B= 1.2+2.3+3.4+...+2009.2010

=>3B=1.2.3+2.3.3+3.4.3+...+2009.2010.3

=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+2009.2010.(2011-2008)

=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+....+2009.2010.2011-2008.2009.2010

=2009.2010.2011

=>B=\(\frac{2009.2010.2011}{3}=2706866330\)

ta có: 1x2+2x3+3x4+....+n(n+1) 
=1x(1+1)+2x(2+1)+3x(3+1)+....­n(n+1) 
=(1^2+2^2+3^2+¡­+n^2)+(1+2+3+....+n) 
=n(n+1)(2n+1)/6+n(n+1)/2 
=[n(n+1)[(2n+1)+3]/6

thay n=2009=> B=\(\frac{2009.\left(2009+1\right).\left(2009.2+1\right)+3}{6}\)=2704847286

=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{2009}\)-\(\frac{1}{2010}\)

=1+(\(\frac{-1}{2}\)+\(\frac{1}{2}\))+(\(\frac{-1}{3}\)+\(\frac{1}{3}\))+...+(\(\frac{-1}{2009}\)+\(\frac{1}{2009}\))-\(\frac{1}{2010}\)

=1+0+0+...+0-\(\frac{1}{2010}\)

=1-\(\frac{1}{2010}\)

=\(\frac{2010}{2010}\)-\(\frac{1}{2010}\)

=\(\frac{2009}{2010}\)

lớp 4 ghê nhỉ đã học bài này rùi tui lớp 6 mà mới học bài này

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

eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee2

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

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

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

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

=> 3A = 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + 99.100.101 - 98.99.100

=> 3A = 99.100.101

=> A = 99.100.101 : 3

=> A = 333300

Đặt S = 1 x 2 + 2 x 3 + ......... + 99 x100

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

3S = 99 x 100 x 101

S = 99 x 100 x 101 : 3 = 333300

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)

\(=1-\frac{1}{2006}\)

\(=\frac{2005}{2006}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)

= \(1-\frac{1}{2006}\)

= \(\frac{2005}{2006}\)

=330

ủng hộ mk nhé

330 là kết quả đúng đó bạn!

\(330\)nha bạn

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

3A = 1x2x3 + 2x3x(4-1) + ... + 99x100x(101-98)

3A = 1x2x3 + 2x3x4 - 1x2x3 + ... + 99x100x101 - 98x99x100

3A = 99x100x101

3A = 999900

A = 333300

Ta có:

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

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

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

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

Suy ra 3A=99x100x101

A=99x100x101/3

A=333300

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

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

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

A . 3 = 1. 2 . 3 + 2 . 3 . (4-1) + 3 .4 . (5 - 2 ) + 4. 5 . (6- 3 )+... + 99 . 100. (101 -98 ) 

A . 3 = 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.

A .  3 = 99 . 100 . 101

A = 99 . 100 . 101 : 3

A = 333 300