ S=\frac{99}{50}\)

để tiện tính toán ta đặt tử số cần tính và mẫu số lần lượt là A và B

xét A:

\(A=1+\left(1+2\right)+\left(1+2+3\right)+\cdots+\left(1+2+3+\cdots+100\right)\)

=> \(A=\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+\cdots+\frac{100.101}{2}\)

\(A=\frac12\left(1\cdot2+2\cdot3+3\cdot4+\cdots+100\cdot101\right)\)

để tính phần trong ngoặc ta nhân riêng 3 với nó:

\(\Rightarrow3\cdot\left(1\cdot2+2\cdot3+\cdots+100\cdot101\right)\)

\(=1\cdot2\cdot3+2\cdot3\cdot3+\cdots+100\cdot101\cdot3\)

\(=\left(1\cdot2\cdot3\right)+\left.2\cdot3\cdot\left(4-1\right)+.\ldots+100\cdot101\cdot\left(102-99\right)\right)\) \(=\left(1\cdot2\cdot3\right)+\left(2\cdot3\cdot4-1\cdot2\cdot3\right)+\cdots+\left(99\cdot100\cdot101-98\cdot99\cdot100\right)+\left(100\cdot101\cdot102-99\cdot100\cdot101\right)\)

\(=100\cdot101\cdot102\)

=> biểu thức trong ngoặc là: \(\frac{100\cdot101\cdot102}{3}\)

=> \(A=\frac12\cdot\frac{100.101.102}{3}\)

CMTT: =>3B=\(99\cdot100\cdot101\)

=> \(B=\frac{99.100.101}{3}\)

=> \(F=\frac{\left(\frac12\cdot\frac{100.101.102}{3}\right)}{\frac{99.100.101}{3}}\)

=> \(F=\frac12\cdot\frac{102}{99}\)

\(F=\frac{17}{33}\)

đặt A=1.2+2.3+...+99.100

=>3A=1.2.3+2.3.3+...+99.100.3

=1.2.3+2.3.(4-1)+...+99.100(101-98)

=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100

=99.100.101=999900

=>A=999900:3=333300

\(a=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Cho A = 1.2 + 2.3 + ...+ 99.100

=> 3A = 1.2 .3 + 2.3.3 + ...+ 99.100.3

3A = 1.2.( 3-0) + 2.3.(4-1) + ....+ 99.100.( 101 - 98)

3A = 1.2.3 + 2.3.4 - 1.2.3 + ...+ 99.100.101 - 98.99.100

3A = ( 1.2.3 + 2.3.4 + 99.100.101) - ( 1.2.3 + ....+ 98.99.100)

3A = 99.100.101

=> A = 99.100.101 . 1/3

thay A vào B

\(B=(\frac{99.100.101.\frac{1}{3}}{99.100.101}):\frac{1}{3}\)

\(B=\frac{1}{3}:\frac{1}{3}\)

\(B=1\)

\(B=\left(\frac{1.2+2.3+...+99.100}{99.100.101}\right)\div\frac{1}{3}\)

\(\text{Đặt}:C=1.2+2.3+...+99.100\)

\(\Rightarrow3C=1.2.3+2.3.3+...+99.100.3\)

\(\Rightarrow3C=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)

\(\Rightarrow3C=1.2.3+2.3.4+...+99.100.101\)

\(\Rightarrow3C=\left(1.2.3+2.3.4+...+99.100.101\right)\)\(-\)\(\left(1.2.3+2.3.4+....+98.99.100\right)\)

\(\Rightarrow3C=99.100.101\)

\(\Rightarrow C=\frac{99.100.101}{3}\)

Thay C vào biểu thức B ta được : 

\(B=\left(\frac{\frac{99.100.101}{3}}{99.100.101}\right)\div\frac{1}{3}=\frac{1}{3}\div\frac{1}{3}=1\)

Vậy B= \(1\)

\(NGU=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

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

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

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

=1/1-1/2+1/3-1/4+..........+1/99-1/100

=1/1-1/100

=99/100