Question

Tính :

(1 + \(\dfrac{1}{1.3}\) ) . ( 1+\(\dfrac{1}{2.4}\) ) . (1+\(\dfrac{1}{3.5}\)) . ... . ( 1+\(\dfrac{1}{2019.2021}\)) 

Answer 1

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:

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

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

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

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