Giải thích các bước giải:
Ta có:
$1+\dfrac{2}{n(n+3)}=\dfrac{n(n+3)+2}{n(n+3)}$
$\to 1+\dfrac{2}{n(n+3)}=\dfrac{n^2+3n+2}{n(n+3)}$
$\to 1+\dfrac{2}{n(n+3)}=\dfrac{n^2+2n+n+2}{n(n+3)}$
$\to 1+\dfrac{2}{n(n+3)}=\dfrac{(n^2+2n)+(n+2)}{n(n+3)}$
$\to 1+\dfrac{2}{n(n+3)}=\dfrac{n(n+2)+(n+2)}{n(n+3)}$
$\to 1+\dfrac{2}{n(n+3)}=\dfrac{(n+1)(n+2)}{n(n+3)}$
Áp dụng công thức trên
Ta có:
$A=(1+\dfrac12)(1+\dfrac15)(1+\dfrac19)....(1+\dfrac2{2017.2020})$
$\to A=(1+\dfrac24)(1+\dfrac2{10})(1+\dfrac2{18})....(1+\dfrac2{2017.2020})$
$\to A=(1+\dfrac2{1.4})(1+\dfrac2{2.5})(1+\dfrac2{3.6})....(1+\dfrac2{2017.2020})$
$\to A=\dfrac{(1+1)(1+2)}{1(1+3)}\cdot \dfrac{(2+1)(2+2)}{2(2+3)}\cdot \dfrac{(3+1)(3+2)}{3(3+3)}\cdots \dfrac{(2017+1)(2017+2)}{2017(2017+3)}$
$\to A=\dfrac{2.3}{1.4}\cdot \dfrac{3.4}{2.5}\cdot \dfrac{4.5}{3.6}\cdots \dfrac{2018.2019}{2017.2020}$
$\to A=\dfrac{2.3.4...2018}{1.2.3..2017}.\dfrac{3.4.5...2019}{4.5.6...2020}$
$\to A=\dfrac{2018.3}{2020}$
$\to A=3.\dfrac{2018}{2020}$
$\to A<3.1$
$\to A<3$
