A= \(-\dfrac{1}{3}+\dfrac{1}{3^2}-...+\dfrac{1}{3^{100}}-\dfrac{1}{3^{101}}\)

=>3A=\(-1+\dfrac{1}{3}-...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)

=>4A=\(-1-\dfrac{1}{3^{101}}\) =>A=\(\dfrac{-1-\dfrac{1}{3^{101}}}{4}\)

Lê Quốc Trần Anh selected this answer.