\(3B=-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(3B+B=\left(-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\right)+\left(\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\right)\)
\(4B=-1-\frac{1}{3^{101}}\)
\(B=\frac{-\left(1+\frac{1}{3^{101}}\right)}{4}\)
\(3B=-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(3B+B=\left(-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\right)+\left(\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\right)\)
\(4B=-1-\frac{1}{3^{101}}\)
\(B=\frac{-\left(1+\frac{1}{3^{101}}\right)}{4}\)