\(\text{Ta có: B= }\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+......+\frac{1}{3^{2017}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2016}}\)
\(\Rightarrow3B-B=1-\frac{1}{3^{2017}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2017}}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^{2017}}}{3}\)