cho mình hỏi, đè này có sai ko?

Bạn ơi , xin lỗi mk ấn nhầm,đề là:    \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2017.2019}\right)\)nha !

giúp mới mình sẽ tích

bn viết rõ đề đi

5435

tích mình nha

luu y : dau /la phan cach giua mau so va tu so

Ta có: \(A=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\cdot\ldots\cdot\left(1+\frac{1}{2017\cdot2019}\right)\)

\(=\left(1+\frac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(1+\frac{1}{\left(3-1\right)\left(3+1\right)}\right)\cdot\ldots\cdot\left(1+\frac{1}{\left(2018-1\right)\left(2018+1\right)}\right)\)

\(=\frac{2^2-1+1}{\left(2-1\right)\left(2+1\right)}\cdot\frac{3^2-1+1}{\left(3-1\right)\left(3+1\right)}\cdot\ldots\cdot\frac{2018^2-1+1}{\left(2018-1\right)\left(2018+1\right)}\)

\(=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\ldots\cdot\frac{2018^2}{2017\cdot2019}=\frac{2\cdot3\cdot\ldots\cdot2018}{1\cdot2\cdot\ldots\cdot2017}\cdot\frac{2\cdot3\cdot\ldots\cdot2018}{3\cdot4\cdot\ldots\cdot2019}\)

\(=\frac{2018}{1}\cdot\frac{2}{2019}=\frac{4036}{2019}<\frac{4038}{2019}\)

=>A<2

Lời giải:
Gọi tích trên là $A$

Xét thừa số tổng quát: $1+\frac{1}{n(n+2)}=\frac{n(n+2)+1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$

Thay $n=1,2,3....,2019$ ta có:

$A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}....\frac{2020^2}{2019.2021}$

$=\frac{2^2.3^2...2020^2}{(1.3)(2.4)(3.5)...(2019.2021)}$

$=\frac{(2.3....2020)(2.3...2020)}{(1.2.3...2019)(3.4...2021)}$

$=2020.\frac{2}{2021}=\frac{4040}{2021}$

\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{2016}{2017}\)

Đề như thế nào ý

2s=2/