\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^7}+\dfrac{1}{3^8}\)
\(3A=3\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^7}+\dfrac{1}{3^8}\right)\)
\(3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}+\dfrac{1}{3^7}\)
\(3A-A=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^6}+\dfrac{1}{3^7}\right)-\left(\dfrac{1}{3}+...+\dfrac{1}{3^7}+\dfrac{1}{3^8}\right)\)
\(2A=1-\dfrac{1}{3^8}\Rightarrow A=\dfrac{1-\dfrac{1}{3^8}}{2}\)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^7}+\dfrac{1}{3^8}\)
\(3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}+\dfrac{1}{3^7}\)
\(\Rightarrow3A-A=\dfrac{3^8-1}{3^8}\)
\(\Rightarrow2A=\dfrac{3^8-1}{3^8}\)
\(\Rightarrow A=\dfrac{3^8-1}{3^8\cdot2}\)
Vậy \(A=\dfrac{3^8-1}{3^8\cdot2}\).
