\(\dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}\\ =\dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\\ =\dfrac{3-1}{2}\cdot\left(\dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\right)\\ =\dfrac{\left(3-1\right)\left(\dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\right)}{2}\\ =\dfrac{3-1+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^5}-\dfrac{1}{3^6}}{2}\\ =\dfrac{3-\dfrac{1}{3^6}}{2}\\ =\dfrac{\dfrac{3^7}{3^6}-\dfrac{1}{3^6}}{2}\\ =\dfrac{2187-1}{729}\cdot\dfrac{1}{2}\\ =\dfrac{2186}{729}\cdot\dfrac{1}{2}\\ =\dfrac{1093}{729}\)
Đặt biểu thức là P , theo bài ra ta có:
\(\dfrac{1}{3}P=\dfrac{1}{3}+\dfrac{1}{3^2}+.......+\dfrac{1}{3^6}+\dfrac{1}{3^7}\)
\(=>P-\dfrac{1}{3}P=\left(1-\dfrac{1}{3^7}\right)\)
\(=>\dfrac{2}{3}P=\dfrac{2186}{2187}\)
\(=>P=\dfrac{2186}{2187}:\dfrac{2}{3}=\dfrac{1093}{729}\)
CHÚC BẠN HỌC TỐT.......
Đặt biểu thức là A , theo đề bài ta có
\(\dfrac{1}{3}A=\dfrac{1}{3}+\dfrac{1}{3^2}+..+\dfrac{1}{3^6}+\dfrac{1}{3^7}\)
=> \(\dfrac{1}{3}A-A=\left(1-\dfrac{1}{3^7}\right)\)
=> \(\dfrac{2}{3}A=\dfrac{2186}{2187}=>A=\dfrac{2186}{2187}:\dfrac{2}{3}=\dfrac{1093}{729}\)
Vậy A = \(\dfrac{1093}{729}\)
