\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
=>\(3B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
=>\(3B-B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-...-\dfrac{1}{3^{99}}\)
=>\(2B=1-\dfrac{1}{3^{99}}\)
=>\(2B=\dfrac{3^{99}-1}{3^{99}}\)
=>\(B=\dfrac{3^{99}-1}{3^{99}\cdot2}\)