Đáp án: $P=\dfrac{2017}{2019}$
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:
$P=2\cdot (\dfrac12-\dfrac13)+2\cdot (\dfrac13-\dfrac14)+...+2\cdot( \dfrac1{2018}-\dfrac1{2019})$
$\to P=2\cdot (\dfrac12-\dfrac13+\dfrac13-\dfrac14+...+\dfrac1{2018}-\dfrac1{2019})$
$\to P=2\cdot (\dfrac12-\dfrac1{2019})$
$\to P=\dfrac{2017}{2019}$
