=333300

mình đoán thế

Đặt A=1.2+2.3+...+99.100

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

A.3=1.2.[3-0]+2.3.[4-1]+...+99.100.[101-98]

A.3=1.2.3+2.3.4-1.2.3+...+99.100.101-99.100.98

A.3=99.100.101

A.3=999900

A=333300

A = 1.2+2.3+3.4+4.5+...+98.99+99.100

3A = 1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)

3A = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

3A = 99.100.101

3A = 999900

A = 333300

nhấn đúng cho mk nha!!!!!!!!!!!!

Cái này trên mạng có            

Ta có : B = 1.2 + 2.3 + 3.4 + ... + 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 = 333300

Vậy B = 333300

           Bài làm :

Ta có :

B= 1.2 + 2.3 + 3.4 + ...+ 99.100

=>3B = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
<=>3B= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
<=>3B= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
<=>3S = 99.100.101

<=> 3S = 999900

<=> B = 999900 : 3 = 333300

Vậy B = 333300

Có: \(3a\left(a+1\right)=\left[\left(a+2\right)-\left(a-1\right)\right].a\left(a+1\right)\)

\(=a\left(a+1\right)\left(a+2\right)-\left(a-1\right)a\left(a+1\right)\)

Xét \(B=1.2+2.3+3.4+...+98.99+99.100\)

\(\Rightarrow3B=3.1.2+3.2.3+3.3.4+...+3.98.99+3.99.100\)

\(=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(=-0.1.2+99.100.101=99.100.101\)

\(\Rightarrow B=33.100.101\)

Đặt A = 1.2 + 2.3 + 3.4 + ... + 99.100

3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 99.100.(101-98)

3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

3A = 99.100.101

A = 33.100.101

A = 333300

\(A=1.2+2.3+3.4+4.5+...+97.98+98.99+99.100\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100\)

\(3A=99.100.101\)

\(A=\frac{99.100.101}{3}=\frac{999900}{3}=333300\)

3A=1.2.3+2.3.(4-1)+.............+98.99.(100-97)+99.100.(101-98)

3A=1.2.3+2.3.4-1.2.3+...........+98.99.100-97.98.99+99.100.101-98.99.100

3A=99.100.101

A=99.100.101:3

A=333300

Ta có : 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + 98.99.3 + 99.100.3

=> 3A = 1.2.( 3 - 0 ) + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + ..... + 98.99.( 100 - 97 ) + 99.100.( 101 - 98 )

=> 3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ..... + 98.99.100 - 97.98.99 + 99.100.101 - 98.99.100

=> 3A = ( 1.2.3 + 2.3.4 + 3.4.5 + ..... + 98.99.100 + 99.100.101 ) - ( 0.1.2 + 1.2.3 + 2.3.4 + ..... + 98.99.100 )

=> 3A = 99.100.101 - 0.1.2

=> 3A = 99.100.101

=> A = 33.100.101

=> A = 333300

Đặt A= 1.2 + 2.3 + 3.4 + ...+ 99.100
 3A = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
3A= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
3A= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
3A = 99.100.101  3S = 3.33.100.101 
 A=33.100.101= 333300

A= 1.2 + 2.3 + 3.4 + ...+ 99.100

3A = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3

3A= 1.2.3+2.3﴾4‐1﴿+3.4﴾5‐2﴿+...+98.99﴾100‐97﴿+99.100﴾101‐98﴿  

3A= 1.2.3+2.3.4‐1.2.3+3.4.5‐2.3.4+...‐97.98.99+99.100.101‐98.99.100

3A = 99.100.101 3S = 3.33.100.101

A=33.100.101= 333300

1.2 là 1 x 1 hay 1,2 vậy

Áp dụng công thức ta có :

\(A=1.2+2.3+3.4+...+99.100=\frac{99.100.101}{3}=333300\)

A=1.2+2.3+3.4+4.5+.....+98.99+99.100 Rút gọn đi ta còn:

A=1+100

=>A=101

Ta có : A = 1.2 + 2.3 + 3.4 + ....... + 99.100

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

=> 3A = 99.100.101

=> A = 99.100.101/3

=> A = 333300

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

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

1/1 - 1/100                                                         =   99/100

98.99/99.100

99/100