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

A=1+1/3^2+1/3^4+...+1/3^100

=>3^2.A=9+1/3+/3^2+...+1/3^98

=>9A-A=(9+1/3+1/3^2+...+1/3^98)-(1+1/3^2+1/3^4+...+1/3^100)

=>8A=9-1/3^100=9-1/3^n

=>1/3^100=1/3^n

=>3^100=3^n

=>n=100

Vậy n=100

Ban nguyen thieu cong thanh dung roi.ket qua bang 100

n= 100

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

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

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

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

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

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

Vậy n = 100

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 \(n=100\)

N=100

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

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

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

Có \(8A=9-\frac{1}{3^n}=9-\frac{1}{3^{98}}\)

\(\Rightarrow n=98\)