Giải thích các bước giải:
Ta có $1+2+3+...+n=\dfrac{n(n+1)}{2}$
$\to \dfrac1{1+2+3+..+n}=\dfrac{2}{n(n+1)}=2\cdot \dfrac{ n+1-n}{n(n+1)}=2\cdot (\dfrac1n-\dfrac1{n+1})$
Áp dụng ta có:
$P=\dfrac1{1+2}+\dfrac1{1+2+3}+...+\dfrac{1}{1+2+3+...+n}$
$P=2\cdot (\dfrac12-\dfrac13)+2\cdot (\dfrac13-\dfrac14)+...+2\cdot( \dfrac1{n-1}-\dfrac1{n})$
$\to P=2\cdot (\dfrac12-\dfrac13+\dfrac13-\dfrac14+...+\dfrac1{n-1}-\dfrac1{n})$
$\to P=2\cdot (\dfrac12-\dfrac1{n})$
$\to P=1-\dfrac{2}{n}$
$\to P=\dfrac{n-2}{n}$
$\to 1+\dfrac1{1+2}+...+\dfrac1{1+2+3+...+n}=1+\dfrac{n-2}{n}=\dfrac{2n-2}{n}$
$\to \dfrac{2n}{1+\dfrac1{1+2}+...+\dfrac1{1+2+3+...+n}}=\dfrac{2n}{\dfrac{2n-2}{n}}=\dfrac{n^2}{n-1}$
$\to \dfrac{n^2}{n-1}=2020$
$\to n^2=2020(n-1)$
$\to n^2=2020n-2020$
$\to n^2-2020n+2020=0$
$\to $Không có nghiệm nguyên
$\to$Không tồn tại $n$ thỏa mãn đề
