Ta có
$A = \dfrac{1}{3^0} + \dfrac{1}{3^1} + \cdots + \dfrac{1}{3^{2005}}$
Khi đó, ta có
$\dfrac{1}{3} A = \dfrac{1}{3^1} + \dfrac{1}{3^2} + \cdots + \dfrac{1}{3^{2006}}$
Khi đó, ta có
$A - \dfrac{1}{3} A = \left( \dfrac{1}{3^0} + \dfrac{1}{3^1} + \cdots + \dfrac{1}{3^{2005}} \right) - \left( \dfrac{1}{3^1} + \dfrac{1}{3^2} + \cdots + \dfrac{1}{3^{2006}} \right)$
$<-> \dfrac{2}{3} A = \dfrac{1}{3^0} - \dfrac{1}{3^{2006}} = \dfrac{3^{2006} - 1}{3^{2006}}$
$<-> A = \dfrac{3^{2006} - 1}{2.3^{2005}}$
