\(A=\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^n-1}{3^n}\)

\(=\frac{3-1}{3}+\frac{9-1}{9}+\frac{27-1}{27}+...+\frac{3^n-1}{3^n}\)

\(=\left(\frac{3}{3}-\frac{1}{3}\right)+\left(\frac{9}{9}-\frac{1}{9}\right)+\left(\frac{27}{27}-\frac{1}{27}\right)+.....+\left(\frac{3^n}{3^n}-\frac{1}{3^n}\right)\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....+\frac{1}{3^n}\right)\)

\(=n-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^n}\right)\)

Bây giờ ta chỉ cần chứng minh:\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^n}< \frac{1}{2}\) là xong!

Thật vậy:\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{n-1}}\)

\(\Rightarrow2B=1-\frac{1}{3^n}\)

\(\Rightarrow B=\frac{1}{2}-\frac{\frac{1}{3^n}}{2}< \frac{1}{2}\) 

Ta có:\(A=n-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^n}\right)\)

\(>n-\frac{1}{2}\left(đpcm\right)\)(bất đẳng thức đổi chiều)

Khó dữ vậy!!!!

Đợi tí , mạng chậm

Ta có:

\(u_1=\dfrac{1}{3^1-1}=\dfrac{1}{2}\\ u_2=\dfrac{2}{3^2-1}=\dfrac{1}{4}\\ u_3=\dfrac{3}{3^3-1}=\dfrac{3}{26}\)

\(\Rightarrow B\)

Chọn B

cm cái j v ạ ?

Chứng minh nó không thuộc Z