Nhầm
\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3^2A=3^2+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow9A-A=\left(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow8A=9-\frac{1}{3^{100}}\)
=> n = 100
Ta có: \(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(\Rightarrow9A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow9A-A=\left(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow8A=9-\frac{1}{3^{100}}\)
\(\Rightarrow9-\frac{1}{3^{100}}=9-\frac{1}{3^n}\)
\(\Rightarrow\frac{1}{3^{100}}=\frac{1}{3^n}\)
\(\Rightarrow3^{100}=3^n\)
\(\Rightarrow n=100\)
Vậy n = 100
\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3^2A=3^2+\frac{1}{3^4}+\frac{1}{3^6}+...+\frac{1}{3^{102}}\)
\(\Rightarrow9A-A=\left(9+\frac{1}{3^4}+\frac{1}{3^6}+...+\frac{1}{3^{102}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow8A=9-\frac{1}{3^{100}}\)
\(\Rightarrow n=100\)
