Question

cho A=1+1/3^2+1/3^4+..............+1/3^100

Biết :8A=9-1/3^n.Vậy n là bao nhiêu

Answer 1

Ta có :\(9A=9+1+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{98}}\)

\(8A=9A-A=9-\dfrac{1}{3^{100}}\)

Mà \(8A=9-\dfrac{1}{3^n}\Rightarrow n=100\)

Vậy n=100

Answer 2

Minh thiếu

\(8A=9A-A=\left(9+1+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{98}}\right)\\ -\left(1+\dfrac{1}{3^2}+\dfrac{1}{3^{\text{4}}}+...+\dfrac{1}{3^{100}}\right)\)

\(8A=9+1+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{98}}-1-\dfrac{1}{3^2}-\dfrac{1}{3^4}-...-\dfrac{1}{3^{100}}\)

\(8A=9+\left(1-1\right)+\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+...+\left(\dfrac{1}{3^{98}}-\dfrac{1}{3^{98}}\right)+\dfrac{1}{3^{100}}\)

\(8A=9-\dfrac{1}{3^{100}}\) Mà theo đề bài \(8A=9-\dfrac{1}{3^n}\)

\(\Rightarrow n=100\)

Vậy n=100