Giải:
Ta có:
\(A=1\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(\Rightarrow9A=9+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow9A-A=\left(9+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow8A=9-\frac{1}{3^{100}}=9-\frac{1}{3^n}\)
\(\Leftrightarrow n=100\)
Vậy \(n=100\)
Đúng 0
Bình luận (1)
