\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

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

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\left(\frac{2018}{2018}-\frac{1}{2018}\right)\)

\(=1+\left(\frac{2017}{2018}\right)\)

\(=\frac{2018}{2018}+\frac{2017}{2018}=\frac{4035}{2018}\)

\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}...+\frac{1}{2017\cdot2018}\)

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

\(=1+\left(1-\frac{1}{2018}\right)\)

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

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

\(=\frac{4035}{2018}\)

\(1+\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{2017x2018}\)

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

\(=1+\left(1-\frac{1}{2018}\right)\)

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

\(=\frac{4035}{2018}\)

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

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

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

\(C=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+.....+\frac{1}{2017x2018}\)

Ta thấy \(\frac{1}{1x2}=\frac{1}{1}-\frac{1}{2}\)

               \(\frac{1}{2x3}=\frac{1}{2}-\frac{1}{3}\)

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

           \(\frac{1}{2017x2018}=\frac{1}{2017}-\frac{1}{2018}\)

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

\(\Rightarrow C=\frac{1}{1}-\frac{1}{2018}\)

\(\Rightarrow C=\frac{2017}{2018}\)

Chúc bạn học tốt nhớ k mình nhá

\(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

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

A = 1X2 +2x3 +...+ 2016x2107

3A = 1x2x3 + 2x3x3 + ...+ 2016x2017x3

3A = 1x2x(3-0) + 2x3x(4-1) + ... + 2016x2017x(2018-1)

3A = 1x2x3 - 1x2x0 +2x3x4 -1x2x3 +...+ 2016x2017x2018 - 2016x2017x2015

Ta loại trừ còn

3A = 2016x2017x2018 - 1x2x0

3A = 2016x2017x2018

A = 2016 x2017 x2018 : 3

A = 1x2 +2x3 +3x4 +...+ 2016 x 2017

3A = 1x2x3 + 2x3x3 +...+2016 x 2017 x3

3A = 1x2x(3-0) + 2x3x(4-1) +...+ 2016x2017x(2018-2015)

Đặt A=1x2+2x3+3x4+...+2016x2017

=>3A=3x1x2+3x2x3+3x3x4+...+3x2016x2017

=>3A=(3-0)x1x2+(4-1)x2x3+(5-2)x3x4+...+(2018-2015)x2016x2017

=>3A=1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+...+2016x2017x2018-2015x2016x2017

=>3A=2016x2017x2018

=>A=\(\frac{2016\times2017\times2018}{3}\)(tự tính nha)

S = 1x2 + 2x3 + 3x4 + 4x5 + ... + 2016x2017

3S = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 2016x2017x(2018-2015)

3S = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 2016x2017x2018 - 2015x2016x2017

3S = 2016x2017x2018

S = 1/3 x 2016x2017x2018.

(2016x2017x2018):3

x= -2018/2017 

Bài làm:

Ta có: \(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2017.2018}=-1\)

\(\Leftrightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)=-1\)

\(\Leftrightarrow x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)=-1\)

\(\Leftrightarrow x\left(1-\frac{1}{2018}\right)=-1\)

\(\Leftrightarrow x.\frac{2017}{2018}=-1\)

\(\Rightarrow x=-\frac{2018}{2017}\)

\(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+...+\frac{x}{2017\cdot2018}=-1\)

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

=> \(\frac{x}{1}-\frac{x}{2018}=-1\)

=> \(\frac{2018x-x}{2018}=-1\)

=> \(\frac{2017x}{2018}=-1\)

=> 2017x = -2018

=> x = -2018/2017

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

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

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

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

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

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

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

\(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2015\times2016}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(=1-\frac{1}{2016}=\frac{2015}{2016}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}+\frac{1}{2016\cdot2017}\)

\(\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+...+\frac{2016-2015}{2015\cdot2016}+\frac{2017-2016}{2016\cdot2017}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)(làm gọn một chút)

\(1-\frac{1}{2017}=\frac{2016}{2017}\)

A = 1.2 + 2.3 + 3.4 + ... + 2017.2018

⇒ 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2017.218.(2019 - 2016)

= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2017.2018.2019 - 2016.2017.2018

= 2017.2018.2019

= 2017.2018.2019

B = 2018³/3 ⇒ 3B = 2018³

Ta có:

2017.2019 = (2018 - 1).(2018 + 1)

= 2018² - 1²

= 2018.2018 - 1 < 2018.2018

⇒ 2017.2018.2019 < 2018.2018.2018

⇒ 3A < 3B

⇒ A < B

Ta có công thức tổng quát là:

\(\frac{n.\left(n+1\right).\left(2n+1\right)}{6}\)

Thay vào sẽ là:

\(\frac{2017.2018.\left(2.2017+1\right)}{6}=2737280785\)

= 2737280785

chỉ nhiu đó thui hả bạn