Question

1/3+1/3²+1/3³+...+1/3⁸

Answer 1

Gọi \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^8}\\ 3D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^7}\\ 3D-D=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^7}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^8}\right)\\ 2D=1-\dfrac{1}{3^8}\\ 2D=1-\dfrac{1}{6561}\\ 2D=\dfrac{6560}{6561}\\ D=\dfrac{6560}{6561}:2\\ D=\dfrac{3280}{6561}\)

Vậy \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^8}=\dfrac{3280}{6561}\)