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25 tháng 2 2020

\(C=\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2017\cdot2018}\right)-\)\(\left(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}\right)\)

Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2017\cdot2018}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)

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

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

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2009}\right)\)

\(\Rightarrow A=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+..+\frac{1}{2017}\)

\(\Rightarrow C=\left(\frac{1}{101}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2018}\right)-\left(\frac{1}{1010}+\frac{1}{1012}+...+\frac{1}{2017}\right)\)

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

9 tháng 4 2020

Đặt S = ( 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/2017.2018 )

Đặt A = ( 1/1.2 + 1/3.4  + ... + 1/2017.2018)

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

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

= ( 1 + 1/2 + ... + 1/2018 ) - 2 ( 1/2 + 1/4 + ... + 1/2018) )

= ( 1 + 1/2 + ... + 1/2018 ) - ( 1 + 1/2 + ... + 1/1009 )

= 1/1010 + 1/1011 + ... + 1/2018

=> A - ( 1/1010 + 1/1011 + ... + 1/2017 ) = 1/2018

=> S = 1/2018

Vậy S = 1/2018

9 tháng 4 2020

thanks bạn nhiều

11 tháng 12 2019

giúp mk vs mn ơi. mình cần gấp chiều mai nộp òi

4 tháng 5 2023

\(A=2017:\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}\right)\)
\(=2017:\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
\(=2017:\left(1-\dfrac{1}{2018}\right)\)
\(=2017:\dfrac{2017}{2018}\)
\(=2017\cdot\dfrac{2018}{2017}\)
\(=2018\)
#NgDat

4 tháng 5 2023

\(A=2017:\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\right)\)

\(A=2017:\left(\dfrac{1}{1}\cdot\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+...+\dfrac{1}{2017}\cdot\dfrac{1}{2018}\right)\)

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

\(A=2017:\left(\dfrac{1}{1}-\dfrac{1}{2018}\right)\)

\(A=2017:\left(\dfrac{2018}{2018}-\dfrac{1}{2018}\right)\)

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

\(A=2018.\)