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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
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
\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(\Rightarrow9A=9\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow9A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow9A-A=\left(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow8A=9-\frac{1}{3^{100}}\Rightarrow n=100\)
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 \(n=100\)
\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(9A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow8A=9-\frac{1}{3^{100}}\)
Có \(8A=9-\frac{1}{3^n}=9-\frac{1}{3^{98}}\)
\(\Rightarrow n=98\)
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
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