$\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{n(n+1)}=\dfrac{2010}{2011}$
$⇔\dfrac{1}{2}(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{n(n+1)})=\dfrac{2010}{4022}$
$⇔ \dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+....+\dfrac{2}{2.n.(n+1)}=\dfrac{2010}{4022}$
$⇔ \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{n.(n+1)}=\dfrac{2010}{4022}$
$⇔\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{2010}{4022}$
$⇔\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{2010}{4022}$
$⇔\dfrac{n+1-2}{2(n+1)}=\dfrac{2010}{4022}$
$⇔\dfrac{n-1}{2(n+1)}=\dfrac{2010}{4022}$
$⇔(n-1).4022=2.(n+1).2010$
$⇔4022n-4022=4020n+4020$
$⇔2n=8042$
$⇔n=4021$
