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