Giải thích các bước giải:
Ta có :
$\dfrac{1}{1+2+3+..+n}=\dfrac{1}{\dfrac{n(n+1)}{2}}=\dfrac{2}{n(n+1)}=\dfrac{2(n+1)-2n}{n(n+1)}=\dfrac{2}{n}-\dfrac{2}{n+1}$
$\to 2020=2x: (1+\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+..+\dfrac{2}{n}-\dfrac{2}{n+1})$
$\to 2020=2x: (2-\dfrac{2}{n+1})$
$\to 2020=2x: 2(1-\dfrac{1}{n+1})$
$\to 2020=2x: (2.\dfrac{n+1-1}{n+1})$
$\to 2020=2x: \dfrac{2n}{n+1}$
$\to 2020=x. \dfrac{n+1}{n}$
$\to x=\dfrac{2020n}{n+1}=2020-\dfrac{2020}{n+1}$
Vì $x\in N^*\to 2020-\dfrac{2020}{n+1}\to 2020\quad\vdots\quad n+1$
$\to n+1\in\{5,10,20,101,202,505,1010,2020\}$ vì $n>4$
$\to x\in\{1616,1818,1919,2000,2010,2016,2018,2019\}$
