Question

\(\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\frac{2}{81}+\frac{2}{243}+\frac{2}{729}\)

Answer 1

\(\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\frac{2}{81}+\frac{2}{243}+\frac{2}{729}=\frac{728}{729}\)

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Answer 2

Đặt A= biểu thức trên

\(A=\frac{2}{3}+\frac{2}{3^2}+\frac{2}{3^3}+...+\frac{2}{3^6}\)

\(3A=3\left(\frac{2}{3}+\frac{2}{3^2}+...+\frac{2}{3^6}\right)\)

\(3A=2+\frac{2}{3}+...+\frac{2}{3^5}\)

\(3A-A=\left(2+\frac{2}{3}+...+\frac{2}{3^5}\right)-\left(\frac{2}{3}+\frac{2}{3^2}+...+\frac{2}{3^6}\right)\)

\(A=\frac{2-\frac{2}{3^6}}{2}\)