Đặt \(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
=>\(3A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
=>\(3A-A=1+\dfrac{1}{3}+...+\dfrac{1}{243}-\dfrac{1}{3}-\dfrac{1}{9}-...-\dfrac{1}{729}\)
=>\(2A=1-\dfrac{1}{729}=\dfrac{728}{729}\)
=>\(A=\dfrac{728}{729}:2=\dfrac{364}{729}\)
`S=1/3+1/9+1/27+1/81+1/243+1/729`
Ta có :
`S_6=(a_1(1-q^n))/(1-q)=(1/3.(1-(1/3)^6))/(1-1/3)=364/729`