ta có :
\(3A=-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(\Rightarrow4A=-1-\frac{1}{3^{101}}\)
\(\Rightarrow4A=\frac{-3^{101}-1}{3^{101}}\)
\(\Rightarrow A=\left(\frac{-3^{101}-1}{3^{101}}\right):4\)
\(A=\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
\(\Rightarrow3A=-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(\Rightarrow3A+A=4A\)
\(=\left(-1+\frac{1}{3}-...-\frac{1}{3^{100}}\right)+\left(\frac{-1}{3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\right)\)
\(=-1+\frac{1}{3}-...-\frac{1}{3^{100}}-\frac{1}{3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
\(=-1-\frac{1}{3^{101}}\)
\(\Rightarrow A=\frac{-1-\frac{1}{3^{101}}}{4}\)
Vậy \(A=\frac{-1-\frac{1}{3^{101}}}{4}\)