Ta có ; A = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.......+\frac{1}{3^{300}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^{199}}\)
\(\Rightarrow3A-A=1-\frac{1}{3^{300}}\)
\(\Rightarrow2A=1-\frac{1}{3^{300}}\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}\)