Bạn tham khảo :

$A = \dfrac{1}{3} + \dfrac{1}{3^2} + \dfrac{1}{3^3}  + ... + \dfrac{1}{3^{2006}}$

$\dfrac{1}{3}A = \dfrac{1}{3} . ( \dfrac{1}{3} + \dfrac{1}{3^2} + \dfrac{1}{3^3}  + ... + \dfrac{1}{3^{2006}})$ 

$\dfrac{1}{3}A  = \dfrac{1}{3^2} + \dfrac{1}{3^3}  + \dfrac{1}{3^4}   ... + \dfrac{1}{3^{2007}}$

$\dfrac{1}{3}A  - A = (\dfrac{1}{3^2} + \dfrac{1}{3^3}  + \dfrac{1}{3^4}   ... + \dfrac{1}{3^{2007}})- ( \dfrac{1}{3} + \dfrac{1}{3^2} + \dfrac{1}{3^3}  + ... + \dfrac{1}{3^{2006}})$

$ \dfrac{-2}{3}A = \dfrac{1}{3^{2007}} -  \dfrac{1}{3}$

$ \dfrac{-2}{3}A = \dfrac{1}{3^{2007}} -  \dfrac{3^{2006}}{3^{2007}}$

$ \dfrac{-2}{3}A = \dfrac{1 - 3^{2006}}{3^{2007}}$

$A = \dfrac{1 - 3^{2006}}{3^{2007}} : \dfrac{-2}{3}$

$A = \dfrac{1 - 3^{2006}}{3^{2007}}. \dfrac{3}{-2}$ 

Đáp án:

$A=\dfrac{3^{2006}-1}{2.3^{2006}}$

Giải thích các bước giải:

Ta có: $A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+.....+\dfrac{1}{3^{2006}}$

⇒ $3A=3.(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+.....+\dfrac{1}{3^{2006}})$

⇔ $3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+.........+\dfrac{1}{3^{2005}}$

Mặt khác:

$3A-A=(1+\dfrac{1}{3}+\dfrac{1}{3^2}+.....+\dfrac{1}{3^{2005}})-(\dfrac{1}{3}+\dfrac{1}{3^2}+.....+\dfrac{1}{3^{2006}})$

⇔ $2A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+......+\dfrac{1}{3^{2005}}-\dfrac{1}{3}-\dfrac{1}{3^2}-....-\dfrac{1}{3^{2006}}$

⇔ $2A=1-\dfrac{1}{3^{2006}}$

⇔ $2A=\dfrac{3^{2006}-1}{3^{2006}}$

⇔ $A=\dfrac{3^{2006}-1}{2.3^{2006}}$

Chúc bạn học tốt !!!!