Question

[yoring]

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

Answer 1

[bảo nam trần]

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

Answer 2

[Nguyễn Huy Tú]

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