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
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 A=1+\(\frac{1}{3^2}\)+\(\frac{1}{3^4}\)+...+\(\frac{1}{3^{100}}\).Biết 8A=9-\(\frac{1}{3^n}\). Vậy n=...
\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\).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=?
\(1+\frac{1}{3^2}+\frac{1}{3^4}+.....+\frac{1}{3^{100}} \)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 A = \(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\) biết 8A = 9--\(\frac{1}{3^n}\)
A=1\(\frac{1}{^{3^2}}+\frac{1}{3^4}+....+\frac{1}{3^{100}}\) Biết 8A=9-\(\frac{1}{3^n}\)
Cho A=1+\(\frac{1}{3^2}\)+\(\frac{1}{3^4}\)+...+\(\frac{1}{3^{100}}\). Biết 8A=9-\(\frac{1}{3^{10}}\)
