A= 1.2 + 2.3 + 3.4 +...+ 2011.2012

=>3A= 1.2.3 + 2.3.3 + 3.4.3 +...+ 2011.2012.3

    3A= 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) +...+ 2011.2012.(2013 - 2010)

    3A= (1.2.3 + 2.3.4 + 3.4.5 +...+ 2011.2012.2013) - (0.1.2 + 1.2.3 + 2.3.4 +...+ 2010.2011.2012)

    3A= (2011.2012.2013) - (0.1.2)

    3A= 8144863716 - 0

\(A=\frac{8144863716}{3}=2714954572\)

Vậy A= 2714954572

\(A=\dfrac{7}{1.2}+\dfrac{7}{2.3}+\dfrac{7}{3.4}+...+\dfrac{7}{2011.2012}\)

\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)\)

\(A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)\)

\(A=7\left(1-\dfrac{1}{2012}\right)=7.\dfrac{2011}{2012}=\dfrac{14077}{2012}\)

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

3A= 3.(1.2 + 2.3 + 3.4 + ..... +99.100)

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

 3A=1.2.3 - 1.2.3 + 2.3.4 - 2.3.4  + .....+99.100.101

3A=99.100.101

A=99.100.101/3=333300

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

\(=4\cdot\dfrac{2014}{2015}=\dfrac{8056}{2015}\)

\(A=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=9\left(1-\dfrac{1}{100}\right)=\dfrac{891}{100}\)

=> 3S = 1.2.3 + 2.3.3 + 3.4.3 + .... + 2011.2012.3

=> 3S = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + .... + 2011.2012.( 2013 - 2010 )

=> 3S = 1.2.3 + 2.3.4 - 1.2.3 + .... + 2011.2012.2013 - 2010.2011.2012

=> 3S = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 2010.2011.2012 - 2010.2011.2012 ) + 2011.2012.2013

=> 3S = 2011.2012.2013

=> S = ( 2011.2012.2013 ) : 3

3S=1.2.3+2.3.(4-1)+...............+2011.2012.(2013-2010)

3S=1.2.3+2.3.4-1.2.3+...............+2011.2012.2013-2010.2011.2012

3S=2011.2012.2013

S=2011.2012.2013:3

S=2714954572

S=2714954572

Trước tiên, chúng ta cần có lý thuyết về biến đổi phân số.

\(\dfrac{b-a}{a\cdot b}=\dfrac{1}{a}-\dfrac{1}{b}\)

Ta có:

\(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\)

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

\(S=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+...-\dfrac{1}{2018}\)

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

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

=1/1.2+1/2.3+1/3.4+...1/2017.2018

=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2017-1/2018

=1-1/2018

=2018/2018-1/2018

=2017/2018

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

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

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

S = \(\dfrac{2017}{2018}\)

\(\dfrac{4}{1.2}+\dfrac{4}{2.3}+...+\dfrac{4}{2021.2022}\\ =4\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2021.2022}\right)\\ =4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\\ =4\left(1-\dfrac{1}{2022}\right)\\ =4.\dfrac{2021}{2022}\\ =\dfrac{4042}{1011}\)

4/1.2 4/2.3 4/3.4 ... 4/2021.4/2022

= 1/4. (1/1- 1/2+ 1/2- 1/3+ 1/3- 1/4+...+1/2021- 1/2022)

=1/4. (1/1- 1/2022)= 1/4. (2022/2022- 1/2022)

= 1/4. 2021/2022

= 2021/8088