\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\frac{1}{3^{100}}\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+.........+\frac{1}{3^{99}}\Rightarrow3B-B=2B=\left(1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\right)=1-\frac{1}{3^{99}}\Rightarrow B=\frac{3^{99}-1}{2.3^{99}}\)