Giải thích các bước giải:
Theo bài ta suy ra :
$S=\dfrac{1}{3}.(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+672})$
Lại có :
$\dfrac{1}{1+2+3+..+n}=\dfrac{1}{\dfrac{n(n+1}{2}}=\dfrac{2}{n(n+1)}=2.\dfrac{n+1-n}{n(n+1)}=\dfrac{2}{n}-\dfrac{2}{n+1}$
$\rightarrow S=\dfrac{1}{3}.(\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+...+\dfrac{2}{672}-\dfrac{2}{673})$
$\rightarrow S=\dfrac{1}{3}.(1-\dfrac{2}{673})$
$\rightarrow S=\dfrac{1}{3}.(\dfrac{673-2}{673})$
$\rightarrow S=\dfrac{671}{2013}$
