\(A=-\dfrac{1}{3}+\dfrac{1}{3^2}-...-\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\)
\(=\dfrac{1}{3}\left(-1+\dfrac{1}{3}\right)+\dfrac{1}{3^3}\left(-1+\dfrac{1}{3}\right)+...+\dfrac{1}{3^{99}}\left(-1+\dfrac{1}{3}\right)\)
\(=\dfrac{-2}{3}\left(\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)
Ta có:
\(B=\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)
\(9B=3+\dfrac{1}{3}+...+\dfrac{1}{3^{97}}\)
\(9B-B=3-\dfrac{1}{3^{99}}\)
\(B=\dfrac{3-\dfrac{1}{3^{99}}}{8}\)
\(A=-\dfrac{2}{3}B=\dfrac{-2}{3}.\dfrac{3-\dfrac{1}{99}}{8}=\dfrac{\dfrac{1}{3^{100}}-1}{4}\)