$B = 1 + \dfrac{1}{3^1} + \dfrac{1}{3^2} + ... + \dfrac{1}{3^{100}}$
$B = \dfrac{3}{3} ( 1 + \dfrac{1}{3^1} + ... + \dfrac{1}{3^{100}} )$
$B = \dfrac{1}{3} ( 1 + \dfrac{1}{3} + \dfrac{1}{3^1} + ... + \dfrac{1}{3^{99}} )$
$2B = \dfrac{1}{3} ( \dfrac{1}{3} + ... + \dfrac{1}{3^{100}} )$
$2B = \dfrac{1}{3} ( \dfrac{1}{3^{100}} - 1 )$
$B = \dfrac{\dfrac{1}{3} ( \dfrac{1}{3^{100}} - 1 )}{2} $