Ta có
$2 = 2.1 = 2.3^0$
$6 = 2.3 = 2.3^1$
$18 = 2.9 = 2.3^2$
$54 = 2.27 = 2.3^3$
$162 = 2.81 = 2.3^4$
$486 = 2.243 = 2.3^5$
Vậy ta tính tổng
$A = \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{18} + \dfrac{1}{54} + \dfrac{1}{162} + \dfrac{1}{486}$
$= \dfrac{1}{2} \left( 1 + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \dfrac{1}{81} + \dfrac{1}{243} \right)$
Ta tính
$B = 1 + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \dfrac{1}{81} + \dfrac{1}{243}$
Khi đó
$\dfrac{1}{3} B = \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \dfrac{1}{81} + \dfrac{1}{243} + \dfrac{1}{729}$
Khi đó
$B - \dfrac{1}{3} B = 1 - \dfrac{1}{729}$
$<-> \dfrac{2}{3} B = \dfrac{728}{729}$
$<-> B = \dfrac{364}{243}$
Suy ra
$A = \dfrac{1}{2} B = \dfrac{182}{243}$.
