Question

\(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+.....+\dfrac{1}{3^8}\)

Answer 1

\(\rightarrow3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^7}\\ \rightarrow3C-C=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^7}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^8}\right)\\ \rightarrow2C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^7}-\dfrac{1}{3}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...-\dfrac{1}{3^8}\\ \rightarrow2C=1-\dfrac{1}{3^8}\\ \rightarrow2C=\dfrac{6550}{6561}\\ \rightarrow C=\dfrac{3280}{6561}\)

Answer 2

=-3280/6561