Ta có: \(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(\Rightarrow\) \(3^2A=3^2+\frac{3^2}{3^2}+\frac{3^2}{3^4}+...+\frac{3^2}{3^{100}}\)
\(\Leftrightarrow9A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(9A-A=9-\frac{1}{3^{100}}\Leftrightarrow8A=9-\frac{1}{3^{100}}\)
\(\Rightarrow\) n = 100
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