Gọi biểu thức trên là A, ta có :

A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100

A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3

A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)

A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.

A x 3 = 99x100x101

A = 99x100x101 : 3

A = 333300

Tổng = 10000 ( mình nghĩ vậy )

A = 2  +  ( 2+ 1).4  +  ( 4 + 1)6  + … +  (98 + 1).100

    = 2  +  2.4 + 4  +  4.6 + 6 + … +  98.100  +  100

    = (2.4 +  4.6 + … +  98.100 )  +  (2  +  4  +  6 + … +  100)

    = 98.100.102 : 6 + 102.50:2

    = 166600 + 2550

    = 169150

= 333300

B x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3

= 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)

= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.

= 99x100x101

B = 99x100x101 : 3

= 333300

nhanh k minh

B= 1x2+3x4+5x6+...+99x100

=> Bx3= 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ...+ 99x100x3

=> Bx3= 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3)+...+99x100x(101-98)

=> Bx3= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 +4x5x6 - 3x4x5 +...+ 99x100x101 - 98x99x100

=> Bx3= 99x100x101

=> B= 99x100x101:3

=> B= 333300

A  = 1 x 2 + 2 x 3 + ....... + 10 x 11

3A = 1 x 2 x 3 + 2 x 3 x 3 + ..........+ 10 x 11 x 3

3A = 1 x 2 x (3-0) + 2 x 3 x (4-1) + .......... + 10 x 11 x (12 -9)

3A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + ........... + 10 x 11 x 12 - 9 x 10 x 11

3A = (1 x 2 x 3 - 1 x 2 x 3) + ( 2 x 3 x 4 - 2 x 3 x 4) +............ + 10 x 11 x 12

3A = 10 x 11 x 12 = 1320

A = 1320 : 3 = 440

Gọi biểu thức trên là A, ta có :

A= 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100

A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3

A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)

A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.

A x 3 = 99x100x101 A = 99x100x101 : 3 A = 333300

 

1x 2 + 2 x 3 + 3 x 4 + ...+ 99 x 100

Ta có: 

1 x 2 x 3                                           = 1 x 2 x 3

2 x 3 x 3     = 2 x 3 x ( 4 - 1)             = 2 x 3 x 4 -  1 x 2 x 3

3 x 4 x 3     =  3 x 4 x ( 5 - 2)            = 3 x 4 x 5 -   2 x 3 x 4 

........................................................= ........................................

99 x 100 x 3 = 99 x 100 x (101 - 98)  = 99 x 100 x 101 - 99 x 100 x 98

Cộng vế với vế ta có:

1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 +...+ 99 x 100 x 3 = 99 x100 x 101

(1 x 2 + 2 x 3 + 3 x 4 +...+ 99 x 100) x 3 =  99 x 100 x 101 

1 x 2 + 2 x 3 + 3 x 4 +...+ 99 x 100 = \(\dfrac{99\times100\times101}{3}\)

1 x 2 + 2 x 3 + 3 x 4 + ....+ 99 x 100 = 333300

333300

Ta có \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

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

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

Lại có B = \(\frac{1}{51.100}+\frac{1}{52.99}+...+\frac{1}{99.52}+\frac{1}{100.51}\)

=> 151B = \(\frac{151}{51.100}+\frac{151}{52.99}+...+\frac{151}{99.52}+\frac{151}{100.51}\)

=> 151B = \(\frac{1}{51}+\frac{1}{100}+\frac{1}{52}+\frac{1}{99}+...+\frac{1}{99}+\frac{1}{52}+\frac{1}{100}+\frac{1}{51}\)

=> 151B = \(2\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)

=> B = \(\frac{2}{151}.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)

Khi đó \(\frac{A}{B}=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{2}{151}\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)}=\frac{1}{\frac{2}{151}}=\frac{151}{2}=75,5\)

A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100

A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3 A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98) ..................................

A x 3 = 99x100x101 A = 333300