đặt A= tổng trên
\(3A=3\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)\)
\(2A=1-\frac{1}{3^6}\)
\(A=\frac{1}{2}-\frac{1}{2\times3^6}\)