\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(\Rightarrow9A=9\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow9A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow9A-A=\left(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow8A=9-\frac{1}{3^{100}}\Rightarrow n=100\)
Vậy n = 100
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