Question

Bài 1 Tính tổng 

A = \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{3^3}\)+ ........+ \(\frac{1}{3^8}\)

 

Answer 1

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)

\(2A=1-\frac{1}{3^8}\)

\(A=\frac{\left(\frac{6560}{6561}\right)}{2}=\frac{3280}{6561}\)

Answer 2

\(A=\frac{3280}{6561}\)

Answer 3

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)

\(3A=3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\) .

\(3A=3.\frac{1}{3}+3.\frac{1}{3^2}+3.\frac{1}{3^3}+...+3.\frac{1}{3^8}\)

\(3A=1+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)

Rồi làm như trên

Answer 4

\(A=\frac{3280}{6561}\)