Đặt A , ta có :

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{999\times1000}+1\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

\(A=2-\frac{1}{1000}\)

\(A=\frac{2000}{1000}-\frac{1}{1000}\)

\(A=\frac{1999}{1000}\)

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}+1=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}+1\)

\(A=1-\frac{1}{1000}+1=\frac{999}{1000}+1=\frac{1999}{1000}\)

Vậy \(A=\frac{1999}{1000}\)

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

A = 1.2 + 2.3 + .... + 999.1000

3A = 1.2.3 + 2.3.(4-1) + .... + 999.1000.(1001 - 998)

3A = 1.2.3 + 2.3.4 - 1.2,3 +..... + 999.1000.1001 - 998.999.1000

3A = 999 . 1000 . 1001

A = 333 x 1000 x 1001 = 333 333 000

3A=1x2x3+2x3x(4-1)+3x4x(5-2)+.............+999x1000x(1001-998)

3A=1x2x3+2x3x4-1x2x3+3x4x5-2x3x4+............+999x1000x1001-998x999x1000

3A=999x1000x1001

A=999x1000x1001:3

A=333333000

\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(S=1-\frac{1}{2018}\)

\(S=\frac{2018}{2018}-\frac{1}{2018}\)

\(S=\frac{2017}{2018}\)

\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}.\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}=\frac{2017}{2018}\)

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

A=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

A=\(1-\frac{1}{50}\)

A=\(\frac{49}{50}\)

\(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)

\(=1-\dfrac{1}{2018}\)

\(=\dfrac{2017}{2018}\)

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

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

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

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

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