Ta có :
$\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{132}$
$=2.(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{132})$
$=2.(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{11.12})$
$=2.(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{11}-\dfrac{1}{12})$
$=2.(\dfrac{1}{2}-\dfrac{1}{12})$
$=2.\dfrac{5}{12} = \dfrac{5}{6}$