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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Gọi B = 1x2 + 2 x 3 + 3 x 4 + ... + 2016 x2017

    3B = 3 x ( 1x2 + 2x3 + 3x4 + ... + 2016x2017)

         = 1x2x3 + 2x3x3 + 3x4x3 + ... + 2016x2017x3 )

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

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

         = 2016 x2017 x2018

      B = 672 x2017 x2018

Mà A = \(\frac{672x2017x2018}{2017x2018}\)

         =  672

Vậy A = 672

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

A=2035153

B=2018

P=1x2+2x3+3x4+...+2017x2018

3P = 1x2x3 + 2x3x3 + 3x4x3 + ... + 2017x2018x3

3P = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + ... +2017x2018x(2019-2016)

3P = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + ... + 2017x2018x2019 - 2016x2017x2018

3P = 2017x2018x2019

P = 2017x2018x2019 : 3

P = 2739315938 

P = 1x2+2x3+3x4+...+2017x2018

3xP = 1x2x3+2x3x3+3x4x3+...+2017x2018x3

3xP = 1x2x3+2x3x(4-1)+3x4x(5-2)+...+2017x2018x(2019-2016)

3xP = 1x2x3+2x3x4-2x3x1+3x4x5-3x4x2+...+2017x2018x2019-2017x2018x2016

3xP = 2017x2018x2019

3xP = 8217947814

P = 8217947814 : 3

P = 2739315938 

Đề đã đầy đủ chưa bạn? Và bạn đang cần làm gì với biểu thức này? 

tih bieu thuc ma

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

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

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

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