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

Ta có công thức : \(\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n\left(n+1\right)}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

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

Ta có :

1/1.2  + 1/2.3   +.............+ 1/99.100

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

=1/1 -1/100 

=99/100

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

A=1-1/100

A=99/100

ai k mk      mk k lai 

A=1-1/100

A=99/100

Nha bạn       

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

A =  1 - 1/100

A = 99/100

Ai k mk mk k lại !

5050 đấy bạn mình cũng không chắc lắm

 S = 1.2 + 2.3 + 3.4 + ..... + 99.100

=> 3S = 1.2.3 + 2.3(4 - 1) + 3.4(5 - 2) + ......... + 99.100(101 - 98)

=> 3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ........ + 99.100.101 - 98.99.100

=> 3S = (1.2.3 + 2.3.4 + 3.4.5 + ..... + 98.99.100 + 99.100.101) - (1.2.3 + 2.3.4 + .......... + 98.99.100)

=> 3S = 99.100.101

=> S = \(\frac{99.100.101}{3}=333300\)

Đặt S = 1 x 2 + 2 x 3 + 3 x 4 +... + 99 x 100

3 S = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + ... + 98 x 99 x 3 + 99 x 100 x 3

3 S = 1 x 2 x 3 + 2 x 3 ( 4 - 1 ) + 3 x 4 ( 5 - 2 ) + ... + 98 x 99 ( 100 - 97 ) + 99 x 100 ( 101 - 98 )

3 S = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + ... - 97 x 98 x 99 + 99 x 100 x 101 - 98 x 99 x 100

3 S = 99 x 100 x 101  3S = 3 x 33 x100 x 101

S = 33 x 100 x 101 = 333 300

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)

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

=1-1/100

=99/100

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

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100

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

= 1 - 1/100

= 99/100

v

Answer:

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

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

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

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

A= 1-2+3-4+4-5+...+99-100

A = ( 1 - 2 ) + ( 2 - 3 ) + ....+ ( 99 - 100 )

A = ( - 1 ) + ( - 1 ) +....+ ( - 1 )

A = ( - 1 ) . 50

A = - 50

B = 1.2 + 2.3 + 3.4 + 4.5 +...+ 99.100 
Nhân cả 2 vế với 3, ta được: 
3A=1.2.3+ 2.3.3+ 3.4.3+ 4.5.3+...... 99.100.3 
= 1.2.3 + 2.3(4-1) + 3.4.(5-2) +...+ 99.100.(101-98) 
= 1.2.3 + 2.3.4 -1.2.3 + 3.4.5-2.3.4 +...+ 99.100.101-98.99.100 
= 99.100.101 
=) B = (99.100.101) :3 
B = 333300  
Vậy  B= 333300 

A= 1-2+3-4+4-5+...+99-100

A = (1-2) + (3-4) + (4-5) + ... + (99-100)

A = (-1) + (-1) + (-1) + ...+ (-1)

A = (-1).50

A = 1

B = 1.2+2.3+3.4+4.5+...+99.100

3B = 1.2.3 +2.3.3+ 3.4.3+...+99.100.3 
3B = 1.2.3 +2.3(4-1)+ 3.4(5-2)+...+99.100(101-98) 
3B = 1.2.3 +2.3.4-1.2.3+ 3.4.5-2.3.4+...+99.100.101-98.99.100 
3B = 99.100.101

3B = 999900 
B = 999900 : 3

B = 333300

hi bn

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

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

1/(1.2)+1/(2.3)+1/(3.4)+...+1/(99.100)

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

=1-1/100

=99/100

tôi không chép bài giang ho đai ca đâu nha.

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

=1-1/100

=99/100