A = \(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)

3²A = \(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

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

8A = \(9-\frac{1}{3^{100}}=9-\frac{1}{3^n}\)

=> n = 100

chào châu chấu là 100 dễ mà

là 100 lấy máy tính mà tính

\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)

\(3^2A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

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

\(8A=9-\frac{1}{3^{100}}\)

=> n = 100

ê mai Phưng câu này mày lấy đâu đấy

tớ đang bí đây nè

de ot 

9A-A=(\(9+1+\frac{1}{3^2}+...+\frac{1}{3^{99}}\))-\(\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)

8A=\(9-\frac{1}{3^{100}}\)

=>n=100

100 là đúng

câu hỏi là gì, cho mỗi điều kiện

Ta có 9A = 9 + 1 + \(\frac{1}{3^2}\) + \(\frac{1}{3^4}\) + ... + \(\frac{1}{3^{98}}\)

 9A  -  A = 9 - \(\frac{1}{3^{100}}\) = 8A

Suy ra : 9 - \(\frac{1}{3^{100}}\) = 9 - \(\frac{1}{3^n}\)

\(\frac{1}{3^{100}}\) =   \(\frac{1}{3^n}\)

Vậy n = 100

\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)

\(\Rightarrow9A=9\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)

\(\Rightarrow9A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

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

\(\Rightarrow8A=9-\frac{1}{3^{100}}\Rightarrow n=100\)

Vậy n = 100