\(A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2017}}\)
\(3A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2016}}\)
\(3A-A=\left(3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2016}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2017}}\right)\)
\(2A=3-\dfrac{1}{3^{2017}}=\dfrac{3^{2018}-1}{3^{2017}}\)
\(\Rightarrow A=\dfrac{3^{2017}\left(3-1\right)}{3^{2017}.2}=\dfrac{3^{2017}.2}{3^{2017}.2}=1\)
A = 1+ \(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)+..................+ \(\dfrac{1}{3^{2017}}\)
\(\dfrac{1}{3}\)A = \(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{3^3}\)+.................+ \(\dfrac{1}{3^{2018}}\)
A - \(\dfrac{1}{3}\)A = 1 - \(\dfrac{1}{3^{2018}}\)
\(\dfrac{2}{3}\)A = \(\dfrac{3^{2018}-1}{3^{2018}}\)
A = \(\dfrac{3^{2018}-1}{3^{2018}}\).\(\dfrac{3}{2}\)
A = \(\dfrac{3^{2019}-3}{3^{2018}.2}\)
