Cho A=1+\(\frac{1}{3^2}\)+\(\frac{1}{3^4}\)+...+\(\frac{1}{3^{100}}\). Biết 8A=9-\(\frac{1}{3^{10}}\)
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
Cho biết
cho A = \(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\) biết 8A = 9--\(\frac{1}{3^n}\)
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
Cho . Biết
. Tìm n
\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(3^2A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(9A-A=\left(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(8A=9-\frac{1}{3^{100}}\)
=> n = 100
ê mai Phưng câu này mày lấy đâu đấy
cho 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\(=9-\frac{1}{3^n}\). 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
cho A=\(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
biết 8A=9-\(\frac{1}{3^n}\)
Nếu như vậy thì bạn hỏi cái gì ???????
Tìm n biết:
8A=9-\(\frac{1}{3^n}\)và A=1+\(\frac{1}{3^2}+\frac{1}{3^4}+....+\frac{1}{3^{100}}\)
Cho A=1+\(\frac{1}{3^2}\)+\(\frac{1}{3^4}\)+...+\(\frac{1}{3^{100}}\).Biết 8A=9-\(\frac{1}{3^n}\). Vậy n=...