đặt \(A=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^n}\)
\(\Rightarrow3A=3+1+\frac{1}{3}+...+\frac{1}{3^{n-1}}\)
\(\Rightarrow3A-A=(3+1+...+\frac{1}{3^{n-1}})-(1+\frac{1}{3}+...+\frac{1}{3^n})\)
\(\Rightarrow2A=3-\frac{1}{3^n}\)
\(\Rightarrow A=(3-\frac{1}{3^n})\div2\)
Đặt A=1 + 1/3 + 1/32 + 1/33+...+ 1/3n
=> 3A= 3 + 1 + 1/3 + 1/32 +...+ 1/3n-1
=> 3A - A = 2A = 3 - 1/3n
=> 2A =(3n+1 - 1) / 3n
=> A= (3n+1 - 1) / 3n.2
K cho mk nha!