Gọi biểu thức trên là A, ta có :

A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 2014x2015

A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 2014x2015x3

A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 2014x2015x(2016-2013)

A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 2014x2015x2016 - 2014x2015x2013.

A x 3 = 2014x2015x2016

A = 2014x2015x2016 : 3

A = 2727117120

1/1*2 + 1/2*3 + 1/3*4 + ... + 1/2013*2014 + 1/2014*2015

= 1 -1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2013 - 1/2014 + 1/2014 - 1/2015

=1-1/2015

=2014/2015

=1-1/2+1/2-1/3+...+1/2014-1/2015

=1-1/2015

=2014/2015

đây là toán lớp 5 mà

vì 1/1*2=1-1/2

   1/2*3=1/2-1/3

.....................

1/2014*2015=1/2014-1/2015

=1-1/2+1/2-1/3+1/3-....+1/2014-1/2015

=1-1/2015

=2014/2115

\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{2014x2015}\)

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

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

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

Cau nay kho

so tan cung cua a la chu so 0

tại sao lại là 0

ta có\(A=\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{2014\cdot2015}\)

             \(=4\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}\right)\)

             \(=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)

               \(=4\left(1-\frac{1}{2015}\right)\)

                \(=4\cdot\frac{2014}{2015}\)

                  \(=\frac{8056}{2015}\)

  VẬY A=\(\frac{8056}{2015}\)

\(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2004.2005}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2004}-\dfrac{1}{2005}\\ =1-\dfrac{1}{2005}\\ =\dfrac{2004}{2005}\)