Giải thích các bước giải:
Ta có:
$1-\dfrac1{1+2+...+n}=1-\dfrac{1}{\dfrac{n(n+1)}{2}}=1-\dfrac2{n(n+1)}=\dfrac{n(n+1)-2}{n(n+1)}=\dfrac{n^2+n-2}{n(n+1)}=\dfrac{(n-1)(n+2)}{n(n+1)}$
$\to B=\dfrac{(2-1)(2+2)}{2(2+1)}.\dfrac{(3-1)(3+2)}{3(3+1)}...\dfrac{(2017-1)(2017+2)}{2017(2017+1)}$
$\to B=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}...\dfrac{2016.2019}{2017.2018}$
$\to B=\dfrac{1.2....2016}{2.3..2017}.\dfrac{4.5..2019}{3.4..2018}$
$\to B=\dfrac1{2017}.\dfrac{2019}{3}$
$\to B=\dfrac{673}{2017}$
