\(1+\frac{1}{3^2}+\frac{1}{3^4}+......+\frac{1}{3^{100}}\).Biết 8A=9-\(\frac{1}{3^n}\)
Vậy n=?
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Cho A= \(1+\frac{1}{^{3^2}}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
Biết 8A=\(9-\frac{1}{3^n}\)
Vậy n=???
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
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Cho A= \(1+\frac{1}{3^2}+\frac{1}{3^4}+..........+\frac{1}{3^{100}}\). Biết 8A\(=9-\frac{1}{3^n}\). Vậy n= ?
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
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\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\).Biết \(8A=9-\frac{1}{3^n}\). Vậy n= ?
\(1+\frac{1}{3^2}+\frac{1}{3^4}+.....+\frac{1}{3^{100}} \)biết\(8A=9-\frac{1}{3^n}\)
Vậy n = bao nhiêu ?
Cho A=1+\(\frac{1}{3^2}\)+\(\frac{1}{3^4}\)+...+\(\frac{1}{3^{100}}\).Biết 8A=9-\(\frac{1}{3^n}\). Vậy n=...
Cho . Biết
.Vậy
?
\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(\Rightarrow9A=9\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow9A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow9A-A=\left(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow8A=9-\frac{1}{3^{100}}\Rightarrow n=100\)
Vậy n = 100
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Cho . Biết
.
Vậy
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 \(n=100\)
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Cho . Biết
.
Vậy
\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(9A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow8A=9-\frac{1}{3^{100}}\)
Có \(8A=9-\frac{1}{3^n}=9-\frac{1}{3^{98}}\)
\(\Rightarrow n=98\)
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Tìm n biết:
8A=9-\(\frac{1}{3^n}\)và A=1+\(\frac{1}{3^2}+\frac{1}{3^4}+....+\frac{1}{3^{100}}\)