Đáp án:
`P=2017/2019`.
Giải thích các bước giải:
`P=1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...+2018)`
`=>`$P=\dfrac13+\dfrac16+\dfrac1{10}\ +\,.\!.\!.+\ \dfrac{1}{\dfrac{(2018+1)\times2018}{2}}$
`=>`$P=\dfrac13+\dfrac16+\dfrac1{10}\ +\,.\!.\!.+\ \dfrac{1}{\dfrac{2018\times2019}{2}}$
`=>P/2=1/6+1/12+1/20+...+1/2018.2019`
`=>P/2=1/(2xx3)+1/(3xx4)+1/(4xx5)+...+1/(2018xx2019`
`=>P/2=1/2-1/3+1/3-1/4+1/4-1/5+...+1/2018-1/2019`
`=>P/2=1/2-1/2019`
`=>P=2xx(1/2-1/2019)`
`=>P=1-2/2019`
`=>P=2017/2019`
Vậy `P=2017/2019`.
