Đặt A=1.2+2.3+3.4+...+99.100

3A=1.2.3+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

A=333300

A = 333 300 nhé bn

333300

**** mik nha bak

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

3S = 1 x 2 x (3-0) + 2 x 3 x (4-1) + ......+99 x 100 x (101 - 98)

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

3S = (1 x 2 x 3 - 1 x 2 x 3) + (2 x 3 x4 - 2 x 3 x4) +....... + (98 x 99 x 100 - 98x 99 x 100) + 99 x 100 x 101

3S = 99 x 100 x 101 = 999900

S = 999900 : 3 = 333300 

Vậy S = 333300

=>3D =1.2.3 + 2.3.3 + 3.4.3 + ..... + 99.100 .3

=> 3D = 1.2.3 - 2.3. ( 4-1) + 3.4. (5-2) + ... + 98.99 (100 - 97 ) + 99 . 100 . ( 101-98)

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

=> 3D= 99.100.101

=> 3D= 999 900 

D= 999 900 .3 = 333 300

=>3D =1.2.3 + 2.3.3 + 3.4.3 + ..... + 99.100 .3

=> 3D = 1.2.3 - 2.3. ( 4-1) + 3.4. (5-2) + ... + 98.99 (100 - 97 ) + 99 . 100 . ( 101-98)

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

=> 3D= 99.100.101

=> 3D= 999 900 

D= 999 900 .3 = 333 300

muốn đụ không ?

A=1/1x2+1/2x3+.....+1/99x100

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

A=1/1 - 1/100

A=99/100

ta có: A=1/1x2+1/2x3+...+1/99x100

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

A=1-1/100

A=99/100

 A=1/1x2+1/2x3+.....+1/99x100

BL:

A=1/1x2+1/2x3+.....+1/99x100

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

A=1/1 - 1/100

A=99/100

Tính tổng:

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

Giải

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

B = \(\dfrac{2}{1\times2}\) + \(\dfrac{2}{2\times3}\)+ \(\dfrac{2}{3\times4}\)+...+ \(\dfrac{2}{99\times100}\)

B = 2 \(\times\) ( \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\)+ \(\dfrac{1}{3\times4}\)+....+ \(\dfrac{1}{99\times100}\))

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

B = 2 \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\))

B = 2 \(\times\) \(\dfrac{99}{100}\)

B = \(\dfrac{99}{50}\)

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

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=100/100-1/100=99/100

99/100

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